10

course mth 173

íÙ«¼XÇîû›«®Ì£ì˜ã•²íassignment #004

004.

08-05-2008

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assignment #005

005.

08-05-2008

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20:10:44

`qNote that there are 9 questions in this assignment.

`q001. We see that the water depth vs. clock time system likely behaves in a much more predictable detailed manner than the stock market. So we will focus for a while on this system. An accurate graph of the water depth vs. clock time will be a smooth curve. Does this curve suggest a constantly changing rate of depth change or a constant rate of depth change? What is in about the curve at a point that tells you the rate of depth change?

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RESPONSE -->

The steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change

confidence assessment: 3

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20:10:58

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RESPONSE -->

self critique assessment: 3

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20:11:55

`q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy.

Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 40 and t = 90?

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RESPONSE -->

At t=10 the depth is y(10) = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm.

At t=20 the depth is y(20) = .01(20^2) - 2(20) + 90 = 4 - 40 + 90 = 54, representing a depth of 54 cm.

At t=90 the depth is y(90) = .01(90^2) - 2(90) + 90 = 81 - 180 + 90 = -9, representing a depth of -9 cm.

confidence assessment: 3

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20:12:02

At t=10 the depth is y(10) = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm.

At t=20 the depth is y(20) = .01(20^2) - 2(20) + 90 = 4 - 40 + 90 = 54, representing a depth of 54 cm.

At t=90 the depth is y(90) = .01(90^2) - 2(90) + 90 = 81 - 180 + 90 = -9, representing a depth of -9 cm.

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RESPONSE -->

self critique assessment: 3

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20:12:56

`q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals?

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RESPONSE -->

From 71 cm to 54 cm is a change of 54 cm - 71 cm = -17 cm; this change takes place between t = 10 sec and t = 20 sec, so the change in clock time is 20 sec - 10 sec = 10 sec. The average rate of change between these to clock times is therefore

ave rate = change in depth / change in clock time = -17 cm / 10 sec = -1.7 cm/s.

From 54 cm to -9 cm is a change of -9 cm - 54 cm = -63 cm; this change takes place between t = 40 sec and t = 90 sec, so the change in clock time is a9 0 sec - 40 sec = 50 sec. The average rate of change between these to clock times is therefore

ave rate = change in depth / change in clock time = -63 cm / 50 sec = -1.26 cm/s.

confidence assessment: 3

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20:13:11

ave rate = change in depth / change in clock time = -17 cm / 10 sec = -1.7 cm/s.

ave rate = change in depth / change in clock time = -63 cm / 50 sec = -1.26 cm/s.

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RESPONSE -->

self critique assessment: 3

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20:14:07

`q004. What is the average rate at which the depth changes between t = 10 and t = 11, and what is the average rate at which the depth changes between t = 10 and t = 10.1?

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RESPONSE -->

At t=10 the depth is y(10) = .01(10^2) - 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm.

At t=11 the depth is y(11) = .01(11^2) - 2(11) + 90 = 1.21 - 22 + 90 = 69.21, representing a depth of 69.21 cm.

The average rate of depth change between t=10 and t = 11 is therefore

change in depth / change in clock time = ( 69.21 - 71) cm / [ (11 - 10) sec ] = -1.79 cm/s.

At t=10.1 the depth is y(10.1) = .01(10.1^2) - 2(10.1) + 90 = 1.0201 - 20.2 + 90 = 70.8201, representing a depth of 70.8201 cm.

The average rate of depth change between t=10 and t = 10.1 is therefore

change in depth / change in clock time = ( 70.8201 - 71) cm / [ (10.1 - 10) sec ] = -1.799 cm/s.

see that for the interval from t = 10 sec to t = 20 sec, then t = 10 s to t = 11 s, then from t = 10 s to t = 10.1 s the progression of average rates is -1.7 cm/s, -1.79 cm/s, -1.799 cm/s. It is important to note that rounding off could have hidden this progression. For example if the 70.8201 cm had been rounded off to 70.82 cm, the last result would have been -1.8 cm and the interpretation of the progression would change. When dealing with small differences it is important not around off too soon.

confidence assessment: 3

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20:14:15

At t=10 the depth is y(10) = .01(10^2) - 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm.

At t=11 the depth is y(11) = .01(11^2) - 2(11) + 90 = 1.21 - 22 + 90 = 69.21, representing a depth of 69.21 cm.

The average rate of depth change between t=10 and t = 11 is therefore

change in depth / change in clock time = ( 69.21 - 71) cm / [ (11 - 10) sec ] = -1.79 cm/s.

At t=10.1 the depth is y(10.1) = .01(10.1^2) - 2(10.1) + 90 = 1.0201 - 20.2 + 90 = 70.8201, representing a depth of 70.8201 cm.

The average rate of depth change between t=10 and t = 10.1 is therefore

change in depth / change in clock time = ( 70.8201 - 71) cm / [ (10.1 - 10) sec ] = -1.799 cm/s.

We see that for the interval from t = 10 sec to t = 20 sec, then t = 10 s to t = 11 s, then from t = 10 s to t = 10.1 s the progression of average rates is -1.7 cm/s, -1.79 cm/s, -1.799 cm/s. It is important to note that rounding off could have hidden this progression. For example if the 70.8201 cm had been rounded off to 70.82 cm, the last result would have been -1.8 cm and the interpretation of the progression would change. When dealing with small differences it is important not around off too soon.

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RESPONSE -->

self critique assessment: 3

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20:15:06

`q005. What do you think is the precise rate at which depth is changing at the instant t = 10?

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RESPONSE -->

The progression -1.7 cm/s, -1.79 cm/s, -1.799 cm/s corresponds to time intervals of `dt = 10, 1, and .1 sec, with all intervals starting at the instant t = 10 sec. That is, we have shorter and shorter intervals starting at t = 10 sec. We therefore expect that the progression might well continue with -1.7999 cm/s, -1.79999 cm/s, etc.. We see that these numbers approach more and more closely to -1.8, and that there is no limit to how closely they approach. It therefore makes sense that at the instant t = 10, the rate is exactly -1.8.

confidence assessment: 3

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20:15:14

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RESPONSE -->

self critique assessment: 3

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20:15:45

`q006. In symbols, what are the depths at clock time t = t1 and at clock time t = t1 + `dt, where `dt is the time interval between the two clock times?

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RESPONSE -->

At clock time t = t1 the depth is y(t1) = .01 t1^2 - 2 t1 + 90 and at clock time t = t1 + `dt the depth is y(t1 + `dt) = .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90.

confidence assessment: 3

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20:15:57

At clock time t = t1 the depth is y(t1) = .01 t1^2 - 2 t1 + 90 and at clock time t = t1 + `dt the depth is y(t1 + `dt) = .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90.

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RESPONSE -->

self critique assessment: 3

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20:16:32

`q007. What is the change in depth between these clock times?

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RESPONSE -->

The change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90)

= .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90)

= .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90)

= .02 t1 `dt + - 2 `dt + .01 `dt^2.

confidence assessment: 3

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20:16:40

The change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90)

= .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90)

= .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90)

= .02 t1 `dt + - 2 `dt + .01 `dt^2.

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RESPONSE -->

self critique assessment: 3

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20:17:35

`q008. What is the average rate at which depth changes between these clock time?

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RESPONSE -->

The average rate is

ave rate = change in depth / change in clock time = ( .02 t1 `dt + - 2 `dt + .01 `dt^2 ) / `dt = .02 t1 - 2 + .01 `dt.

confidence assessment: 3

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20:17:45

The average rate is

ave rate = change in depth / change in clock time = ( .02 t1 `dt + - 2 `dt + .01 `dt^2 ) / `dt = .02 t1 - 2 + .01 `dt.

Note that as `dt shrinks to 0 this expression approaches .02 t1 - 2.

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RESPONSE -->

self critique assessment: 3

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20:18:23

`q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results?

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RESPONSE -->

At t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10.

confidence assessment: 3

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20:18:31

At t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10.

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RESPONSE -->

self critique assessment: 3

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©Šã¿Ó™Gþ›Þ˜©™Øäøz~UÌ›¦

assignment #006

006. goin' the other way

08-05-2008

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20:35:49

`qNote that there are 7 questions in this assignment.

`q001. If the water and a certain cylinder is changing depth at a rate of -4 cm/sec at the t = 20 second instant, at which instant the depth is 80 cm, then approximately what do you expect the depth will be at the t = 21 second instant?

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RESPONSE -->

At a rate of -4 cm/s, for the 1-second interval between t = 20 s and t = 21 s the change in depth would be -4 cm/s * 1 sec = -4 cm. If the depth was 80 cm at t = 20 sec, the depth at t = 21 sec would be 80 cm - 4 cm/ = 76 cm.

confidence assessment: 3

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20:36:00

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RESPONSE -->

self critique assessment: 3

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20:36:37

`q002. Using the same information, what the you expect the depth will be depth at the t = 30 sec instant? Do you think this estimate is more or less accurate than the estimate you made for the t = 21 second instant?

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RESPONSE -->

At - 4 cm/s, during the 10-second interval between t = 20 sec and t = 30 sec we would expect a depth change of -4 cm/sec * 10 sec = -40 cm, which would result in a t = 30 sec depth of 80 cm - 40 cm = 40 cm.

confidence assessment: 3

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20:36:45

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RESPONSE -->

self critique assessment: 3

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20:37:11

`q003. If you know that the depth in the preceding example is changing at the rate of -3 cm/s at the t = 30 sec instant, how will this change your estimate for the depth at t = 30 seconds--i.e., will your estimate be the same as before, will you estimate a greater change in depth or a lesser change in depth?

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RESPONSE -->

Since the rate of depth change has changed from -4 cm / s at t = 20 s to -3 cm / s at t = 30 s, we conclude that the depth probably wouldn't change as much has before.

confidence assessment: 3

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20:37:18

Since the rate of depth change has changed from -4 cm / s at t = 20 s to -3 cm / s at t = 30 s, we conclude that the depth probably wouldn't change as much has before.

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RESPONSE -->

self critique assessment: 3

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20:37:49

`q004. What is your specific estimate of the depth at t = 30 seconds?

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RESPONSE -->

Knowing that at t = 20 sec the rate is -4 cm/s, and at t = 30 sec the rate is -3 cm/s, we could reasonably conjecture that the approximate average rate of change between these to clock times must be about -3.5 cm/s. Over the 10-second interval between t = 20 s and t = 30 s, this would result in a depth change of -3.5 cm/s * 10 sec = -35 cm, and a t = 30 s depth of 80 cm - 35 cm = 45 cm.

confidence assessment: 3

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20:37:57

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RESPONSE -->

self critique assessment: 3

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20:38:25

`q005. If we have a uniform cylinder with a uniformly sized hole from which water is leaking, so that the quadratic model is very nearly a precise model of what actually happens, then the prediction that the depth will change and average rate of -3.5 cm/sec is accurate. This is because the rate at which the water depth changes will in this case be a linear function of clock time, and the average value of a linear function between two clock times must be equal to the average of its values at those to clock times.

If y is the function that tells us the depth of the water as a function of clock time, then we let y ' stand for the function that tells us the rate at which depth changes as a function of clock time.

If the rate at which depth changes is accurately modeled by the linear function y ' = .1 t - 6, with t in sec and y in cm/s, verify that the rates at t = 20 sec and t = 30 sec are indeed -4 cm/s and -3 cm/s.

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RESPONSE -->

At t = 20 sec, we evaluate y ' to obtain y ' = .1 ( 20 sec) - 6 = 2 - 6 = -4, representing -4 cm/s.

At t = 30 sec, we evaluate y' to obtain y' = .1 ( 30 sec) - 6 = 3 - 6 = -3, representing -3 cm/s.

confidence assessment:

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20:38:34

At t = 20 sec, we evaluate y ' to obtain y ' = .1 ( 20 sec) - 6 = 2 - 6 = -4, representing -4 cm/s.

At t = 30 sec, we evaluate y' to obtain y' = .1 ( 30 sec) - 6 = 3 - 6 = -3, representing -3 cm/s.

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RESPONSE -->

self critique assessment: 3

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20:39:03

`q006. For the rate function y ' = .1 t - 6, at what clock time does the rate of depth change first equal zero?

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RESPONSE -->

The rate of depth change is first equal to zero when y ' = .1 t - 6 = 0. This equation is easily solved to see that t = 60 sec.

confidence assessment: 3

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20:39:09

The rate of depth change is first equal to zero when y ' = .1 t - 6 = 0. This equation is easily solved to see that t = 60 sec.

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RESPONSE -->

self critique assessment: 3

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20:39:35

`q007. How much depth change is there between t = 20 sec and the time at which depth stops changing?

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RESPONSE -->

The rate of depth change at t = 20 sec is - 4 cm/s; at t = 60 sec the rate is 0 cm/s. The average rate at which depth changes during this 40-second interval is therefore the average of -4 cm/s and 0 cm/s, or -2 cm/s.

At an average rate of -2 cm/s for 40 s, the depth change will be -80 cm. Starting at 80 cm when t = 20 sec, we see that the depth at t = 60 is therefore 80 cm - 80 cm = 0.

confidence assessment: 3

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20:39:43

At an average rate of -2 cm/s for 40 s, the depth change will be -80 cm. Starting at 80 cm when t = 20 sec, we see that the depth at t = 60 is therefore 80 cm - 80 cm = 0.

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RESPONSE -->

self critique assessment: 3

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}‘BöàhÍ•e¦ãÒ|òm´´¦‰±àD´ñ„

assignment #007

007. Depth functions and rate functions.

08-05-2008

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21:03:26

`qNote that there are 9 questions in this assignment.

`q001. The function y = .05 t^2 - 6 t + 100 is related to the rate function y ' = .1 t - 6 in that if y = .05 t^2 - 6 t + 100 represents the depth, then the depth change between any two clock times t is the same as that predicted by the rate function y ' = .1 t - 6. We saw before that for y ' = .1 t - 6, the depth change between t = 20 and t = 30 had to be 35 cm. Show that for the depth function y = .05 t^2 - 6t + 100, the change in depth between t = 20 and t = 30 is indeed 35 cm.

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RESPONSE -->

The depth at t = 20 will be .05(20^2) - 6(20) + 100 = 20 - 120 + 100 = 0.

The depth at t = 30 will be .05(30^2) - 6(30) + 100 = 45 - 180 + 100 = -35.

Thus the depth changes from 0 cm to -35 cm during the 10-second time interval between t = 20 s and t = 30 s. This gives us and average rate of

ave rate = change in depth / change in clock time = -35 cm / (10 sec) = -3.5 cm/s.

confidence assessment: 3

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21:03:34

The depth at t = 20 will be .05(20^2) - 6(20) + 100 = 20 - 120 + 100 = 0.

The depth at t = 30 will be .05(30^2) - 6(30) + 100 = 45 - 180 + 100 = -35.

Thus the depth changes from 0 cm to -35 cm during the 10-second time interval between t = 20 s and t = 30 s. This gives us and average rate of

ave rate = change in depth / change in clock time = -35 cm / (10 sec) = -3.5 cm/s.

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RESPONSE -->

self critique assessment: 3

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21:04:00

`q002. What depth change is predicted by the rate function y ' = .1 t - 6 between t = 30 and t = 40? What is the change in the depth function y = .05 t^2 - 6 t + 100 between t = 30 and t = 40? How does this confirm the relationship between the rate function y ' and the depth function y?

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RESPONSE -->

At t = 30 and t = 40 we have y ' = .1 * 30 - 6 = -3 and y ' = .1 * 40 - 6 = -2. The average of the two corresponding rates is therefore -2.5 cm/s. During the 10-second interval between t = 30 and t = 40 we therefore predict the depth change of

predicted depth change based on rate function = -2.5 cm/s * 10 s = -25 cm.

At t = 30 the depth function was previously seen to have value -35, representing -35 cm. At t = 40 sec we evaluate the depth function and find that the depth is -60 cm. The change in depth is therefore

depth change has predicted by depth function = -60 cm - (-35 cm) = -25 cm.

The relationship between the rate function and the depth function is that both should predict a same change in depth between the same two clock times. This is the case in this example.

confidence assessment: 3

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21:04:07

predicted depth change based on rate function = -2.5 cm/s * 10 s = -25 cm.

depth change has predicted by depth function = -60 cm - (-35 cm) = -25 cm.

The relationship between the rate function and the depth function is that both should predict a same change in depth between the same two clock times. This is the case in this example.

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RESPONSE -->

self critique assessment: 3

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21:04:41

`q003. Show that the change in the depth function y = .05 t^2 - 6 t + 30 between t = 20 and t = 30 is the same as that predicted by the rate function y ' = .1 t - 6. Show the same for the time interval between t = 30 and t = 40. Note that the predictions for the y ' function have already been made.

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RESPONSE -->

The prediction from the rate function is a depth change of -35 cm, and has already been made in a previous problem. Evaluating the new depth function at t = 20 we get y = .05(20^2) - 6(20) + 30 = -70, representing -70 cm. Evaluating the same function at t = 30 we get y = -105 cm. This implies the depth change of -105 cm - (-70 cm) = -35 cm.

Evaluating the new depth function at t = 40 sec we get y = depth = -130 cm. Thus the change from t = 30 to t = 40 is -130 cm - (-105 cm) = -25 cm. This is identical to the change predicted in the preceding problem for the given depth function.

confidence assessment: 3

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21:05:03

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RESPONSE -->

self critique assessment: 3

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21:05:30

`q004. Why is it that the depth functions y = .05 t^2 - 6 t + 30 and y = .05 t^2 - 6 t + 100 give the same change in depth between two given clock times?

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RESPONSE -->

The only difference between the two functions is a constant number at the end. One function and with +30 and the other with +100. The first depth function will therefore always be 70 units greater than the other. If one changes by a certain amount between two clock times, the other, always being exactly 70 units greater, must also change by the same amount between those two clock times.

confidence assessment: 3

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21:05:38

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RESPONSE -->

self critique assessment: 3

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21:06:15

`q005. We saw earlier that if y = a t^2 + b t + c, then the average rate of depth change between t = t1 and t = t1 + `dt is 2 a t1 + b + a `dt. If `dt is a very short time, then the rate becomes very clost to 2 a t1 + b. This can happen for any t1, so we might as well just say t instead of t1, so the rate at any instant is y ' = 2 a t + b. So the functions y = a t^2 + b t + c and y ' = 2 a t + b are related by the fact that if the function y represents the depth, then the function y ' represents the rate at which depth changes. If y = .05 t^2 - 6 t + 100, then what are the values of a, b and c in the form y = a t^2 + b t + c? What therefore is the function y ' = 2 a t + b?

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RESPONSE -->

If y = .05 t^2 - 6 t + 100 is of form y = a t^2 + b t + c, then a = .05, b = -6 and c = 100. The function y ' is 2 a t + b. With the given values of a and b we see that y ' = 2 ( .05) t + (-6), which simplifies to y ' = .1 t - 6

confidence assessment: 3

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21:06:23

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RESPONSE -->

self critique assessment: 3

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21:07:03

`q006. For the function y = .05 t^2 - 6 t + 30, what is the function y ' ?

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RESPONSE -->

The values of a, b and c are respectively .05, -6 and 30. Thus y ' = 2 a t + b = 2(.05) t + (-6) = .1 t - 6.

This is identical to the y ' function in the preceding example. The only difference between the present y function and the last is the constant term c at the end, 30 in this example and 100 in the preceding. This constant difference has no effect on the derivative, which is related to the fact that it has no effect on the slope of the graph at a point.

confidence assessment: 3

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21:07:10

The values of a, b and c are respectively .05, -6 and 30. Thus y ' = 2 a t + b = 2(.05) t + (-6) = .1 t - 6.

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RESPONSE -->

self critique assessment: 3

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21:07:41

`q007. For some functions y we can find the rate function y ' using rules which we will develop later in the course. We have already found the rule for a quadratic function of the form y = a t^2 + b t + c. The y ' function is called the derivative of the y function, and the y function is called an antiderivative of the y ' function. What is the derivative of the function y = .05 t^2 - 6 t + 130? Give at least two new antiderivative functions for the rate function y ' = .1 t - 6.

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RESPONSE -->

The derivative of y = .05 t^2 - 6 t + 130 is .1 t - 6; as in the preceding problem this function has a = .05 and b = -6, and differs from the preceding two y functions only by the value of c. Since c has no effect on the derivative, the derivative is the same as before.

If y ' = .1 t - 6, then a = 1 / 2 ( .1) = .05 and b = -6 and we see that the function y is y = .05 t^2 - 6 t + c, where c can be any constant. We could choose any two different values of c and obtain a function which is an antiderivative of y ' = .1 t - 6. Let's use c = 17 and c = -54 to get the functions y = .05 t^2 - 6 t + 17 and y = .05 t^2 - 6 t - 54.

confidence assessment: 3

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21:07:49

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RESPONSE -->

self critique assessment: 3

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21:09:13

`q008. For a given function y, there is only one derivative function y '. For a given rate function y ', there is more than one antiderivative function. Explain how these statements are illustrated by the preceding example.

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RESPONSE -->

The derivative function give the rate at which the original function changes for every value of t, and there can be only one rate for a given t. Thus the values of the derivative function are completely determined by the original function. In the previous examples we saw several different functions with the same derivative function. This occurred when the derivative functions differed only by the constant number at the end.

However, for a given derivative function, if we get one antiderivative, we can add any constant number to get another antiderivative. y = .05 t^2 - 6 t +17, y = .05 t^2 - 6 t + 30, and y = .05 t^2 - 6 t + 100, etc. are all antiderivatives of y ' = .1 t - 6.

confidence assessment: 3

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21:09:20

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RESPONSE -->

self critique assessment: 3

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21:09:43

`q009. What do all the antiderivative functions of the rate function y ' = .1 t - 6 have in common? How do they differ? How many antiderivative functions do you think there could be for the given rate function?

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RESPONSE -->

All antiderivatives must contain .05 t^2 - 6 t. They may also contain a nonzero constant term, such as -4, which would give us y = .05 t^2 - 6 t - 4. We could have used any number for this last constant.

confidence assessment: 3

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21:09:50

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RESPONSE -->

self critique assessment: 3

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çˆ¯Æ§¦ÑÓÉ~ÞøªÚùoêg¹¡í÷Ç‡Ïù

assignment #008

008. Approximate depth graph from the rate function

08-05-2008

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22:01:09

`q001. Note that there are 5 questions in thie assignment.

Sketch a graph of the function y ' = .1 t - 6 for t = 0 to t = 100. Describe your graph.

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RESPONSE -->

The graph of this function has an intecept on the y' axis at (0,-6) and an intercept on the x axis at (60,0). The graph is a straight line with slope .1. At t = 100 the graph point is (100,4).

confidence assessment: 3

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22:01:15

The graph of this function has an intecept on the y' axis at (0,-6) and an intercept on the x axis at (60,0). The graph is a straight line with slope .1. At t = 100 the graph point is (100,4).

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RESPONSE -->

self critique assessment: 3

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22:02:09

`q002. Now sketch the graph of a function y which starts at the point (0, 100), and whose slope at every t is equal to the value of y ' at t.

Thus, for example, at t = 0 we have y ' = .1 * 0 - 6 = -6, so our graph of y will start off a the t = 0 point (0,100) with a slope of -6, and the graph begins by decreasing rather rapidly.

But the slope won't remain at -6. By the time we get to t = 10 the slope will be y ' = .1 * 10 - 6 = -5, and the graph is decreasing less rapidly than before.

Then by the time we get to t = 20 the slope will be y ' = . 1 * 20 - 6 = -4, etc..

If you sketch a graph of y vs. t with these calculated slopes at the corresponding t values, what does the graph look like? Will it be increasing or decreasing, and will it be doing so at an increasing, decreasing or constant rate? Is the answer to this question different for different parts of the graph? If so over what intervals of the graph do the different answers apply?

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RESPONSE -->

The graph will have slopes -6 at t = 0, -5 at t = 10, -4 at t = 9, etc.. At least for awhile, the slope will remain negative so the graph will be decreasing. The negative slopes will however become less and less steep. So the graph be decreasing at a decreasing rate.

It's obvious that the slopes will eventually reach 0, and since y' is the slope of the y vs. t graph it's clear that this will happen when y' = .1 t - 6 becomes 0. Setting .1 t - 6 = 0 we get t = 60. Note, and think about about the fact, that this coincides with the x-intercept of the graph of y' vs. t. At this point the slope will be 0 and the graph will have leveled off at least for an instant.

Past t = 60 the values of y' will be positive, so the slope of the y vs. t graph will be positive and graph of y vs. t will therefore be increasing. The values of y' will also be increasing, so that the slopes of the y vs. t graph will be increasing, and we can say that the graph will be increasing at an increasing rate.

confidence assessment: 3

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22:02:15

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RESPONSE -->

self critique assessment: 3

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22:03:06

`q003. The graph of y vs. t corresponding to the given rate function y ' = .1 t - 6 has slope -6 at the point (0,100). This slope immediately begins changing, and becomes -5 by the time t = 10. However, let us assume that the slope doesn't change until we get to the t = 10 point. This assumption isn't completely accurate, but we're going to see how it works out.

If the slope remained -6 for t = 0 to t = 10, then starting at (0, 100) what point would we reach when t = 10?

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RESPONSE -->

The slope of the graph is the ratio slope = rise / run. If the slope remains at -6 from t = 0 to t = 10, then the difference between 10 is the run. Thus the run is 10 and the slope is -6, so the rise is

rise = slope * run = -6 * 10 = -60.

The y coordinate of the graph therefore changes by -60, from y = 100 to y = 100 + (-60) = 40. The corresponding point is (10, 40).

confidence assessment: 3

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22:03:13

rise = slope * run = -6 * 10 = -60.

The y coordinate of the graph therefore changes by -60, from y = 100 to y = 100 + (-60) = 40. The corresponding point is (10, 40).

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RESPONSE -->

self critique assessment: 3

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22:03:58

`q004. We see that we reach the point (10, 40) by assuming a slope of -6 from t = 0 to t = 10. We have seen that at t = 10 the slope will be y ' = .1 * 10 - 6 = -5. If we maintain this slope for the interval t = 10 to t = 20, what will be the coordinates of the t = 20 point?

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RESPONSE -->

The run from t = 10 to t = 20 is 10. With a slope of -5 this implies a rise of

rise = slope * run = -5 * 10 = -50.

Starting from point (10,40), a rise of -50 and a run of 10 takes the graph to the point (20, -10).

confidence assessment: 3

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22:04:07

The run from t = 10 to t = 20 is 10. With a slope of -5 this implies a rise of

rise = slope * run = -5 * 10 = -50.

Starting from point (10,40), a rise of -50 and a run of 10 takes the graph to the point (20, -10).

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RESPONSE -->

self critique assessment: 3

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22:04:32

`q005. Continue this process up to the t = 70 point, using a 10-unit t interval for each approximation. Describe the graph that results.

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RESPONSE -->

The slope at t = 20 is y ' = .1 * 20 - 6 = -4. From t = 20 to t = 30 the run is 10, so the rise is rise = slope * run = -4 * 10 = -40. Starting from (20,-10) a rise of -40 and a run of 10 takes us to (30, -50).

The slope at t = 30 is y ' = .1 * 30 - 6 = -3. From t = 30 to t = 40 the run is 10, so the rise is rise = slope * run = -3 * 10 = -30. Starting from (30,-50) a rise of -30 and a run of 10 takes us to (40, -80).

The slope at t = 40 is y ' = .1 * 40 - 6 = -2. From t = 40 to t = 50 the run is 10, so the rise is rise = slope * run = -2 * 10 = -20. Starting from (40,-80) a rise of -20 and a run of 10 takes us to (50, -100).

The slope at t = 50 is y ' = .1 * 50 - 6 = -1. From t = 50 to t = 60 the run is 10, so the rise is rise = slope * run = -1 * 10 = -10. Starting from (50,-100) a rise of -10 and a run of 10 takes us to (60, -110).

The slope at t = 60 is y ' = .1 * 70 - 6 = -0. From t = 60 to t = 70 the run is 10, so the rise is rise = slope * run = -0 * 10 = 0. Starting from (60,-110) a rise of and a run of 10 takes us to (70, -110).

confidence assessment: 3

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22:04:39

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RESPONSE -->

self critique assessment: 3

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ÙÑ…œÃÊúá´Šº}”ÜýÆ“½·ÜãÙOÂ

assignment #009

009. Finding the average value of the rate using a predicted point

08-05-2008

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22:19:53

`qNote that there are 9 questions in this assignment.

`q001. The process we used in the preceding series of exercises to approximate the graph of y corresponding to the graph of y ' can be significantly improved. Recall that we used the initial slope to calculate the change in the y graph over each interval.

For example on the first interval we used the initial slope -6 for the entire interval, even though we already knew that the slope would be -5 by the time we reached the t = 10 point at the end of that interval. We would have been more accurate in our approximation if we had used the average -5.5 of these two slopes.

Using the average of the two slopes, what point would we end up at when t = 10?

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RESPONSE -->

If the average value slope = -5.5 is used between t = 0 and t = 10, we find that the rise is -55 and we go from (0,100) to (10,45).

confidence assessment: 3

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22:20:02

If the average value slope = -5.5 is used between t = 0 and t = 10, we find that the rise is -55 and we go from (0,100) to (10,45).

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RESPONSE -->

self critique assessment: 3

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22:20:42

`q002. We find that using the average of the two slopes we reach the point (10, 45). At this point we have y ' = -5.

By the time we reach the t = 20 point, what will be the slope? What average slope should we therefore use on the interval from t = 10 to t = 20? Using this average slope what point will we end up with when t = 20?

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RESPONSE -->

The slope at t = 20 is y ' = -4. Averaging this with the slope y ' = -5 at t = 10 we estimate an average slope of (-5 + -4) / 2 = -4.5 between t = 10 and t = 20. This would imply a rise of -45, taking the graph from (10,45) to (20, 0).

confidence assessment: 3

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22:21:29

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RESPONSE -->

self critique assessment: 3

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22:22:13

`q003. We found that we reach the point (20, 0), where the slope is -4. Following the process of the preceding two steps, what point will we reach at t = 30?

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RESPONSE -->

The slope at t = 20 is y ' = -3. Averaging this with the slope y ' = -4 at t = 20 we estimate an average slope of (-4 + -3) / 2 = -4.5 between t =20 and t = 30. This would imply a rise of -35, taking the graph from (20,0) to (30, -35).

confidence assessment: 3

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22:22:21

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RESPONSE -->

self critique assessment: 3

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22:22:36

`q004. Continue this process up through t = 70. How does this graph differ from the preceding graph, which was made without averaging slopes?

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RESPONSE -->

The average slopes over the next four intervals would be -2.5, -1.5, -.5 and +.5. These slopes would imply rises of -25, -15, -5 and +5, taking the graph from (30,-35) to (40, -60) then to (50, -75) then to (60, -80) and finally to (70, -75). Note that the graph decreases at a decreasing rate up to the point (60, -80) before turning around and increasing to the ponit (70, -75).

confidence assessment: 3

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22:22:44

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RESPONSE -->

self critique assessment: 3

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22:23:07

`q005. Now suppose that y = -.2 t^2 + 5 t + 100 represents depth vs. clock time. What is the corresponding rate function y ' ?

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RESPONSE -->

The rate function corresponding to y = a t^2 + b t + c is y ' = 2 a t + b. In this example a = -.2, b = 5 and c = 100 so y ' = 2(-.2) t + 5, or y ' = -.4 t + 5.

confidence assessment: 3

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22:23:16

The rate function corresponding to y = a t^2 + b t + c is y ' = 2 a t + b. In this example a = -.2, b = 5 and c = 100 so y ' = 2(-.2) t + 5, or y ' = -.4 t + 5.

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RESPONSE -->

self critique assessment: 3

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22:23:53

`q006. From the depth function and the rate function we can find the coordinates of the t = 30 point of the graph, as well as the rate of change of the function at that point. The rate of change of the function is identical to the slope of the graph at that point, so we can find the slope of the graph.

What are the coordinates of the t = 30 point and what is the corresponding slope of the graph? Sketch a graph of a short line segment through that point with that slope.

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RESPONSE -->

At t = 30 we have y = -.2 * 30^2 + 5 * 30 + 100 = 70, and y ' = -.4 * 30 + 5 = -7. The graph will therefore consist of the point (30, 70) and a short line segment with slope -7.

confidence assessment: 3

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22:24:02

At t = 30 we have y = -.2 * 30^2 + 5 * 30 + 100 = 70, and y ' = -.4 * 30 + 5 = -7. The graph will therefore consist of the point (30, 70) and a short line segment with slope -7.

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RESPONSE -->

self critique assessment: 3

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22:24:24

`q007. What is the equation of the straight line through the t = 30 point of the depth function of the preceding example, having a slope matching the t = 30 rate of change of that function?

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RESPONSE -->

A straight line through (30, 70) with slope -7 has equation

y - 70 = -7 ( x - 30),

found by the point-slope form of a straight line.

This equation is easily rearranged to the form

y = -7 x + 280.

confidence assessment: 3

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22:24:30

A straight line through (30, 70) with slope -7 has equation

y - 70 = -7 ( x - 30),

found by the point-slope form of a straight line.

This equation is easily rearranged to the form

y = -7 x + 280.

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RESPONSE -->

self critique assessment: 3

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22:25:16

`q008. Evaluate the depth function y at t = 30, 31, 32. Evaluate the y coordinate of the straight line of the preceding question at t = 30, 31, 32. What are your results?

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RESPONSE -->

Plugging t = 30, 31, 32 into the original function y = -.2 t^2 + 5 t + 100 we get y = 70, 62.8, 55.2 respectively.

Plugging t = 30, 31, 32 into the straight-line equation y = -7 t + 280 we get y = 70, 63, 56 respectively.

confidence assessment: 3

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22:25:29

Plugging t = 30, 31, 32 into the original function y = -.2 t^2 + 5 t + 100 we get y = 70, 62.8, 55.2 respectively.

Plugging t = 30, 31, 32 into the straight-line equation y = -7 t + 280 we get y = 70, 63, 56 respectively.

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RESPONSE -->

self critique assessment: 3

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22:25:48

`q009. By how much does the straight-line function differ from the actual depth function at each of the three points? How would you describe the pattern of these differences?

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RESPONSE -->

At t = 30 the two functions are identical.

At t = 31 the values are 62.8 and 63; the function is .2 units below the straight line.

At t = 32 the values are 55.2 and 56; the function is now .8 units below the straight line.

The pattern of differences is therefore 0, -.2, -.8. The function falls further and further below the straight line as we move away from the point (30, 70).

confidence assessment: 3

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22:26:01

At t = 30 the two functions are identical.

At t = 31 the values are 62.8 and 63; the function is .2 units below the straight line.

At t = 32 the values are 55.2 and 56; the function is now .8 units below the straight line.

The pattern of differences is therefore 0, -.2, -.8. The function falls further and further below the straight line as we move away from the point (30, 70).

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RESPONSE -->

self critique assessment: 3

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°¢RþðMêò™ÁàØñ|UñŸ÷Æ²êØf®µ€¶²ÊÝ

assignment #010

010. Income Streams

08-05-2008

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22:35:06

`qNote that there are 11 questions in this assignment.

`q001. Suppose that you expect for the next 6 years to receive a steady stream of extra income, at the rate of $20,000 per year. This income is expected to 'flow in' at a constant rate, day by day. Suppose furthermore that you expect that your money will grow at a constant annual rate of 8%.

Assuming you do not use any of the money or interest, the question we want to answer is how much you would therefore expect to have at the end of 6 years. This problem is complicated by the fact that the money that goes in, say, today will earn interest over a longer period than the money you earn tomorrow. In fact, the money you receive in one minute will earn interest for a different length of time than the money you receive in a different minute.

To begin to answer the problem you could start out by saying that while some of the money will earn interest for 6 years, some will go in at the very end of the 6-year period and will therefore not earn any interest at all, so the average time period for earning interest will be 3 years. You could then calculate the interest on the full amount for 3 years, and get a fairly good idea of the final amount. As we will see, you can't get a precise estimate this way, but it gives you a reasonable starting estimate.

What would be the simple interest on the total 6-year amount of money flow for 3 years, and what would be the final amount?

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RESPONSE -->

The total money flow would be $20,000 per year for 6 years, or $120,000. 8% simple interest for 3 years would be 3 * 8% = 24%, and 24% of $120,000 is $28,800, so your first estimate would be $28,800. You therefore expect to end up with something not too far from $148,800.

confidence assessment: 3

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22:35:19

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RESPONSE -->

self critique assessment: 3

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22:36:40

`q002. Money usually doesn't earn simple interest. Interest is almost always compounded. If the interest on $120,000 was compounded annually at 8% for 3 years, what would be the final amount?

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RESPONSE -->

There is a more efficient way to calculate this, but we'll see that shortly.

8% of $120,000 is $9600, so after 1 year we will have $120,000 + $9600 = $129,600.

confidence assessment: 3

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22:36:50

There is a more efficient way to calculate this, but we'll see that shortly.

8% of $120,000 is $9600, so after 1 year we will have $120,000 + $9600 = $129,600.

Note that this result could have been obtained by multiplying $120,000 by 1.08. We will use this strategy for the next two years.

After another year we will have $129,600 * 1.08 = $139,968. If you wish you can take 8% of $129,600 and add it to $129,600; you will still get $139,968.

After a third year you will have $139,968 * 1.08 = $151,165. Note that this beats the $148,800 you calculated with simple interest. This is because each year the interest is applied to a greater amount than before; previously the interest was just applied to the starting amount.

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RESPONSE -->

self critique assessment: 3

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22:37:21

`q003. Money can earn interest that is compounded annually, quarterly, weekly, daily, hourly, or whatever. For given interest rate the maximum interest will be obtained if the money is compounded continuously.

When initial principle P0 is compounded continuously at rate r for t interest periods (e.g., at rate 8% = .08 for t years), the amount at the end of the time is P = P0 * e^(r t).

If the $120,000 was compounded continuously at 8% annual interest for 3 years, what would be the amount at the end of that period?

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RESPONSE -->

would have P0 = $120,000, r = .08 and t = 3. So the amount would be

P = $120,000 * e^(.08 * 3) = $120,000 * e^(.24) = $152,549.

This beats the annual compounding by over $1,000.

confidence assessment: 3

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22:37:29

We would have P0 = $120,000, r = .08 and t = 3. So the amount would be

P = $120,000 * e^(.08 * 3) = $120,000 * e^(.24) = $152,549.

This beats the annual compounding by over $1,000.

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RESPONSE -->

self critique assessment: 3

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22:38:01

`q004. This keeps getting better and better. We end up with more and more money. Now let's see how much money we would really end up with, if the money started compounding continuously as soon as it 'flowed' into the account.

As a first step, we note that 6 years is 72 months. About how much would the money flowing into the account during the 6th month be worth at the end of the 72 months?

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RESPONSE -->

Flowing into the account at a constant rate of $20,000 per year, we see that the monthly flow is $20,000 / 12 = $1,666.67.

The $1,666,67 that flows in during the 6th month will have about 72 - 5.5 = 66.5 months to grow.

confidence assessment: 3

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22:38:08

Flowing into the account at a constant rate of $20,000 per year, we see that the monthly flow is $20,000 / 12 = $1,666.67.

The $1,666,67 that flows in during the 6th month will have about 72 - 5.5 = 66.5 months to grow.

[ Note that we use the 5.5 instead of 6 because the midpoint of the 6th month is 5.5 months after the beginning of the 72-month period (the first month starts at month 0 and ends up at month 1, so its midway point is month .5; the nth month starts at the end of month n-1 and ends at the end of month n so its midway point is (n - 1) + .5). ]

66.5 months is 66.5/12 = 5.54 years, approximately. So the $1,666,67 would grow at the continuous rate of 8 percent for 5.54 years. This would result in a principle of

$1,666,67 * e^(.08 * 5.54) = $2596.14.

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RESPONSE -->

self critique assessment: 3

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22:38:28

`q005. How much will the money flowing into the account during the 66th month be worth at the end of the 72 months?

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RESPONSE -->

Again $1,666,67 would flow into the account during this month. The midpoint of this month is 65.5 months after the start of the 72-month period, so the money will have an average of about 6.5 months, or 6.5 / 12 = .54 years, approx., to grow. Thus the money that flows in during the 66th month will grow to $1,666,67 * e^(.08 * .54) = $1737, approx..

confidence assessment: 3

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22:38:35

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RESPONSE -->

self critique assessment: 3

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22:39:00

`q006. We could do a month-by-month calculation and add up all the results to get a pretty accurate approximation of the amount at the end of the 72 months. We would end up with $152,511, not quite as much as our last estimate for the entire 72 months.

Of course this approximation still isn't completely accurate, because the money that comes in that the beginning of a month earns interest for longer than the money that comes in at the end of the month. We could chop up the 72 months into over 2000 days and calculate the value of the money that comes in each day. But even that wouldn't be completely accurate.

We now develop a model that will be completely accurate. We first imagine a short time span near some point in the 72 months, and calculate the value of the income flow during that time span. We are going to use symbols because our calculation asked to apply to any time span at anytime during the 72 months.

We will use t for the time since beginning of the 6-year period, to the short time span we are considering; in previous examples t was the midpoint of the specified month: t = 5.5 months in the first calculation and 66.5 months in the second.

We will use `dt (remember that this stands for the symbolic expression delta-t) for the duration of the time interval; in the previous examples `dt was 1 month.

For a time interval of length `dt, how much money flows? Assume `dt is in years. If this money flow occurs at t years from the beginning of the 6-year period, then how long as it have to grow?

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RESPONSE -->

The amount in 1 year is $20,000 so the amount in `dt years is $20,000 * `dt.

Money that flows in near the time t years from the beginning of the 6-year period will have (6-t) years to grow.

confidence assessment: 3

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22:39:07

The amount in 1 year is $20,000 so the amount in `dt years is $20,000 * `dt.

Money that flows in near the time t years from the beginning of the 6-year period will have (6-t) years to grow.

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RESPONSE -->

self critique assessment: 3

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22:39:28

`q007. How much will the $20,000 `dt amount received at t years from the start grow to in the remaining (6-t) years?

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RESPONSE -->

Amount P0 will grow to P0 e^( r t) in t years, so in 6-t years amount $20,000 `dt will grow to $20,000 `dt e^(.08 (6-t) ).

confidence assessment: 3

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22:39:37

Amount P0 will grow to P0 e^( r t) in t years, so in 6-t years amount $20,000 `dt will grow to $20,000 `dt e^(.08 (6-t) ).

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RESPONSE -->

self critique assessment: 3

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22:39:56

`q008. So for a contribution at t years from the beginning of the 6-year period, at what rate would we say that money is being contributed to the final t = 6 year value? Let y stand for the final value of the money, and `dy for the contribution from the interval `dt.

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RESPONSE -->

The money that flows in during time interval `dt will grow to

`dy = $20,000 `dt * e^(.08 (6-t) ),

so the rate is about

`dy / `dt = [ $20,000 `dt * e^(.08 (6-t) ) ] / `dt = $20,000 e^(.08 ( 6-t) ).

As `dt shrinks to 0, this expression approaches the exact rate y ' = dy / dt at which money is being contributed to the final value.

confidence assessment: 3

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22:40:03

The money that flows in during time interval `dt will grow to

`dy = $20,000 `dt * e^(.08 (6-t) ),

so the rate is about

`dy / `dt = [ $20,000 `dt * e^(.08 (6-t) ) ] / `dt = $20,000 e^(.08 ( 6-t) ).

As `dt shrinks to 0, this expression approaches the exact rate y ' = dy / dt at which money is being contributed to the final value.

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RESPONSE -->

self critique assessment: 3

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22:40:21

`q009. As we just saw the rate at which money accumulates is y' = dy / dt = 20,000 e^(.08(6-t)).

How do we calculate the total quantity accumulated given the rate function and the time interval over which it accumulates?

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RESPONSE -->

The rate function is the derivative of the quantity function. An antiderivative of a rate-of-change function is a change-in-quantity function. We therefore calculate the total change in a quantity from the rate function by finding the change in this antiderivative.

confidence assessment: 3

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22:40:28

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RESPONSE -->

self critique assessment: 3

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22:40:51

`q010. If the money that flows into the account t years along the 6-year cycle adds to the final value of the money at the previously mentioned rate y' = dy / dt = 20,000 e^(.08(6-t)), then what function describes the final value of the money accumulated through time t?

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RESPONSE -->

The function must be an antiderivative of the rate function y ' = $20,000 e^(.08 ( 6-t) ).

The y ' function can be expressed as y ' = $20,000 e^(.08 * 6) e^(-.08 t)

= $20,000 * 1.616 e^(-.08 t)

= $32,321 e^(-.08 t).

The general antiderivative of e^(-.08 t) is -1/.08 e^(-.08 t) + c = -12.5 e^(-.08 t) + c.

The general antiderivative of y ' = $32,321 e^(-.08 t) is y = $32,321 * -12.5 e^(-.08 t) = -404,160 e^(-.08 t) + c.

The value of the constant c could be determine if we knew, for example, the amount of money present at t = 0. However as we will see in the next step, the value of c doesn't affect the change in the quantity we are considering in this problem.

confidence assessment: 3

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22:41:00

The function must be an antiderivative of the rate function y ' = $20,000 e^(.08 ( 6-t) ).

The y ' function can be expressed as y ' = $20,000 e^(.08 * 6) e^(-.08 t)

= $20,000 * 1.616 e^(-.08 t)

= $32,321 e^(-.08 t).

The general antiderivative of e^(-.08 t) is -1/.08 e^(-.08 t) + c = -12.5 e^(-.08 t) + c.

The general antiderivative of y ' = $32,321 e^(-.08 t) is y = $32,321 * -12.5 e^(-.08 t) = -404,160 e^(-.08 t) + c.

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RESPONSE -->

self critique assessment: 3

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22:41:23

`q011. How much money accumulates during the 6 years?

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RESPONSE -->

y represents the money accumulated through t years. We subtract the money assumulated at 0 years from the money accumulated at 6 years to get the amount accumulated during the time interval from t = 0 to t = 6 years.

The result is -404,160 e^(-.08 * 6 ) + c - [ -404,160 e^(-.08 * 0 ) + c ] = $154,072.

confidence assessment: 3

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22:41:30

The result is -404,160 e^(-.08 * 6 ) + c - [ -404,160 e^(-.08 * 0 ) + c ] = $154,072.

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RESPONSE -->

self critique assessment: 3

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22:42:16

**** redo the last few steps -- we need to be able to make the step from f(t) `dt to the integral, though the steps taken here can illuminate the Fundamental Theorem

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RESPONSE -->

self critique assessment:

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Í¬îJìÄ½¥k“µ¾Q÷™{²jw„ù€¯ç

assignment #011

011. Rules for calculating derivatives of some functions.

08-05-2008

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22:49:41

`qNote that there are 9 questions in this assignment.

`q001. The most basic functions you studied precalculus were:

the power functions y = x^n for various values of n,

the exponential function y = e^x,

the natural logarithm function y = ln(x), and

the sine and cosine functions y = sin(x) and y = cos(x).

We have fairly simple rules for finding the derivative functions y ' corresponding to each of these functions. Those rules are as follows:

If y = x^n for any n except 0, then y ' = n x^(n-1).

If y = e^x then y ' = e^x (that's right, the rate of change of this basic exponential function is identical to the value of the function).

If y = ln(x) then y ' = 1/x. If y = sin(x) then y ' = cos(x).

If y = cos(x) then y ' = - sin(x).

There are also some rules for calculating the derivatives of combined functions like the product function x^5 * sin(x), the quotient function e^x / cos(x), or the composite function sin ( x^5).

We will see these rules later, but for the present we will mention one easy rule, that if we multiply one of these functions by some constant number the derivative function will be the derivative of that function multiply by the same constant number.

Thus for example,

since the derivative of sin(x) is cos(x), the derivative of 5 sin(x) is 5 cos(x); or

since the derivative of ln(x) is 1 / x, the derivative of -4 ln(x) is -4 (1/x) = -4 / x.

Using these rules, find the derivatives of the functions y = -3 e^x, y = .02 ln(x), y = 7 x^3, y = sin(x) / 5.

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RESPONSE -->

If t = -2, then z = t^2 = (-2)^2 = 4, so y = e^z = e^4 = 55, approx..

If t = -1, then z = t^2 = (-1 )^2 = 1, so y = e^z = e^1 = 2.72, approx..

If t = -.5, then z = t^2 = (-.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx.

If t = 0, then z = t^2 = ( 0 )^2 = 0, so y = e^z = e^0 = 1.

If t = .5, then z = t^2 = (.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx.

If t = 1, then z = t^2 = ( 1 )^2 = 1, so y = e^z = e^1 = 2.72, approx..

If t = 2, then z = t^2 = ( 2 )^2 = 4, so y = e^z = e^4 = 55, approx..

confidence assessment: 3

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22:49:54

The derivative of y = -3 e^x is -3 times the derivative of y = e^x.

Since by the given rules the derivative of e^x is e^x, the derivative of y = - 3 e^x is y ' = - 3 e^x.

The derivative of y = .02 ln(x) is .02 times the derivative of y = ln(x).

Since the derivative of ln(x) is 1 / x, the derivative of y = .02 ln(x) is y ' = .02 * 1 / x = .02 / x.

The derivative of y = 7 x^3 is 7 times the derivative of x^3. Since the derivative of x^n is n x^(n-1), the derivative of x^3 is 3 x^(3-1), or 3 x^3.

The derivative of y = 7 x^3 is therefore y ' = 7 ( 3 x^2) = 21 x^2.

The derivative of y = sin(x) / 5 is 1/5 the derivative of sin(x). The derivative of sin(x), according to the rules given above, is cos(x). Thus the derivative of y = sin(x) / 5 is y ' = 1/5 cos(x), or y ' = cos(x) / 5.

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RESPONSE -->

self critique assessment: 3

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22:50:22

`q002. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = 5 * ln(t), then at what rate is water rising in the container when t = 10?

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RESPONSE -->

If t = -2, then y = e^(t^2) = e^(-2)^2 = e^4 = 55, approx.

If t = -1, then y = e^(t^2) = e^(-1)^2 = e^1 = 2.72, approx.

If t = -.5, then y = e^(t^2) = e^(-.5)^2 = e^.25 = 1.28, approx.

If t = 0, then y = e^(t^2) = e^(0)^2 = e^0 = 1.

If t = .5, then y = e^(t^2) = e^(.5)^2 = e^.25 = 1.28, approx.

If t = 1, then y = e^(t^2) = e^(1)^2 = e^1 = 2.72, approx.

If t = 2, then y = e^(t^2) = e^(2)^2 = e^2 = 55, approx.

confidence assessment: 3

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22:50:27

The time rate at which water is rising is the derivative y ' = dy / dt of the depth function y. Since the derivative of ln(t) is 1 / t, we have

rate = y ' = 5 * 1 / t = 5 / t.

Since the rate is y ' = 5 / t, when t = 10 the water is rising at rate y ' = 5 / 10 = .5.

If y is depth in cm and t is clock time in seconds, then the rate is y ' = dy / dt = .5 cm/sec.

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RESPONSE -->

self critique assessment: 3

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22:50:53

`q003. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = e^t / 10, then at what rate is water rising in the container when t = 2?

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RESPONSE -->

The first function encountered by the variable x is the ln(x) function, so we will say that z = ln(x). The result of this calculation is then entered into the cosine function, so we say that y = cos(z).

Thus we have y = cos(z) = cos( ln(x) ).

We also say that the function y(x) is the composite of the functions cosine and natural log functions, i.e., the composite of y = cos(z) and z = ln(x).

confidence assessment: 3

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22:50:59

The time rate at which water is rising is the derivative y ' = dy / dt of the depth function y. Since the derivative of e^t is e^t, we have

rate = y ' = e^t / 10.

Since the rate is y ' = e^t / 10, when t = 2 the water is rising at rate y ' = e^2 / 10 = .73, approx.

If y is depth in cm and t is clock time in seconds, then the rate is y ' = dy / dt = .73 cm/sec.

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RESPONSE -->

self critique assessment: 3

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22:51:22

`q004. If the altitude of a certain rocket is given as a function of clock time t by y = 12 * t^3, then what function gives the rate of altitude change, and at what rate is the altitude changing when t = 15?

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RESPONSE -->

The first function encountered by the variable t is ln(t), so we say that z = ln(t). This value is then squared so we say that y = z^2.

Thus we have y = z^2 = (ln(t))^2.

We also say that we have here the composite of the squaring function and the natural log function, i.e., the composite of y = z^2 and z = ln(t).

confidence assessment: 3

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22:51:30

The time rate at which altitude is changing is the derivative y ' = dy / dt of the depth function y. Since the derivative of t^3 is 3 t^2, we have

rate = y ' = 12 * (3 t^2) = 36 t^2.

Since the rate is y ' = 36 t^2, when t = 15 the altitude is changing at rate y ' = 36 * 15^2 = 8100, approx.

If y is altitude in feet and t is clock time in seconds, then the rate is y ' = dy / dt = 8100 ft/sec.

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RESPONSE -->

self critique assessment: 3

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22:51:57

`q005. If the position of a certain pendulum is given relative to its equilibrium position by the function y = .35 sin(t), then what function gives the corresponding rate of position change, and what rate is position changing when t = 0, when t = `pi/2, and when t = 4?

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RESPONSE -->

The first function encountered by the variable is cos(x), so we say that z = cos(x). We then take the natural log of this function, so we say that y = ln(z).

Thus we have y = ln(z) = ln(cos(x)).

This function is the composite of the natural log and cosine function, i.e., the composite of y = ln(z) and z = cos(x).

confidence assessment: 3

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22:52:03

The time rate at which position is changing is the derivative y ' = dy / dt of the position function y. Since the derivative of sin(t) is cos(t), we have

rate = y ' = .35 cos(t).

Since the rate is y ' = .35 cos(t),

When t = 0 the position is changing at rate y ' = .35 cos(0) = .35.

When t = `pi/2 the position is changing at rate y ' = .35 cos(`pi/2) = 0.

When t = 4 the position is changing at rate y ' = .35 cos(4) = -.23.

If y is position in cm and t is clock time in seconds, then the rates are .35 cm/s (motion in the positive direction), -.35 cm/s (motion in the negative direction), and -.23 cm/s (motion in the negative direction but not quite as fast).

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RESPONSE -->

self critique assessment: 3

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22:52:52

`q006. Another rule is not too surprising: The derivative of the sum of two functions is the sum of the derivatives of these functions. What are the derivatives of the functions y = 4 x^3 - 7 x^2 + 6 x, y = 4 sin(x) + 8 ln(x), and y = 5 e^x - 3 x^-5?

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RESPONSE -->

confidence assessment: 3

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22:52:58

Since y = 4 x^3 - 7 x^2 + 6 x is the sum of the functions 4 x^3, -7 x^2 and 6x, whose derivatives are 12 x^2, -14 x and 6, respectively, we see that y ' is the sum of these derivatives:

y ' = 12 x^2 - 14 x + 6.

Since y = 4 sin(x) + 8 ln(x) is the sum of the functions 4 sin(x) and 8 ln(x), whose derivatives are , respectively, 4 cos(x) and 8 / x, we see that y ' is the sum of these derivatives:

y ' = 4 cos(x) + 8 / x

Since y = 5 e^x - 3 x^-5 is the sum of the functions 5 e^x and 3 x^-5, whose derivatives are, respectively, 5 e^x and -15 x^-6, we see that y ' is the sum of these derivatives:

y ' = 5 e^x + 15 x^-6.

Note that the derivative of x^-4, where n = -4, is n x^(n-1) = -4 x^(-4-1) = -4 x^-5.

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RESPONSE -->

self critique assessment: 3

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22:53:20

`q007. The rule for the product of two functions is a bit surprising: The derivative of the product f * g of two functions is f ' * g + g ' * f. What are the derivatives of the functions y = x^3 * sin(x), y = e^t cos(t), and y = ln(z) * z^-3?

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RESPONSE -->

For this composite we have f(z) = ln(z) and g(x) = 5 x^7. Thus

f ' (z) = 1 / z and g ' (x) = 35 x^6.

We see that f ' (g(x)) = 1 / g(x) = 1 / ( 5 x^7).

So the derivative of y = ln( 5 x^7) is

y ' = g ' (x) * f ' (g(x)) = 35 x^6 * [ 1 / ( 5 x^7 ) ].

confidence assessment: 3

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22:53:27

The derivative of y = x^3 * sin(x), which is of form f * g if f = x^3 and g = sin(x), is

f ' g + g ' f = (x^3) ' sin(x) + x^3 (sin(x)) '

= 3x^2 sin(x) + x^3 cos(x).

The derivative of y = e^t cos(t), which is of form f * g if f = e^t and g = cos(t), is

f ' g + g ' f = (e^t) ' cos(t) + e^t (cos(t) ) '

= e^t cos(t) + e^t (-sin(t)) = e^t [ cos(t) - sin(t) ].

The derivative of y = ln(z) * z^-3, which is of form f * g if f = ln(z) and g = z^-3, is

f ' g + g ' f = (ln(z)) ' z^-3 +ln(z) ( z^-3) '

= 1/z * z^-3 + ln(z) * (-3 z^-4) =

z^-4 - 3 ln(z) * z^-4 = z^-4 (1 - 3 ln(z)).

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RESPONSE -->

self critique assessment: 3

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22:53:51

`q008. The rule for the quotient of two functions is perhaps even more surprising:

The derivative of the quotient f / g of two functions is [ f ' g - g ' f ] / g^2.

What are the derivatives of the functions y = e^t / t^5, y = sin(x) / cos(x) and y = ln(x) / sin(x)?

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RESPONSE -->

This function is the composite of f(z) = e^z and g(t) = t^2. We see right away that f ' (z) = e^z and g ' (t) = 2t.

Thus the derivative of y = e^(t^2) is y ' = g ' (t) * f ' (g(t)) = 2 t * e^(t^2).

confidence assessment: 3

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22:53:57

The derivative of y = e^t / t^5, which is of form f / g if f = e^t and g = t^5, is

(f ' g - g ' f) / g^2= ( (e^t) ' t^5 - e^t (t^5) ' ) / (t^5)^2

= (e^t * t^5 - e^t * 5 t^4) / (t^5)^2

= t^4 * e^t ( t - 5) / t^10

= e^t (t-5) / t^6..

The derivative of y = sin(x) / cos(x), which is of form f / g if f = sin(x) and g = cos(x), is

(f ' g + g ' f) / g^2 =( (sin(x)) ' cos(x) - sin(x) (cos(x)) ' ) / (cos(x))^2 '

= (cos(x) * cos(x) - sin(x) * -sin(x) ) / (cos(x))^2 =

( (cos(x))^2 + (sin(x))^2 ) / (cos(x))^2 =

1 / cos(x)^2.

Note that we have used the Pythagorean identity (sin(x))^2 + (cos(x))^2 = 1.

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RESPONSE -->

self critique assessment: 3

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22:54:25

The derivative of y = e^t / t^5, which is of form f / g if f = e^t and g = t^5, is

(f ' g - g ' f) / g^2= ( (e^t) ' t^5 - e^t (t^5) ' ) / (t^5)^2

= (e^t * t^5 - e^t * 5 t^4) / (t^5)^2

= t^4 * e^t ( t - 5) / t^10

= e^t (t-5) / t^6..

The derivative of y = sin(x) / cos(x), which is of form f / g if f = sin(x) and g = cos(x), is

(f ' g + g ' f) / g^2 =( (sin(x)) ' cos(x) - sin(x) (cos(x)) ' ) / (cos(x))^2 '

= (cos(x) * cos(x) - sin(x) * -sin(x) ) / (cos(x))^2 =

( (cos(x))^2 + (sin(x))^2 ) / (cos(x))^2 =

1 / cos(x)^2.

Note that we have used the Pythagorean identity (sin(x))^2 + (cos(x))^2 = 1.

The derivative of y = ln(x) / sin(x), which is of form f / g if f = ln(x) and g = sin(x), is

(f ' g + g ' f) / g^2 =( (ln(x)) ' sin(x) - ln(x) (sin(x)) ' ) / (sin(x))^2 =

(sin(x) * 1/x - ln(x) * cos(x) ) / (sin(x))^2 =

( sin(x) / x - ln(x) cos(x) ) / (sin(x))^2 =

1 / ( x sin(x)) - ln(x) cos(x) / (sin(x))^2. Further simplification using the tangent function is possible, but the answer here will be left in terms of the sine and cosine functions.

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RESPONSE -->

the composite of f(z) = cos(z) and z(t) = e^t, with derivatives f ' (z) = -sin(z) and g ' (t) = e^t.

Thus the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t).

self critique assessment: 3

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22:54:42

`q009. Combining the above rules find the derivatives of the following functions: y =4 ln(x) / sin(x) - sin(x) * cos(x); y = 3 e^t / t + 6 ln(t), y = -5 t^5 / ln(t) + sin(t) / 5.

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RESPONSE -->

the composite of f(z) = cos(z) and z(t) = e^t, with derivatives f ' (z) = -sin(z) and g ' (t) = e^t.

Thus the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t).

confidence assessment: 3

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22:55:01

Since the derivative of sin(x) / ln(x) is 1 / ( x sin(x)) + ln(x) cos(x) / (sin(x))^2 ), as just seen, and the derivative of sin(x) * cos(x) is easily seen by the product rule to be -(sin(x))^2 + (cos(x))^2, we see that the derivative of y = 4 sin(x) / ln(x) - sin(x) * cos(x) is y ' = 4 [ 1 / ( x sin(x) - ln(x) cos(x) / (sin(x))^2 ] - ( -(sin(x))^2 + (cos(x))^2 ) =

4 / ( x sin(x) ) - 4 ln(x) cos(x) / ( sin(x))^2 + (sin(x))^2 - (cos(x))^2.

Further rearrangement is possible but will not be done here.

The derivative of 3 e^t / t is found by the quotient rule to be ( 3 e^t * t - 3 e^t * 1 ) / t^2 = 3 e^t ( t - 1) / t^2, the derivative of 6 ln(t) is 6 / t, so the derivative of y = 3 e^t / t + 6 ln(t) is therefore

y ' = 3 e^t ( t - 1) / t^2 + 6 / t.

Since the derivative of -5 t^5 / ln(t) is found by the quotient rule to be ( -25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2, and the derivative of sin(t) / 5 is cos(t) / 5, we see that the derivative of y = -5 t^5 / ln(t) + sin(t) / 5 is

y ' = (-25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2 + cos(t) / 5 =

-25 t^4 ln(t) + 5 t^4 / (ln(t))^2 + cos(t) / 5.

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RESPONSE -->

self critique assessment: 3

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•æådÇæSçÈÏ¼‡àÂRƒÄÎÃqµÚ·¯êÎÙ

assignment #012

012. The Chain Rule

08-05-2008

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22:56:57

`qNote that there are 12 questions in this assignment.

`q001. When we form the composite of two functions, we first apply one function to the variable, then we apply the other function to the result.

We can for example first apply the function z = t^2 to the variable t, then we can apply the function y = e^z to our result.

If we apply the functions as specified to the values t = -2, -1, -.5, 0, .5, 1, 2, what y values to we get?

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RESPONSE -->

the composite of f(z) = cos(z) and z(t) = e^t, with derivatives f ' (z) = -sin(z) and g ' (t) = e^t.

Thus the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t).

confidence assessment: 3

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22:57:12

If t = -2, then z = t^2 = (-2)^2 = 4, so y = e^z = e^4 = 55, approx..

If t = -1, then z = t^2 = (-1 )^2 = 1, so y = e^z = e^1 = 2.72, approx..

If t = -.5, then z = t^2 = (-.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx.

If t = 0, then z = t^2 = ( 0 )^2 = 0, so y = e^z = e^0 = 1.

If t = .5, then z = t^2 = (.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx.

If t = 1, then z = t^2 = ( 1 )^2 = 1, so y = e^z = e^1 = 2.72, approx..

If t = 2, then z = t^2 = ( 2 )^2 = 4, so y = e^z = e^4 = 55, approx..

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RESPONSE -->

self critique assessment: 3

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22:57:31

`q002. If we evaluate the function y = e^(t^2) at t = -2, -1, -.5, 0, .5, 1, 2, what y values do we get?

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RESPONSE -->

If t = -2, then y = e^(t^2) = e^(-2)^2 = e^4 = 55, approx.

If t = -1, then y = e^(t^2) = e^(-1)^2 = e^1 = 2.72, approx.

If t = -.5, then y = e^(t^2) = e^(-.5)^2 = e^.25 = 1.28, approx.

If t = 0, then y = e^(t^2) = e^(0)^2 = e^0 = 1.

If t = .5, then y = e^(t^2) = e^(.5)^2 = e^.25 = 1.28, approx.

If t = 1, then y = e^(t^2) = e^(1)^2 = e^1 = 2.72, approx.

If t = 2, then y = e^(t^2) = e^(2)^2 = e^2 = 55, approx.

confidence assessment: 3

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22:57:38

If t = -2, then y = e^(t^2) = e^(-2)^2 = e^4 = 55, approx.

If t = -1, then y = e^(t^2) = e^(-1)^2 = e^1 = 2.72, approx.

If t = -.5, then y = e^(t^2) = e^(-.5)^2 = e^.25 = 1.28, approx.

If t = 0, then y = e^(t^2) = e^(0)^2 = e^0 = 1.

If t = .5, then y = e^(t^2) = e^(.5)^2 = e^.25 = 1.28, approx.

If t = 1, then y = e^(t^2) = e^(1)^2 = e^1 = 2.72, approx.

If t = 2, then y = e^(t^2) = e^(2)^2 = e^2 = 55, approx.

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RESPONSE -->

self critique assessment: 3

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23:00:54

`q003. We see from the preceding two examples that that the function y = e^(t^2) results from the 'chain' of simple functions z = t^2 and y = e^z.

What would be the 'chain' of simple functions for the function y = cos ( ln(x) )?

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RESPONSE -->

y'=(sin(ln(x)))/x

confidence assessment: 3

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23:01:27

Thus we have y = cos(z) = cos( ln(x) ).

We also say that the function y(x) is the composite of the functions cosine and natural log functions, i.e., the composite of y = cos(z) and z = ln(x).

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RESPONSE -->

self critique assessment: 3

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23:01:50

`q004. What would be the chain of functions for y = ( ln(t) ) ^ 2?

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RESPONSE -->

The first function encountered by the variable t is ln(t), so we say that z = ln(t). This value is then squared so we say that y = z^2.

Thus we have y = z^2 = (ln(t))^2.

We also say that we have here the composite of the squaring function and the natural log function, i.e., the composite of y = z^2 and z = ln(t).

confidence assessment: 3

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23:01:56

The first function encountered by the variable t is ln(t), so we say that z = ln(t). This value is then squared so we say that y = z^2.

Thus we have y = z^2 = (ln(t))^2.

We also say that we have here the composite of the squaring function and the natural log function, i.e., the composite of y = z^2 and z = ln(t).

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RESPONSE -->

self critique assessment: 3

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23:02:33

`q005. What would be the chain of functions for y = ln ( cos(x) )?

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RESPONSE -->

The first function encountered by the variable is cos(x), so we say that z = cos(x). We then take the natural log of this function, so we say that y = ln(z).

Thus we have y = ln(z) = ln(cos(x)).

This function is the composite of the natural log and cosine function, i.e., the composite of y = ln(z) and z = cos(x).

confidence assessment: 3

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23:02:40

The first function encountered by the variable is cos(x), so we say that z = cos(x). We then take the natural log of this function, so we say that y = ln(z).

Thus we have y = ln(z) = ln(cos(x)).

This function is the composite of the natural log and cosine function, i.e., the composite of y = ln(z) and z = cos(x).

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RESPONSE -->

self critique assessment: 3

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23:04:07

`q006. The rule for the derivative of a chain of functions is as follows:

The derivative of the function y = f ( g ( x) ) is y ' = g' ( x ) * f ' ( g ( x) ).

For example if y = cos ( x^2 ) then we see that the f function is f(z) = cos(z) and the g function is the x^2 function, so that f ( g ( x) ) = f ( x^2 ) = cos ( x^2 ) . By the chain rule the derivative of this function will be

(cos(x^2)) ' = g ' ( x) * f ' ( g ( x) ) .

g(x) = x^2 so g'(x) = 2 x.

f ( z ) = cos ( z) so f ' ( z ) = - sin( z ), so f ' ( g ( x ) ) = - sin ( g ( x ) ) = - sin ( x^2).

Thus we obtain the derivative

(cos(x^2)) ' = g ' ( x ) * f ' ( g ( x ) ) =

2 x * ( - sin ( x^2 ) ) =

- 2 x sin ( x^2).

Apply the rule to find the derivative of y = sin ( ln ( x ) ) .

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RESPONSE -->

that y = sin ( ln(x) ) is the composite f(g(x)) of f(z) = sin(z) and g(x) = ln(x). The derivative of this composite is g ' (x) * f ' ( g(x) ).

Since g(x) = ln(x), we have g ' (x) = ( ln(x) ) ' = 1/x.

Since f(z) = sin(z) we have f ' (z) = cos(z).

Thus the derivative of y = sin( ln (x) ) is

y ' = g ' (x) * f ' (g(x)) = 1 / x * cos ( g(x) ) =

1 / x * cos( ln(x) ).

confidence assessment: 3

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23:04:15

We see that y = sin ( ln(x) ) is the composite f(g(x)) of f(z) = sin(z) and g(x) = ln(x). The derivative of this composite is g ' (x) * f ' ( g(x) ).

Since g(x) = ln(x), we have g ' (x) = ( ln(x) ) ' = 1/x.

Since f(z) = sin(z) we have f ' (z) = cos(z).

Thus the derivative of y = sin( ln (x) ) is

y ' = g ' (x) * f ' (g(x)) = 1 / x * cos ( g(x) ) =

1 / x * cos( ln(x) ).

Note how the derivative of the 'inner function' g(x) = ln(x) appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the sine function.

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RESPONSE -->

self critique assessment: 3

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23:04:35

`q007. Find the derivative of y = ln ( 5 x^7 ) .

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RESPONSE -->

For this composite we have f(z) = ln(z) and g(x) = 5 x^7. Thus

f ' (z) = 1 / z and g ' (x) = 35 x^6.

We see that f ' (g(x)) = 1 / g(x) = 1 / ( 5 x^7).

So the derivative of y = ln( 5 x^7) is

y ' = g ' (x) * f ' (g(x)) = 35 x^6 * [ 1 / ( 5 x^7 ) ].

confidence assessment: 3

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23:04:40

For this composite we have f(z) = ln(z) and g(x) = 5 x^7. Thus

f ' (z) = 1 / z and g ' (x) = 35 x^6.

We see that f ' (g(x)) = 1 / g(x) = 1 / ( 5 x^7).

So the derivative of y = ln( 5 x^7) is

y ' = g ' (x) * f ' (g(x)) = 35 x^6 * [ 1 / ( 5 x^7 ) ].

Note again how the derivative of the 'inner function' g(x) = 5 x^7 appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the reciprocal or 1 / z function.

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RESPONSE -->

self critique assessment: 3

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23:04:59

`q008. Find the derivative of y = e ^ ( t ^ 2 ).

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RESPONSE -->

This function is the composite of f(z) = e^z and g(t) = t^2. We see right away that f ' (z) = e^z and g ' (t) = 2t.

Thus the derivative of y = e^(t^2) is y ' = g ' (t) * f ' (g(t)) = 2 t * e^(t^2).

confidence assessment: 3

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23:05:13

This function is the composite of f(z) = e^z and g(t) = t^2. We see right away that f ' (z) = e^z and g ' (t) = 2t.

Thus the derivative of y = e^(t^2) is y ' = g ' (t) * f ' (g(t)) = 2 t * e^(t^2).

Note once more how the derivative of the 'inner function' g(t) = t^2 appears 'out in front' of the derivative of the 'outer' function.

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RESPONSE -->

self critique assessment: 3

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23:05:38

`q009. Find the derivative of y = cos ( e^t ).

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RESPONSE -->

the composite of f(z) = cos(z) and z(t) = e^t, with derivatives f ' (z) = -sin(z) and g ' (t) = e^t.

Thus the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t).

confidence assessment: 3

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23:05:45

We have the composite of f(z) = cos(z) and z(t) = e^t, with derivatives f ' (z) = -sin(z) and g ' (t) = e^t.

Thus the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t).

Note how the 'inner function' is unchanged, as it has been in previous examples, and how its derivative appears in front of the derivative of the 'outer' function.

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RESPONSE -->

self critique assessment: 3

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23:06:30

`q010. Find the derivative of y = ( ln ( t ) ) ^ 9.

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RESPONSE -->

have y = f(g(t)) with f(z) = z^9 and g(t) = ln(t). f ' (z) = 9 z^8 and g ' (t) = 1 / t. Thus

y ' = g ' (t) * f ' (g(t)) = 1/t * 9 ( ln(t) )^8.

confidence assessment: 3

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23:06:42

We have y = f(g(t)) with f(z) = z^9 and g(t) = ln(t). f ' (z) = 9 z^8 and g ' (t) = 1 / t. Thus

y ' = g ' (t) * f ' (g(t)) = 1/t * 9 ( ln(t) )^8.

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RESPONSE -->

self critique assessment: 3

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23:07:03

`q011. Find the derivative of y = sin^4 ( x ).

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RESPONSE -->

The composite here is y = f(g(x)) with f(z) = z^4 and g(x) = sin(x). Note that the notation sin^4 means to raise the value of the sine function to the fourth power.

We see that f ' (z) = 4 z^3 and g ' (x) = cos(x). Thus

y ' = g ' (x) * f ' (g(x)) = cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x).

confidence assessment: 3

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23:07:09

The composite here is y = f(g(x)) with f(z) = z^4 and g(x) = sin(x). Note that the notation sin^4 means to raise the value of the sine function to the fourth power.

We see that f ' (z) = 4 z^3 and g ' (x) = cos(x). Thus

y ' = g ' (x) * f ' (g(x)) = cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x).

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RESPONSE -->

self critique assessment: 3

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23:07:39

`q012. Find the derivative of y = cos ( 3x ).

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RESPONSE -->

This is the composite y = f(g(x)) with f(z) = cos(z) and g(x) = 3x. We obtain f ' (z) = - sin(z) and g ' (x) = 3. Thus

y ' = g ' (x) * f ' (g(x)) = 3 * (-sin (3x) ) = - 3 sin(3x).

confidence assessment: 3

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23:07:47

This is the composite y = f(g(x)) with f(z) = cos(z) and g(x) = 3x. We obtain f ' (z) = - sin(z) and g ' (x) = 3. Thus

y ' = g ' (x) * f ' (g(x)) = 3 * (-sin (3x) ) = - 3 sin(3x).

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RESPONSE -->

self critique assessment: 3

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¼ËwJ{®~ï‡sÎšß÷~…»åÓ–®|Þ÷NÚ¨Ó

assignment #013

013. Applications of the Chain Rule

08-05-2008

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23:19:47

`qNote that there are 4 questions in this assignment.

`q001. The Fahrenheit temperature T of a potato just taken from the oven is given by the function T(t) = 70 + 120 e^(-.1 t), where t is the time in minutes since the potato was removed from the oven. At what rate is the temperature changing at t = 5?

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RESPONSE -->

The rate of temperature change is given by the derivative function ( T ( t ) ) ', also written T ' (t).

Since T(t) is the sum of the constant function 70, whose derivative is zero, and 120 times the composite function e^(-.1 t), whose derivative is -.1 e^(-.1 t), we see that T ' (t) = 120 * ( -.1 e^(-.1 t) ) = -12 e^(-.1 t).

confidence assessment: 3

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23:19:54

The rate of temperature change is given by the derivative function ( T ( t ) ) ', also written T ' (t).

Note that e^(-.1 t) is the composite of f(z) = e^z and g(t) = -.1 t, and that its derivative is therefore found using the chain rule.

When t = 5, we have T ' (5) = -12 e^(-.1 * 5 ) = -12 e^-.5 = -7.3, approx.. This represents rate = change in T / change in t in units of degrees / minute, so at t = 5 minutes the temperature is changing by -7.3 degrees/minute.

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RESPONSE -->

self critique assessment: 3

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23:20:32

`q002. The weight in grams of a growing plant is closely modeled by the function W(t) = .01 e^(.3 t ), where t is the number of days since the seed germinated. At what rate is the weight of the plant changing when t = 10?

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RESPONSE -->

The rate of change of weight is given by the derivative function ( W ( t ) ) ', also written W ' (t).

Since W(t) is .01 times the composite function e^(.3 t), whose derivative is .3 e^(.3 t), we see that W ' (t) = .01 * ( .3 e^(.3 t) ) = .003 e^(.3 t).

confidence assessment: 3

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23:20:40

The rate of change of weight is given by the derivative function ( W ( t ) ) ', also written W ' (t).

Since W(t) is .01 times the composite function e^(.3 t), whose derivative is .3 e^(.3 t), we see that W ' (t) = .01 * ( .3 e^(.3 t) ) = .003 e^(.3 t).

Note that e^(.3 t) is the composite of f(z) = e^z and g(t) = .3 t, and that its derivative is therefore found using the chain rule.

When t = 10 we have W ' (10) = .003 e^(.3 * 10) = .03 e^(3) = .06. Since W is given in grams and t in days, W ' will represent change in weight / change in clock time, measured in grams / day.

Thus at t = 10 days the weight is changing by .06 grams / day.

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RESPONSE -->

self critique assessment: 3

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23:22:24

`q003. The height above the ground, in feet, of a child in a Ferris wheel is given by y(t) = 6 + 40 sin ( .2 t - 1.6 ), where t is clock time in seconds. At what rate is the child's height changing at the instant t = 10?

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RESPONSE -->

The rate of change of altitude is given by the derivative function ( y ( t ) ) ', also written y ' (t).

Since y(t) is the sum of the constant term 6, with derivative zero, and 40 times the composite function sin (.2 t - 1.6), whose derivative is .2 cos(.2 t - 1.6), we see that y ' (t) = 40 * ( .2 cos(.2 t - 1.6) ) = 8 cos(.2 t - 1.6).

confidence assessment: 3

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23:22:31

The rate of change of altitude is given by the derivative function ( y ( t ) ) ', also written y ' (t).

Note that sin(.2t - 1.6) is the composite of f(z) = sin(z) and g(t) = .2 t - 1.6, and that its derivative is therefore found using the chain rule.

Thus at t = 10 seconds we have rate y ' (10) = 8 cos( .2 * 10 - 1.6) = 8 cos( .4) = 7.4, approx.. Since y represents altitude in feet and t represents clock time in seconds, this represents 7.4 feet per second. The child is rising at 7.4 feet per second when t = 10 sec.

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RESPONSE -->

self critique assessment: 3

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23:23:54

`q004. The grade point average of a certain group of students seems to be modeled as a function of weekly study time by G(t) = ( 10 + 3t ) / (20 + t ) + `sqrt( t / 60 ). At what rate does the grade point average go up as study time is added for a typical student who spends 40 hours per week studying? Without calculating G(40.5), estimate how much the grade point average for this student would go up if she spend another 1/2 hour per week studying.

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RESPONSE -->

The rate of change of grade point average is given by the derivative function ( G ( t ) ) ', also written G ' (t).

Since G(t) is the sum of the quotient function (10 + 3 t ) / ( 20 + t), with derivative 50 / ( 20 + t ) ^ 2, and the composite function `sqrt( t / 60) , whose derivative is 1 / (120 `sqrt( t / 60) ), we see that G ' (t) = 50 / ( 20 + t ) ^ 2 + 1 / (120 `sqrt( t / 60) ).

confidence assessment: 3

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23:24:02

The rate of change of grade point average is given by the derivative function ( G ( t ) ) ', also written G ' (t).

Note that `sqrt(t / 60) is the composite of f(z) = `sqrt(z) and g(t) = t / 60, and that its derivative is therefore found using the chain rule.

Thus if t = 40 we have rate G ' (40) = 50 / ( 20 + 40 ) ^ 2 + 1 / (120 `sqrt( 40 / 60) ) = .024, approx.. Since G represents grade point and t represents weekly study time in hours, this represents .024 grade points per hour of weekly study time. The grade point is rising by .024 per additional hour of study.

To estimate G(40.5) we assume that the .024 grade point rise per additional hour of study time remains valid as we increase study time from 40 to 40.5 hours. This is in increase of .5 hours in weekly study time so we would expect the grade point to go up by

grade point change = .5 hours * .024 points / hour = .012 points.

Since G(40) = ( 10 + 3 * 40) / (20 + 40) + `sqrt( 40 / 60) = 2.97 approx, the additional half-hour per week will tend to raise this by .012 to around 2.97. If the student is aiming for a 3-point, a couple more hours would do but the .5 hours won't.

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RESPONSE -->

self critique assessment: 3

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¼|ÍçOžpxºŽù¯ÞÏ¢ìÏx•

assignment #014

014. Tangent Lines and Tangent Line Approximations

08-05-2008

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23:33:58

`q001. What are the coordinates of the x = 5 point and the slope at that point of the graph of the function y = .3 x^2?

What is the equation of the line through the point and having that slope?

Sketch the line and the curve between x = 3 and x = 7 and describe your sketch.

How close, in the vertical direction, is the line to the graph of the y = .3 x^2 function when x = 5.5, 6, 6.5 and 7?

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RESPONSE -->

The x = 5 point has y coordinate y = .3 * 5^2 = 7.5, so the point lies at (5, 7.5).

The slope that that point is found by evaluating the derivative at that point. The derivative is y ' = .6 x, so the slope is y ' = .6 * 5 = 3.

The line therefore passes through the point (5,7.5) and has slope 3. Its equation is therefore y - 7.5 = 3 ( x - 5), which simplifies to y = 3 x - 7.5.

When x = 5.5, 6, 6.5 and 7, the y coordinates of the line are

y = 3 ( 5.5 ) - 7.5 = 9,

y = 3 (6) -7.5 = 10.5,

y = 3 (6.5) -7.5 = 12,

y = 3 (7) -7.5 = 13.5.

At the same x coordinates the function y =.3 x ^ 2 takes values

y = .3 (5.5) ^ 2 = 9.075,

y =.3 (6) ^ 2 = 10.8,

y =.3 (6.5) ^ 2 = 12.675, and

y =.3 (7) ^ 2 = 14.7.

The straight line is therefore lower than the curve by

9.075 - 9 = .075 units when x = 5.5,

10.8-10.5 = .3 when x = 6,

12.675-12 = .675 when x = 6.5, and

14.7-13.5 = 1.2 when x = 7.

We see that as we move away from the common point (5,7.5), the line moves away from the curve more and more rapidly.

Your sketch should show a straight line tangent to the parabolic curve y =.3 x^2, with the curve always above the line except that the point tangency, and moving more and more rapidly away from the line the further is removed from the point (5,7.5).

The line you have drawn is called the line tangent to the curve at the point (5,7.5).

confidence assessment: 3

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23:34:06

The x = 5 point has y coordinate y = .3 * 5^2 = 7.5, so the point lies at (5, 7.5).

The slope that that point is found by evaluating the derivative at that point. The derivative is y ' = .6 x, so the slope is y ' = .6 * 5 = 3.

The line therefore passes through the point (5,7.5) and has slope 3. Its equation is therefore y - 7.5 = 3 ( x - 5), which simplifies to y = 3 x - 7.5.

When x = 5.5, 6, 6.5 and 7, the y coordinates of the line are

y = 3 ( 5.5 ) - 7.5 = 9,

y = 3 (6) -7.5 = 10.5,

y = 3 (6.5) -7.5 = 12,

y = 3 (7) -7.5 = 13.5.

At the same x coordinates the function y =.3 x ^ 2 takes values

y = .3 (5.5) ^ 2 = 9.075,

y =.3 (6) ^ 2 = 10.8,

y =.3 (6.5) ^ 2 = 12.675, and

y =.3 (7) ^ 2 = 14.7.

The straight line is therefore lower than the curve by

9.075 - 9 = .075 units when x = 5.5,

10.8-10.5 = .3 when x = 6,

12.675-12 = .675 when x = 6.5, and

14.7-13.5 = 1.2 when x = 7.

We see that as we move away from the common point (5,7.5), the line moves away from the curve more and more rapidly.

The line you have drawn is called the line tangent to the curve at the point (5,7.5).

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RESPONSE -->

self critique assessment: 3

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23:34:40

`q002. What is the equation of the line tangent to the curve y = 120 e^(-.02 t) at the t = 40 point?

If we follow the tangent line instead of the curve, what will be the y coordinate at t = 40.3?

How close will this be to the y value predicted by the original function?

Will the tangent line be closer than this or further from the original function at t = 41.2?

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RESPONSE -->

The tangent line at the x = 40 point will have y coordinate y = 120 e^(-.02 * 40) = 53.9194, accurate to six significant figures. The derivative of the function is y ' = -2.4 e^(-.02 t) and at x = 40 has value y ' = -2.4 e^(-.02 * 40) = -1.07838, accurate to six significant figures.

The tangent line will therefore pass through the point (40, 53.9194) and will have slope 1.07838. Its equation will therefore be

y - 53.9194 = -1.07838 ( x - 40), which we can solve for y to obtain

y = -1.07838 x + 97.0546.

The values of the function at x = 40.3 and x = 41.2 are 53.5969 and 52.6408.

The y values corresponding to the tangent line are 53.5958 and 52.6253. We see that the tangent line at x = 40.3 is .0011 units lower than the curve of the function, and at x = 41.2 the tangent line is .0155 units lower--about 14 times as far from the curve as at the x = 40.3 point.

confidence assessment: 3

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23:34:48

The tangent line will therefore pass through the point (40, 53.9194) and will have slope 1.07838. Its equation will therefore be

y - 53.9194 = -1.07838 ( x - 40), which we can solve for y to obtain

y = -1.07838 x + 97.0546.

The values of the function at x = 40.3 and x = 41.2 are 53.5969 and 52.6408.

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RESPONSE -->

self critique assessment: 3

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23:35:16

`q003. Sketch a graph of this curve and the t = 40 tangent line and describe how the closeness of the tangent line to the curve changes as we move away from the t = 40 point.

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RESPONSE -->

show the gradually curving exponential function very close to the tangent line. The tangent line is straight while the slope of the exponential function is always increasing. At the t - 40 point the two meet for an instant, but as we move both to the right and to the left the exponential curve moves away from the tangent line, moving away very gradually at first and then with increasing rapidity.

confidence assessment: 3

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23:35:23

Your sketch should show the gradually curving exponential function very close to the tangent line. The tangent line is straight while the slope of the exponential function is always increasing. At the t - 40 point the two meet for an instant, but as we move both to the right and to the left the exponential curve moves away from the tangent line, moving away very gradually at first and then with increasing rapidity.

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RESPONSE -->

self critique assessment: 3

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23:35:44

`q004. What is the equation of the line tangent to the curve y = 20 ln( 5 x ) at the x = 90 point?

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RESPONSE -->

At x = 90, we have y = 20 ln (5 * 90) = 20 ln(450) = 20 * 6.1 = 126, approx..

So we are looking for a straight line through the point (90, 126). The slope of this straight line should be equal to the derivative of y = 20 ln(5 x) at x = 90.

The derivative of ln(5x) is, by the chain rule, (5x) ' * 1 / (5x) = 5 / (5x) = 1/x. So y ' = 20 * 1 / x = 20 / x.

At x = 90 then we have y ' = 20 / 90 = .22, approx.. Thus the tangent line passes through (90, 126) and has slope .22. The equation of this line is

( y - 126 ) / (x - 90) = .22, which is easily rearranged to the form

y = .22 x + 106.

confidence assessment: 3

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23:35:54

At x = 90, we have y = 20 ln (5 * 90) = 20 ln(450) = 20 * 6.1 = 126, approx..

So we are looking for a straight line through the point (90, 126). The slope of this straight line should be equal to the derivative of y = 20 ln(5 x) at x = 90.

The derivative of ln(5x) is, by the chain rule, (5x) ' * 1 / (5x) = 5 / (5x) = 1/x. So y ' = 20 * 1 / x = 20 / x.

At x = 90 then we have y ' = 20 / 90 = .22, approx.. Thus the tangent line passes through (90, 126) and has slope .22. The equation of this line is

( y - 126 ) / (x - 90) = .22, which is easily rearranged to the form

y = .22 x + 106.

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RESPONSE -->

self critique assessment: 3

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23:36:15

`q005. Using the tangent line and the value of the function at x = 3, estimate the value of f(x) = x^5 and x = 3.1. Compare with the actual value.

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RESPONSE -->

f ' (x) = 5 x^4. When x = 3, f(x) = 3^5 = 243 and f ' (x) = 5 * 3^4 = 405.

Thus tangent line passes through (3, 243) and has slope 405. Its equation is therefore

(y - 243) / (x - 3) = 405, which simplifies to

y = 405 x - 972.

This function could be evaluated at x = 3.1. However a more direct approach simply uses the differential.

In this approach we note that at x = 3 the derivative, which gives the slope of the tangent line, is 405. This means that between (3, 243) and the x = 3.1 point the rise/run ratio should be close to 405. The run from x = 3 to x = 3.1 is 1, so a slope of 405 therefore implies a rise equal to slope * run = 40.5.

A rise of about 40.5 means that y = 3.1^5 should change by about 40.5, from 243 to about 283.5.

The actual value of 3.1^5 is about 286.3. The discrepancy between the straight-line approximation and the actual value is due to the fact that y = x^5 is concave up, meaning that the slope is actually increasing.

The closely-related idea of the differential, which will be developed more fully in the next section, is that if we know the rate at which the function is changing at a given point, we can use this rate to estimate its change as we move to nearby points.

confidence assessment: 3

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23:36:21

f ' (x) = 5 x^4. When x = 3, f(x) = 3^5 = 243 and f ' (x) = 5 * 3^4 = 405.

Thus tangent line passes through (3, 243) and has slope 405. Its equation is therefore

(y - 243) / (x - 3) = 405, which simplifies to

y = 405 x - 972.

This function could be evaluated at x = 3.1. However a more direct approach simply uses the differential.

A rise of about 40.5 means that y = 3.1^5 should change by about 40.5, from 243 to about 283.5.

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RESPONSE -->

self critique assessment: 3

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23:36:41

`q006. Using the differential and the value of the function at x = e, estimate the value of ln(2.8). Compare with the actual value.

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RESPONSE -->

At x = e we have ln(x) = ln(e) = 1 (this is because of the definition of the natural log function as the inverse of the exponential function).

The derivative of ln(x) is 1/x, so at x = e the rate at which financial log function is changing is 1 / e.

{}Since e = 2.718, approx., between x = e and x = 2.8 x changes by about 2.8 - 2.718 = .082. Since the rate at which x is changing in this vicinity remains close to x = 1/e, the change in y is approximately

`dy = rate * `dx = 1/e * .082 = .082 / 2.718 = .030.

Thus at x = 2.8, y = ln(x) takes value approximately 1 + .030 - 1.030.

The actual value of ln(2.8) to six significant figures is 1.02961, which when rounded off to four significant figures is 1.030, in agreement with our approximation.

confidence assessment: 3

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23:36:48

At x = e we have ln(x) = ln(e) = 1 (this is because of the definition of the natural log function as the inverse of the exponential function).

The derivative of ln(x) is 1/x, so at x = e the rate at which financial log function is changing is 1 / e.

`dy = rate * `dx = 1/e * .082 = .082 / 2.718 = .030.

Thus at x = 2.8, y = ln(x) takes value approximately 1 + .030 - 1.030.

The actual value of ln(2.8) to six significant figures is 1.02961, which when rounded off to four significant figures is 1.030, in agreement with our approximation.

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RESPONSE -->

self critique assessment: 3

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23:37:10

`q007. Using the tangent line verify that the square root of a number close to 1 is twice as close to 1 as the number. Hint: Find the tangent line approximation for the function f(x) = `sqrt(x) at an appropriate point.

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RESPONSE -->

The function f(x) = `sqrt(x) takes value 1 when x = 1. The derivative of f(x) = `sqrt(x) is 1 / ( 2 * `sqrt(x) ), which takes value 1/2 when x = 1.

The tangent-line approximation to f(x) is therefore a straight line through (1, 1) having slope 1/2. This line has equation (y - 1) / (x - 1) = 1/2, or y = 1/2 ( x - 1) + 1.

This number differs from 1 by 1/2 (x-1), which is half the difference between 1 and x. This approximation works for x values near 1.

Thus if x is close to 1, then y is approximately twice as close to 1 as is x.

confidence assessment: 3

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23:37:16

The function f(x) = `sqrt(x) takes value 1 when x = 1. The derivative of f(x) = `sqrt(x) is 1 / ( 2 * `sqrt(x) ), which takes value 1/2 when x = 1.

The tangent-line approximation to f(x) is therefore a straight line through (1, 1) having slope 1/2. This line has equation (y - 1) / (x - 1) = 1/2, or y = 1/2 ( x - 1) + 1.

This number differs from 1 by 1/2 (x-1), which is half the difference between 1 and x. This approximation works for x values near 1.

Thus if x is close to 1, then y is approximately twice as close to 1 as is x.

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RESPONSE -->

self critique assessment: 3

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23:37:49

`q008. Using the tangent line approximaton verify that the square of a number close to 1 is twice as far from 1 as the number.

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RESPONSE -->

The tangent-line approximation to f(x) = x^2, near x = 1, is y = 2(x-1) + 1. This number is twice as far from 1 as is x.

confidence assessment: 3

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23:37:57

The tangent-line approximation to f(x) = x^2, near x = 1, is y = 2(x-1) + 1. This number is twice as far from 1 as is x.{}{}STUDENT COMMENT: I am not really understanding this part. INSTRUCTOR RESPONSE: *&*& You should understand the part about the tangent-line approximation near x = 1 being y = 2(x-1) + 1.{}{}Graph y = 2(x-1) + 1. If you move horizontally over from the point (1, 1) you are changing x. If you then move vertically up to the graph from this point you will find the change in y. For this function that change will be double the change in x. *&*&.

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RESPONSE -->

self critique assessment: 3

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è×…¡°¯â›¶©~ä{·ß¿”Ä

assignment #004

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-05-2008

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20:01:13

query modeling project #2 #5. $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300?

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RESPONSE -->

growth rate is r=0.1

growth factor is 1.1

7.27 year

4.25 year

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20:01:24

** The growth rate is .10 and the growth factor is 1.10 so the amount function is $200 * 1.10^t.

This graph passes through the y axis at (0, 200), increases at an increasing rate and is asymptotic to the negative x axis.

For t=0, 1, 2 you get p(t) values $200, $220 and $242--you can already see that the rate is increasing since the increases are $20 and $22.

Note that you should know from precalculus the characteristics of the graphs of exponential, polynomial an power functions (give that a quick review if you don't--it will definitely pay off in this course).

$400 is double the initial $200. We need to find how long it takes to achieve this.

Using trial and error we find that $200 * 1.10^tDoub = $400 if tDoub is a little more than 7. So doubling takes a little more than 7 years. The actual time, accurate to 5 significant figures, is 7.2725 years. If you are using trial and error it will take several tries to attain this accuracy; 7.3 years is a reasonable approximation for trail and error.

To reach $300 from the original $200, using amount function $200 * 1.10^t, takes a little over 4.25 years (4.2541 years to 5 significant figures); again this can be found by trial and error.

The amount function reaches $600 in a little over 11.5 years (11.527 years), so to get from $300 to $600 takes about 11.5 years - 4.25 years = 7.25 years (actually 7.2725 years to 5 significant figures). **

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RESPONSE -->

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20:03:31

At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?

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RESPONSE -->

the t = 20 value is $200 * 1.1^20 = $1340, approx.

Half the t = 20 value is therefore $1340/2 = $670 approx..

By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx..

For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45).ÿ At 12.75=674.20 so it would probably be about12.72.ÿ

This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr.

This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end.

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20:03:43

** The t = 20 value is $200 * 1.1^20 = $1340, approx.

Half the t = 20 value is therefore $1340/2 = $670 approx..

By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx..

This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr.

This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. **

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RESPONSE -->

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20:04:32

query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%

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RESPONSE -->

for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double.

for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double.

Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double.

The final 4-year amount increases by more and more with each 10% increase in interest rate.

The doubling time decreases, but by less and less with each 10% increase in interest rate

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20:04:40

** We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double.

The final 4-year amount increases by more and more with each 10% increase in interest rate.

The doubling time decreases, but by less and less with each 10% increase in interest rate. **

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RESPONSE -->

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20:05:39

query #11. equation for doubling time

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RESPONSE -->

P0 * (1+r)^t = 2 P0

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20:05:46

** the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The resulting equation is therefore

P0 * (1+r)^t = 2 P0.

Note that this simplifies to

(1 + r)^ t = 2,

and that this result depends only on the interest rate, not on the initial amount P0. **

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RESPONSE -->

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20:06:52

Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%.

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RESPONSE -->

dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get

1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2].

This can be written as

1.08^2 * 1.08^doublingtime = 2 * 1.08^2.

Dividing both sides by 1.08^2 we obtain

1.08^doublingtime = 2

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20:07:14

**dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get

1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2].

This can be written as

1.08^2 * 1.08^doublingtime = 2 * 1.08^2.

Dividing both sides by 1.08^2 we obtain

1.08^doublingtime = 2.

We can then use trial and error to find the doubling time that works. We get something like 9 years. **

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RESPONSE -->

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20:08:15

Desribe how on your graph how you obtained an estimate of the doubling time.

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RESPONSE -->

you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis.

The interval from t = 0 to the clock time found on the horizontal axis is the doubling time

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20:08:21

**In this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis.

The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. **

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RESPONSE -->

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í™¤‚Ïé®}Žûº¶}s–¦ùK¤‡

assignment #005

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-05-2008

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20:20:50

Growth rate and growth factor: Describe the difference between growth rate and growth factor and give a short example of how each might be used

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RESPONSE -->

When multiplied by a quantity the growth rate tells us how much the quantity will change over a single period.

When multiplied by the quantity the growth factor gives us the new quantity at the end of the next period.

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20:21:05

** Specific statements:

When multiplied by a quantity the growth rate tells us how much the quantity will change over a single period.

When multiplied by the quantity the growth factor gives us the new quantity at the end of the next period. **

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RESPONSE -->

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20:21:42

Class notes #05 trapezoidal representation.

Explain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented

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RESPONSE -->

The slope of the trapezoids will indicate rise over run

or the slope will represent a change in depth / time interval

thus an average rate of change of depth with respect to time

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20:21:49

** GOOD ANSWER BY STUDENT WITH INSTRUCTOR COMMENTS:

The slope of the trapezoids will indicate rise over run

or the slope will represent a change in depth / time interval

thus an average rate of change of depth with respect to time

INSTRUCTOR COMMENTS:

More detail follows:

** To explain the meaning of the slope you have to reason the question out in terms of rise and run and slope.

For this example rise represents change in depth and run represent change in clock time; rise / run therefore represents change in depth divided by change in clock time, which is the average rate of change. **

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RESPONSE -->

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20:22:33

Explain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval.

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RESPONSE -->

The area of a rate vs. time graph rep. the change in quantity.

Calculating the area under the graph is basically integration

The accumulated area of all the trapezoids for a range will give us thetotal change in quantity.

The more trapezoids used the more accurate the approx.

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20:22:38

**STUDENT RESPONSE WITH INSTRUCTOR COMMENTS:

The area of a rate vs. time graph rep. the change in quantity.

Calculating the area under the graph is basically integration

The accumulated area of all the trapezoids for a range will give us thetotal change in quantity.

The more trapezoids used the more accurate the approx.

INSTRUCTOR COMMENTS: All very good but the other key point is that the average altitude represents the average rate, which when multiplied by the width which represents time interval gives the change in quantity

You have to reason this out in terms of altitudes, widths and areas.

For the rate of depth change example altitude represents rate of depth change so average altitude represents average rate of depth change, and width represents change in clock time.

average altitude * width therefore represents ave rate of depth change * duration of time interval = change in depth.

For the rate of change of a quantity other than depth, the reasoning is identical except you'll be talking about something besides depth. **

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RESPONSE -->

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20:23:21

ÿÿÿ #17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What is the function Q(t)?

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RESPONSE -->

Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have

Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or

Q(t)=550(.89)^tÿ

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20:23:27

** Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have

Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or

Q(t)=550(.89)^tÿ **

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RESPONSE -->

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20:24:04

How much antibiotic is present at 3:00 p.m.?

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RESPONSE -->

3:00 p.m. is 5 hours after the initial time so at that time there will be

Q(5) = 550 mg * .89^5 = 307.123mg

in the blood

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20:24:10

** 3:00 p.m. is 5 hours after the initial time so at that time there will be

Q(5) = 550 mg * .89^5 = 307.123mg

in the blood **

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RESPONSE -->

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20:24:48

Describe your graph and explain how it was used to estimate half-life.

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RESPONSE -->

Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point.

The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down.

The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines.

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20:24:55

** Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point.

The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. **

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RESPONSE -->

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20:25:36

What is the equation to find the half-life?ÿ What is its most simplified form?

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RESPONSE -->

Q(doublingTime) = 1/2 Q(0)or

550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have

.89^doublingTime = .5

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20:25:45

** Q(doublingTime) = 1/2 Q(0)or

550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have

.89^doublingTime = .5.

We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). **

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RESPONSE -->

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20:26:11

#19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0.

For what values of t did Q(t) lie between .005 Q0 and .01 Q0?

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RESPONSE -->

Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0.

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20:26:19

** Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0.

Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0.

Solving Q(t) = .05 Q0 we rewrite this as

Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get

1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get

t = -31.4 approx.

Solving Q(t) = .1 Q0 we rewrite this as

Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get

1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get

t = -24.2 approx.

(The solution for .005 Q0 is about -55.6, for .01 is about -48.3

For this solution any value between about t = -48.3 and t = -55.6 will work). **

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RESPONSE -->

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20:27:13

explain why the negative t axis is a horizontal asymptote for this function.

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RESPONSE -->

The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero.

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20:27:18

** The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero. **

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RESPONSE -->

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20:27:48

#22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )?

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RESPONSE -->

12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx.

So this function is of the form y = A b^x for b = .61 approx.

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20:27:53

** 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx.

So this function is of the form y = A b^x for b = .61 approx.. **

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RESPONSE -->

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20:28:42

what is b for the function y = .007 ( e^(.71 x) )?

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RESPONSE -->

12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx.

So this function is of the form y = A b^x for b = 2.041 approx.

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20:28:48

** 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx.

So this function is of the form y = A b^x for b = 2.041 approx.. **

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RESPONSE -->

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20:29:15

what is b for the function y = -13 ( e^(3.9 x) )?

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RESPONSE -->

12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx.

So this function is of the form y = A b^x for b = 49.4 approx.

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20:29:20

** 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx.

So this function is of the form y = A b^x for b = 49.4 approx.. **

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RESPONSE -->

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20:29:54

List these functions, each in the form y = A b^x.

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RESPONSE -->

y=12(.6065^x)

y=.007(2.03399^x) and

y=-13(49.40244^x)

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20:29:59

** The functions are

y=12(.6065^x)

y=.007(2.03399^x) and

y=-13(49.40244^x) **

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RESPONSE -->

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20:30:39

query text problem 1.1 #23 dolphin energy prop cube of vel

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RESPONSE -->

A proportionality to the cube would be E = k v^3.

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20:30:44

** A proportionality to the cube would be E = k v^3. **

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RESPONSE -->

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20:31:16

query text problem 1.1 #32 temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts

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RESPONSE -->

The interpretation would be that the vertical intercept represents the temperature at clock time t = 0, while the horizontal intercept represents the clock time at which the temperature reaches zero.

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20:31:23

** The interpretation would be that the vertical intercept represents the temperature at clock time t = 0, while the horizontal intercept represents the clock time at which the temperature reaches zero. **

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RESPONSE -->

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20:31:47

what is the meaning of the equation H(30) = 10?

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RESPONSE -->

This means that when clock time t is 30, the temperature H is 10.

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20:31:52

** This means that when clock time t is 30, the temperature H is 10. **

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RESPONSE -->

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20:32:17

What is the meaning of the vertical intercept?

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RESPONSE -->

This is the value of H when t = 0--i.e., the temperature at clock time 0.

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20:32:24

** This is the value of H when t = 0--i.e., the temperature at clock time 0. **

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RESPONSE -->

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20:32:51

What is the meaning of the horizontal intercept?

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RESPONSE -->

This is the t value when H = 0--the clock time when temperature reaches 0

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20:32:55

** This is the t value when H = 0--the clock time when temperature reaches 0 **

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20:33:47

query text problem 1.1.31. Water freezes 0 C, 32 F; boils 100 C, 212 F. Give your solution to problem 1.1.32.

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RESPONSE -->

The graph contains points (0, 32) and (100, 212). The slope is therefore (212-32) / (100-0) = 1.8.

The y-intercept is 32 so the equation of the line is

y = 1.8 x + 32, or using F and C

F = 1.8 C + 32.

To find the Fahrenheit temp corresponding to 20 C we substitute C = 20 into F = 1.8 C + 32 to get

F = 1.8 * 20 + 32 = 36 + 32 = 68

The two temperatures will be equal when F = C. Substituting C for F in F = 1.8 C + 32 we get

C = 1.8 C + 32. Subtracting 1.8 C from both sides we have

-.8 C = 32 or

C = 32 / (-.8) = -40.

The scales read the same at -40 degrees.

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20:33:52

** The graph contains points (0, 32) and (100, 212). The slope is therefore (212-32) / (100-0) = 1.8.

The y-intercept is 32 so the equation of the line is

y = 1.8 x + 32, or using F and C

F = 1.8 C + 32.

To find the Fahrenheit temp corresponding to 20 C we substitute C = 20 into F = 1.8 C + 32 to get

F = 1.8 * 20 + 32 = 36 + 32 = 68

The two temperatures will be equal when F = C. Substituting C for F in F = 1.8 C + 32 we get

C = 1.8 C + 32. Subtracting 1.8 C from both sides we have

-.8 C = 32 or

C = 32 / (-.8) = -40.

The scales read the same at -40 degrees. **

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„…®»¬‰wÕ³‚–•Äôî„Ìšµ

assignment #006

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Calculus I

08-05-2008

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20:41:04

Query class notes #06 If x is the height of a sandpile and y the volume, what proportionality governs geometrically similar sandpiles? Why should this be the proportionality?

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RESPONSE -->

the proportionality is y = k x^3. Any proportionality of volumes is a y = k x^3 proportionality because volumes can be filled with tiny cubes; surface areas are y = k x^2 because surfaces can be covered with tiny squares.

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20:41:10

** the proportionality is y = k x^3. Any proportionality of volumes is a y = k x^3 proportionality because volumes can be filled with tiny cubes; surface areas are y = k x^2 because surfaces can be covered with tiny squares. **

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20:42:31

If x is the radius of a spherical balloon and y the surface area, what proportionality governs the relationship between y and x? Why should this be the proportionality?

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RESPONSE -->

Just as little cubes can be thought of as filling the volume to any desired level of accuracy, little squares can be thought of as covering any smooth surface. Cubes 'scale up' in three dimensions, squares in only two. So the proportionality is y = k x^2.

Surfaces can be covered as nearly as we like with tiny squares (the more closely we want to cover a sphere the tinier the squares would have to be). The area of a square is proportional to the square of its linear dimensions. Radius is a linear dimension. Thus the proportionality for areas is y = k x^2.

By contrast, for volumes or things that depend on volume, like mass or weight, we would use tiny cubes to fill the volume. Volume of a cube is proportional to the cube of linear dimensions. Thus the proportionality for a volume would be y = k x^3.

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20:42:37

** Just as little cubes can be thought of as filling the volume to any desired level of accuracy, little squares can be thought of as covering any smooth surface. Cubes 'scale up' in three dimensions, squares in only two. So the proportionality is y = k x^2.

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20:44:17

Explain how you would use the concept of the differential to find the volume of a sandpile of height 5.01 given the volume of a geometrically similar sandpile of height 5, and given the value of k in the y = k x^3 proportionality between height and volume.

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RESPONSE -->

The class notes showed you that the slope of the y = k x^3 graph is given by the rate-of-change function y' = 3 k x^2. Once you have evaluated k, using the given information, you can evaluate y' at x = 5. That gives you the slope of the line tangent to the curve, and also the rate at which y is changing with respect to x. When you multiply this rate by the change in x, you get the change in y.

The differential is 3 k x^2 `dx and is approximately equal to the corresponding `dy. Since `dy / `dx = 3 k x^2, the differential looks like a simple algebraic rearrangement `dy = 3 k x^2 `dx, though what's involved isn't really simple algebra. The differential expresses the fact that near a point, provided the function has a continuous derivative, the approximate change in y can be found by multiplying the change in x by the derivative). That is, `dy = derivative * `dx (approx)., or `dy = slope at given point * `dx (approx), or `dy = 3 k x^2 `dx (approx).

The idea is that the derivative is the rate of change of the function. We can use the rate of change and the change in x to find the change in y.

The differential uses the fact that near x = 5 the change in y can be approximated using the rate of change at x = 5.

Our proportionality is y = k x^3. Let y = f(x) = k x^3. Then y' = f'(x) = 3 k x^2. When x = 5 we have y' = f'(5) = 75 k, whatever k is. To estimate the change in y corresponding to the change .01 in x, we will multiply y ' by .01, getting a change of y ' `dx = 75 k * .01.

}

SPECIFIC EXAMPLE: We don't know what k is for this specific question. As a specific example suppose our information let us to the value k = .002, so that our proportionality is y = .002 x^3. Then the rate of change when x is 5 would be f'(5) = 3 k x^2 = 3 k * 5^2 = 75 k = .15 and the value of y would be y = f(5) = .002 * 5^3 = .25. This tells us that at x = 5 the function is changing at a rate of .15 units of y for each unit of x.

Thus if x changes from 5 to 5.01 we expect that the change will be

change in y = (dy/dx) * `dx =

rate of change * change in x (approx) =

.15 * .01 = .0015,

so that when x = 5.01, y should be .0015 greater than it was when x was 5. Thus y = .25 + .0015 = .2515. This is the differential approximation. It doesn't take account of the fact that the rate changes slightly between x=5 and x = 5.01. But we don't expect it to change much over that short increment, so we expect that the approximation is pretty good.

Now, if you evaluate f at x = 5.01 you get .251503. This is a little different than the .2515 approximation we got from the differential--the differential is off by .000003. That's not much, and we expected it wouldn't be much because the derivative doesn't change much over that short interval. But it does change a little, and that's the reason for the discrepancy.

The differential works very well for decently behaved functions (ones with smooth curves for graphs) over sufficiently short intervals.

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20:44:24

** The class notes showed you that the slope of the y = k x^3 graph is given by the rate-of-change function y' = 3 k x^2. Once you have evaluated k, using the given information, you can evaluate y' at x = 5. That gives you the slope of the line tangent to the curve, and also the rate at which y is changing with respect to x. When you multiply this rate by the change in x, you get the change in y.

The idea is that the derivative is the rate of change of the function. We can use the rate of change and the change in x to find the change in y.

The differential uses the fact that near x = 5 the change in y can be approximated using the rate of change at x = 5.

}

Thus if x changes from 5 to 5.01 we expect that the change will be

change in y = (dy/dx) * `dx =

rate of change * change in x (approx) =

.15 * .01 = .0015,

The differential works very well for decently behaved functions (ones with smooth curves for graphs) over sufficiently short intervals.**

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20:46:41

What would be the rate of depth change for the depth function y = .02 t^2 - 3 t + 6 at t = 30? (instant response not required)

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RESPONSE -->

the rate of change for depth function y = a t^2 + b t + c is y ' = 2 a t + b.

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20:46:51

** You saw in the class notes and in the q_a_ that the rate of change for depth function y = a t^2 + b t + c is y ' = 2 a t + b. This is the function that should be evaluated to give you the rate.

Evaluating the rate of depth change function y ' = .04 t - 3 for t = 30 we get y ' = .04 * 30 - 3 = 1.2 - 3 = -1.8.

COMMON ERROR: y = .02(30)^2 - 2(30) + 6 =-36 would be the rate of depth change

INSTRUCTOR COMMENT: This is the depth, not the rate of depth change. **

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20:47:53

modeling project 3 problem a single quarter-cup of sand makes a cube 1.5 inches on a side. How many quarter-cups would be required to make a cube with twice the scale, 3 inches on a side? Explain how you know this.

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RESPONSE -->

You can think of stacking single cubes--to double the dimensions of a single cube you would need 2 layers, 2 rows of 2 in each layer.

Thus it would take 8 cubes 1.5 inches on a side to make a cube 3 inches on a side.

Since each 1.5 inch cube containts a quarter-cup, a 3 inch cube would contain 8 quarter-cups.

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20:48:39

** You can think of stacking single cubes--to double the dimensions of a single cube you would need 2 layers, 2 rows of 2 in each layer.

Thus it would take 8 cubes 1.5 inches on a side to make a cube 3 inches on a side.

Since each 1.5 inch cube containts a quarter-cup, a 3 inch cube would contain 8 quarter-cups.

COMMON ERROR:

It would take 2 quarter-cups.

INSTRUCTOR COMMENT: 2 quarter-cups would make two 1.5 inch cubes, which would not be a 3-inch cube but could make a rectangular solid with a square base 1.5 inches on a side and 3 inches high. **

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20:50:55

What value of the parameter a would model this situation? How many quarter-cups does this model predict for a cube three inches on a side? How does this compare with your previous answer?

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RESPONSE -->

y = a x^3,

with y = 1 (representing one quarter-cup) when x = 1.5. So we have

1 = a * 1.5^3, so that

a = 1 / 1.5^3 = .296 approx.

So the model is y = .2963 x^3.

Therefore if x = 3 we have

y = .296 * 3^3 = 7.992, which is the same as 8 except for roundoff error.

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20:51:00

** The proportionality would be

y = a x^3,

with y = 1 (representing one quarter-cup) when x = 1.5. So we have

1 = a * 1.5^3, so that

a = 1 / 1.5^3 = .296 approx.

So the model is y = .2963 x^3.

Therefore if x = 3 we have

y = .296 * 3^3 = 7.992, which is the same as 8 except for roundoff error. **

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20:51:31

What would be the side measurement of a cube designed to hold 30 quarter-cups of sand? What equation did you solve to get this?

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RESPONSE -->

30 = .296 x^3 so that

x^3 = 30 / .296 = 101, approx, and

x = 101^(1/3) = 4.7, approx.

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20:51:37

** You are given the number of quarter-cups, which corresponds to y. Thus we have

30 = .296 x^3 so that

x^3 = 30 / .296 = 101, approx, and

x = 101^(1/3) = 4.7, approx..**

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20:52:09

query problem 2. Someone used 1/2 cup instead of 1/4 cup. The best-fit function was y = .002 x^3. What function would have been obtained using 1/4 cup?

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RESPONSE -->

In this case, since it takes two quarter-cups to make a half-cup, the person would need twice as many quarter-cups to get the same volume y.

He would have obtained half as many half-cups as the actual number of quarter-cups.

To get the function for the number of quarter-cups he would therefore have to double the value of y, so the function would be y = .004 x^3.

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20:52:13

** In this case, since it takes two quarter-cups to make a half-cup, the person would need twice as many quarter-cups to get the same volume y.

He would have obtained half as many half-cups as the actual number of quarter-cups.

To get the function for the number of quarter-cups he would therefore have to double the value of y, so the function would be y = .004 x^3. **

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20:52:46

query problem 4. number of swings vs. length data. Which function fits best?

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RESPONSE -->

If you try the different functions, then for each one you can find a value of a corresponding to every data point. For example if you use y = a x^-2 you can plug in every (x, y) pair and solve to see if your values of a are reasonably consistent. Try this for the data and you will find that y = a x^-2 does not give you consistent a values-every (x, y) pair you plug in will give you a very different value of a.

The shape of the graph gives you a pretty good indication of which one to try, provided you know the shapes of the basic graphs.

For this specific situation the graph of the # of swings vs. length decreases at a decreasing rate.

The graphs of y = a x^.p for p = -.3, -.4, -.5, -.6 and -.7 all decrease at a decreasing rate. In this case you would find that the a x^-.5 function works nicely, giving a nearly constant value of a.

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20:52:51

** If you try the different functions, then for each one you can find a value of a corresponding to every data point. For example if you use y = a x^-2 you can plug in every (x, y) pair and solve to see if your values of a are reasonably consistent. Try this for the data and you will find that y = a x^-2 does not give you consistent a values-every (x, y) pair you plug in will give you a very different value of a.

The shape of the graph gives you a pretty good indication of which one to try, provided you know the shapes of the basic graphs.

For this specific situation the graph of the # of swings vs. length decreases at a decreasing rate.

The graphs of y = a x^.p for p = -.3, -.4, -.5, -.6 and -.7 all decrease at a decreasing rate. In this case you would find that the a x^-.5 function works nicely, giving a nearly constant value of a. **

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20:53:14

problem 7. time per swing model. For your data what expression represents the number of swings per minute?

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RESPONSE -->

The model that best fits the data is a x^-.5, and with accurate data we find that a is close to 55.

The model is pretty close to

# per minute frequency = 55 x^-.5.

As a specific example let's say we obtained counts of 53, 40, 33 and 26 cycles in a minute at lengths of 1, 2, 3 and 4 feet, then using y = a x^-.5 gives you a = y * x^.5. Evaluating a for y = 53 and x = 1 gives us a = 53 * 1^.5 = 53; for y = 40 and x = 2 we would get a = 40 * 2^.5 = 56; for y = 34 and x = 3 we get a = 33 * 3^.5 = 55; for y = 26 and x = 4 we get a = 26 * 4^.5 = 52. Since our value of a are reasonably constant the y = a x^.5 model works pretty well, with a value of a around 54.

The value of a for accurate data turns out to be about 55.

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20:53:19

** The model that best fits the data is a x^-.5, and with accurate data we find that a is close to 55.

The model is pretty close to

# per minute frequency = 55 x^-.5.

The value of a for accurate data turns out to be about 55.**

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20:53:50

If the time per swing in seconds is y, then what expression represents the number of swings per minute?

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RESPONSE -->

To get the number of swings per minute you would divide 60 seconds by the number of seconds in a swing (e.g., if a swing takes 2 seconds you have 30 swings in a minute). So you would have f = 60 / y, where f is frequency in swings per minute.

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20:53:56

** To get the number of swings per minute you would divide 60 seconds by the number of seconds in a swing (e.g., if a swing takes 2 seconds you have 30 swings in a minute). So you would have f = 60 / y, where f is frequency in swings per minute.

COMMON ERROR: y * 60

INSTRUCTOR COMMENT: That would give more swings per minute for a greater y. But greater y implies a longer time for a swing, which would imply fewer swings per minute. This is not consistent with your answer. **

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20:54:23

If the time per swing is a x ^ .5, for the value determined previously for the parameter a, then what expression represents the number of swings per minute? How does this expression compare with the function you obtained for the number of swings per minute vs. length?

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RESPONSE -->

Time per swing turns out to be a x^.5--this is what you would obtain if you did the experiment very accurately and correctly determined the power function. For x in feet a will be about 1.1.

Since the number of swings per minute is 60/(time per swing), you have f = 60 / (a x^.5), where f is frequency in swings / minute.

Simplifying this gives f = (60 / a) * x^.5.

60/a is just a constant, so the above expression is of form f = k * x^-.5, consistent with earlier statements.

60 / a = 60 / 1.1 = 55, approx., confirming our frequency model F = 55 x^-.5.

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20:54:29

** Time per swing turns out to be a x^.5--this is what you would obtain if you did the experiment very accurately and correctly determined the power function. For x in feet a will be about 1.1.

Since the number of swings per minute is 60/(time per swing), you have f = 60 / (a x^.5), where f is frequency in swings / minute.

Simplifying this gives f = (60 / a) * x^.5.

60/a is just a constant, so the above expression is of form f = k * x^-.5, consistent with earlier statements.

60 / a = 60 / 1.1 = 55, approx., confirming our frequency model F = 55 x^-.5. **

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20:55:28

query problem 8. model of time per swing what are the pendulum lengths that would result in periods of .1 second and 100 seconds?

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RESPONSE -->

According to the model T = 1.1 x^.5 , where T is period in seconds and x is length in feet, we have periods T = .1 and T = 100. So we solve for x:

For T = .1 we get:

.1 = 1.2 x^.5 which gives us

x ^ .5 = .1 / 1.2 so that

x^.5 = .083 and after squaring both sides we get

x = .083^2 = .0069 approx., representing .0069 feet.

We also solve for T = 100:

100 = 1.2 x^.5, obtaining

x^.5 = 100 / 1.2 = 83, approx., so that

x = 83^2 = 6900, approx., representing a pendulum 6900 ft (about 1.3 miles) long.

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20:55:33

** You would use your own model here.

This solution uses T = 1.1 x^.5. You can adapt the solution to your own model.

According to the model T = 1.1 x^.5 , where T is period in seconds and x is length in feet, we have periods T = .1 and T = 100. So we solve for x:

For T = .1 we get:

.1 = 1.2 x^.5 which gives us

x ^ .5 = .1 / 1.2 so that

x^.5 = .083 and after squaring both sides we get

x = .083^2 = .0069 approx., representing .0069 feet.

We also solve for T = 100:

100 = 1.2 x^.5, obtaining

x^.5 = 100 / 1.2 = 83, approx., so that

x = 83^2 = 6900, approx., representing a pendulum 6900 ft (about 1.3 miles) long. **

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20:57:50

query problem 9. length ratio x2 / x1.

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RESPONSE -->

If lengths are x2 and x1, you would substitute x2 and x1 for L in the frequency relationship f = 60 / (1.1 `sqrt(L)) to get 60 / (1.1 `sqrt(x1) ) and 60 / (1.1 `sqrt(x2)).

Alternative form is f = 55 L^-.5. Substituting would give you 55 * x1^-.5 and 55 * x2^-.5.

If you just had f = a L^-.5 (same as y = a x^-.5) you would get f1 = a x1^-.5 and f2 = a x2^-.5

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20:58:05

What expressions, in terms of x1 and x2, represent the frequencies (i.e., number of swings per minute) of the two pendulums?

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20:58:16

** The solution is to be in terms of x1 and x2.

If lengths are x2 and x1, you would substitute x2 and x1 for L in the frequency relationship f = 60 / (1.1 `sqrt(L)) to get 60 / (1.1 `sqrt(x1) ) and 60 / (1.1 `sqrt(x2)).

Alternative form is f = 55 L^-.5. Substituting would give you 55 * x1^-.5 and 55 * x2^-.5.

If you just had f = a L^-.5 (same as y = a x^-.5) you would get f1 = a x1^-.5 and f2 = a x2^-.5 **

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20:59:38

What expression, in terms of x1 and x2, represents the ratio of the frequencies of the two pendulums?

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RESPONSE -->

We need to do this in terms of the symbols x1 and x2. If f = a x^-.5 then f1 = a x1^-.5 and f2 = a x2^-.5.

With these expressions we would get

f2 / f1 = a x2^-.5 / (a x1^-.5) =

x2^-.5 / x1^-.5 =

(x2 / x1)^-.5 =

1 / (x2 / x1)^.5 =

(x1 / x2)^.5.

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20:59:43

** We need to do this in terms of the symbols x1 and x2. If f = a x^-.5 then f1 = a x1^-.5 and f2 = a x2^-.5.

With these expressions we would get

f2 / f1 = a x2^-.5 / (a x1^-.5) =

x2^-.5 / x1^-.5 =

(x2 / x1)^-.5 =

1 / (x2 / x1)^.5 =

(x1 / x2)^.5.

Note that it doesn't matter what a is, since a quickly divides out of our quotient. For example if a = 55 we get

f2 / f1 = 55 x2^-.5 / (55 x1^-.5) =

x2^-.5 / x1^-.5 =

(x2 / x1)^-.5 =

1 / (x2 / x1)^.5 =

(x1 / x2)^.5.

This is the same result we got when a was not specified. This shouldn't be surprising, since the parameter a divided out in the third step. **

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21:00:16

query problem Challenge Problem for Calculus-Bound Students: how much would the frequency change between lengths of 2.4 and 2.6 feet

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RESPONSE -->

I worked to get the frequency at 2.4 and 2.6

y = 55.6583(2.4^-.5) = 35.9273 and y = 55.6583(2.6^-.5)= 34.5178.

subtracted to get -1.40949 difference between 2.4 and 2.6.

This, along with the change in length of .2, gives average rate -1.409 cycles/min / (.2 ft) = -7.045 (cycles/min)/ft , based on the behavior between 2.4 ft and 2.6 ft.

This average rate would predict a change of -7.045 (cycles/min)/ft * 1 ft = -7/045 cycles/min for the 1-foot increase between 2 ft and 3 ft.

The change obtained by evaluating the model at 2 ft and 3 ft was -7.2221 cycles/min.

The answers are different because the equation is not linear and the difference between 2.4 and 2.6 does not take into account the change in the rate of frequency change between 2 and 2.4 and 2.6 and 3

for 4.4 and 4.6

y = 55.6583(4.4^-.5) y = 55.6583(4.6^-.5)

y = 26.5341 y = 25.6508

Dividing difference in y by change in x we get -2.9165 cycles/min / ft, compared to the actual change -2.938 obtained from the model.

The answers between 4-5 and 2-3 are different because the equation is not linear and the frequency is changing at all points.

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21:00:22

** STUDENT SOLUTION: Note that we are using frequency in cycles / minute.

I worked to get the frequency at 2.4 and 2.6

y = 55.6583(2.4^-.5) = 35.9273 and y = 55.6583(2.6^-.5)= 34.5178.

subtracted to get -1.40949 difference between 2.4 and 2.6.

This, along with the change in length of .2, gives average rate -1.409 cycles/min / (.2 ft) = -7.045 (cycles/min)/ft , based on the behavior between 2.4 ft and 2.6 ft.

This average rate would predict a change of -7.045 (cycles/min)/ft * 1 ft = -7/045 cycles/min for the 1-foot increase between 2 ft and 3 ft.

The change obtained by evaluating the model at 2 ft and 3 ft was -7.2221 cycles/min.

for 4.4 and 4.6

y = 55.6583(4.4^-.5) y = 55.6583(4.6^-.5)

y = 26.5341 y = 25.6508

Dividing difference in y by change in x we get -2.9165 cycles/min / ft, compared to the actual change -2.938 obtained from the model.

The answers between 4-5 and 2-3 are different because the equation is not linear and the frequency is changing at all points. **

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21:00:59

query problem 1.2.19 formula for exponential function through left (1,6) and (2,18)

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RESPONSE -->

An exponential function has one of several forms, including y = A * b^x and y = A * e^(kx).

Using y = A * b^t and substituting the t and y coordinates of the two points gives us

6 = A * b^1

18 = A * b^2.

Dividing the second equation by the first we get

3 = b^(2-1) or b = 3.

Substituting this into the first equation we get

6 = A * 3^1 so

A = 2.

Thus the model is y = 2 * 3^t .

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21:01:03

** An exponential function has one of several forms, including y = A * b^x and y = A * e^(kx).

Using y = A * b^t and substituting the t and y coordinates of the two points gives us

6 = A * b^1

18 = A * b^2.

Dividing the second equation by the first we get

3 = b^(2-1) or b = 3.

Substituting this into the first equation we get

6 = A * 3^1 so

A = 2.

Thus the model is y = 2 * 3^t . **

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{á³Â¢×}éœéú¾ÛÐYãL}œÅôÄùËÞŽ

assignment #007

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-05-2008

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21:13:53

Query class notes #07Explain how we obtain the tangent line to a y = k x^3 function at a point on its graph, and explain why this tangent line gives a good approximation to the function near that point.

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If we know that y=kx^3, as in the sandpile model, we can find the derivative as y = 3kx^2.

This derivative will tell us the rate at which the volume changes with respect to the diameter of the pile.

On a graph of the y = k x^3 curve the slope of the tangent line is equal to the derivative.

Through the given point we can sketch a line with the calculated slope; this will be the tangent line.

Knowing the slope and the change in x we easily find the corresponding rise of the tangent line, which is the approximate change in the y = k x^3 function.

In short you use y' = 3 k x^2 to calculate the slope, which you combine with the change `dx in x to get a good estimate of the change `dy in y.

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21:13:58

** If we know that y=kx^3, as in the sandpile model, we can find the derivative as y = 3kx^2.

This derivative will tell us the rate at which the volume changes with respect to the diameter of the pile.

On a graph of the y = k x^3 curve the slope of the tangent line is equal to the derivative.

Through the given point we can sketch a line with the calculated slope; this will be the tangent line.

Knowing the slope and the change in x we easily find the corresponding rise of the tangent line, which is the approximate change in the y = k x^3 function.

In short you use y' = 3 k x^2 to calculate the slope, which you combine with the change `dx in x to get a good estimate of the change `dy in y. **

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21:17:09

Query class notes #08What equation do we get from the statement 'the rate of temperature change is proportional to the difference between the temperature and the 20 degree room temperature'? What sort of graph do we get from this equation and why?

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Would it be y = x-20 degrees., with y being the rate of temperature change and x being the temperature?You get a graph with a straight line and a slope of -20?

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21:17:16

STUDENT SOLUTION AND INSTRUCTOR COMMENT:

Would it be y = x-20 degrees., with y being the rate of temperature change and x being the temperature?You get a graph with a straight line and a slope of -20?

INSTRUCTOR COMMENT:

Not a bad attempt. However, you wouldn't use y to represent a rate, but rather dy /dt or y'. An in this example I'm going to use T for temperature, t for clock time. Read further.

We need a graph of temperature vs. clock time, not rate of change of temperature vs. clock time.

The difference between temperature and room temperature is indeed (T - 20). The rate of change of the temperature would be dT / dt. To say that these to our proportional is to say that dT / dt = k ( T - 20). To solve the situation we would need the proportionality constant k, just as with sandpiles and other examples you have encountered.

Thus the relationship is dT / dt = k ( T - 20). Since dT / dt is the rate of change of T with respect to t, it tells you the slope of the graph of T vs. t. So the equation tells you that the slope of the graph is proportional to T - 20. Thus, for example, if T starts high, T - 20 will be a relatively large positive number. We might therefore expect k ( T - 20) to be a relatively large positive number, depending on what k is.

For positive k this would give our graph a positive slope, and the temperature would move away from room temperature. If we are talking about something taken from the oven, this wouldn't happen--the temperature would move closer to room temperature. This could be accomplished using a negative value of k.

As the temperature moves closer to room temperature, T - 20 becomes smaller, and the steepness of the graph will decrease--i.e., as temperature approaches room temperature, it will do so more and more slowly.

So the graph approaches the T = 20 value more and more slowly, approaching as an asymptote. **

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21:17:57

Query Inverse Functions and Logarithms, Problem 7. Construct table for the squaring function f(x) = x^2, using x values between 0 and 2 with a step of .5. Reverse the columns of this table to form a partial table for the inverse function.

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We get the following ordered pairs:

Table 1-- (0,0),(.5,.25),(1,1),(1.5,2.25),(2,4)

Table2--(0,0),(.25,.5),(1,1),(2.25,1.5),(4,2).

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21:18:02

STUDENT SOLUTION:

We get the following ordered pairs:

Table 1-- (0,0),(.5,.25),(1,1),(1.5,2.25),(2,4)

Table2--(0,0),(.25,.5),(1,1),(2.25,1.5),(4,2).

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21:18:27

Plot the points corresponding to the table of the squaring function, and plot the points corresponding to the table of its inverse. Sketch a smooth curve corresponding to each function.

The diagonal line on the graph is the line y = x. Connect each point on the graph of the squaring function to the corresponding point on the graph of its inverse function. How are these pairs of points positioned with respect to the y = x line?

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The segments connecting the graph points for function and for its inverse will cross the y = x line at a right angle, and the graph points for the function and for the inverse will lie and equal distances on either side of this line. The graph of the inverse is therefore a reflection of the graph of the original function through the line y = x.

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21:18:35

** The segments connecting the graph points for function and for its inverse will cross the y = x line at a right angle, and the graph points for the function and for the inverse will lie and equal distances on either side of this line. The graph of the inverse is therefore a reflection of the graph of the original function through the line y = x. **

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21:32:34

**** 8. If we reversed the columns of the 'complete' table of the squaring function from 0 to 12, precisely what table would we get?

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would get a table of the square root function with the first column running from 0 to 144, the second column consisting of the square roots of these numbers, which run from 0 to 12.

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21:32:41

** We would get a table of the square root function with the first column running from 0 to 144, the second column consisting of the square roots of these numbers, which run from 0 to 12. **

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21:33:39

Sketch the graphs of the functions described by both tables. 9. If we could construct the 'complete' table of the squaring function from 0 to infinity, listing all possible positive numbers in the x column, then why would we be certain that every possible positive number would appear exactly one time in the second column?

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The table you constructed had only some of the possible x and y values. A complete table, which couldn't actually be written down but can to an extent be imagined, would contain all possible x values.

We could be sure because every number is the square of some other number.

If the function was, for example, x / (x^2 + 1) there would be a great many positive numbers that wouldn't appear in the second column. But this is not the case for the squaring function.

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21:33:44

** The table you constructed had only some of the possible x and y values. A complete table, which couldn't actually be written down but can to an extent be imagined, would contain all possible x values.

We could be sure because every number is the square of some other number.

If the function was, for example, x / (x^2 + 1) there would be a great many positive numbers that wouldn't appear in the second column. But this is not the case for the squaring function. **

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21:36:48

What number would appear in the second column next to the number 4.31 in the first column?

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In the original table the second column would read 18.57, approx..

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21:36:53

** In the original table the second column would read 18.57, approx.. This is the square of 4.31. **

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21:38:07

What number would appear in the second column next to the number `sqrt(18) in the first column?

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18 would appear in the second column because the square of sqrt(18) is 18.

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21:38:12

** 18 would appear in the second column because the square of sqrt(18) is 18. **

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21:38:36

What number would appear in the second column next to the number `pi in the first column?

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The number would be `pi^2

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21:38:42

** The number would be `pi^2 **

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21:39:15

What would we obtain if we reversed the columns of this table?

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We would obtain the inverse, the square roots of the squares being in the y colume and the squared numbers being in the x column.

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21:39:23

STUDENT ANSWER: We would obtain the inverse, the square roots of the squares being in the y colume and the squared numbers being in the x column.

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21:39:56

What number would appear in the second column next to the number 4.31 in the first column of this table?

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This number would be 4.31 squared,18.5761.

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21:40:02

This number would be 4.31 squared,18.5761.

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21:40:34

What number would appear in the second column next to the number

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pi^2 in the first column of this table

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21:40:39

`pi^2 in the first column of this table?

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21:40:48

STUDENT ANSWER: This number would be the square root, 'pi

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21:41:22

What number would appear in the second column next to the number -3 in the first column of this table?

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The number -3 doesn't appear in the second column of the original table so it won't appear in the first column of the inverted table.

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21:41:28

** The number -3 doesn't appear in the second column of the original table so it won't appear in the first column of the inverted table.

Note that sqrt(-3) is not a real number, since the square of a real number must be positive. **

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21:42:59

13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible:

2 ^ x = 18

2 ^ (4x) = 12

5 * 2^x = 52

2^(3x - 4) = 9.

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b^x = c is translated into logarithmic notation as log{base b}(c) = x. So:

2^x = 18 translates directly to log{bas3 2}(18) = x.

For 5 * 2^x = 52, divide both sides by 5 to get

2^x = 10.4. Now take logs:

2x = log{base 2}(10.4) so

x = 1/2 log{base 2}(10.4). Evaluate on your calculator.

2^(3x-4) = 9 translates to log{base 2}(9) = 3x - 4.

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21:43:04

b^x = c is translated into logarithmic notation as log{base b}(c) = x. So:

2^x = 18 translates directly to log{bas3 2}(18) = x.

For 5 * 2^x = 52, divide both sides by 5 to get

2^x = 10.4. Now take logs:

2x = log{base 2}(10.4) so

x = 1/2 log{base 2}(10.4). Evaluate on your calculator.

2^(3x-4) = 9 translates to log{base 2}(9) = 3x - 4.

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21:45:28

14. Solve 2^(3x-5) + 4 = 0

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2^(3x-5) + 4 = 0 rearranges to

2^(3x-5) =-4, which we translate as

3x-5 = log {base 2}(-4) = log(-4) / log (2).

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21:45:57

2^(3x-5) + 4 = 0 rearranges to

2^(3x-5) =-4, which we translate as

3x-5 = log {base 2}(-4) = log(-4) / log (2).

However log(-4) doesn't exist. When you invert the 10^x table you don't end up with any negative x values. So there is no solution to this problem.

Be sure that you thoroughly understand the following rules:

10^x = b translates to x = log(b), where log is understood to be the base-10 log.

e^x = b translates to x = ln(b), where ln is the natural log.

a^x = b translates to x = log{base a} (b), where log{base a} would be written in your text as log with subscript a.

log{base a}(b) = log(b) / log(a), where log is the base-10 log. It also works with the natural log: log{base a}(b) = ln(b) / ln(a).

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21:47:12

Solve 2^(1/x) - 3 = 0

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2^(1/x) = 3. Then take log of both sides:

log(2^(1/x) ) = log(3). Use properties of logs:

(1/x) log(2) = log(3). Solve for x:

x = log(2) / log(3).

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21:47:17

** Rearrange to

2^(1/x) = 3. Then take log of both sides:

log(2^(1/x) ) = log(3). Use properties of logs:

(1/x) log(2) = log(3). Solve for x:

x = log(2) / log(3). **

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21:47:50

Solve 2^x * 2^(1/x) = 15

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2^x * 2^(1/x) = 15. By the laws of exponents we get

2^(x +1/x) = 15 so that

x + 1/x = log {base2}(15) or

x + 1/x =log(15) / log(2). Multiply both sides by x to get

x^2 + 1 = [log(15) / log(2) ] * x.

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21:48:04

** 2^x * 2^(1/x) = 15. By the laws of exponents we get

2^(x +1/x) = 15 so that

x + 1/x = log {base2}(15) or

x + 1/x =log(15) / log(2). Multiply both sides by x to get

x^2 + 1 = [log(15) / log(2) ] * x.

This is a quadratic equation.

}Rearrange to get

x^2 - [ log(15) / log(2) ] * x + 1 = 0 or

x^2 - 3.91 * x + 1 = 0. Solve using the quadratic fomula. **

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21:48:37

Solve (2^x)^4 = 5

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log( (2^x)^4 ) = log(5). Using laws of logarithms

4 log(2^x) = log(5)

4 * x log(2) = log(5)

4x = log(5) / log(2)

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21:48:42

** log( (2^x)^4 ) = log(5). Using laws of logarithms

4 log(2^x) = log(5)

4 * x log(2) = log(5)

4x = log(5) / log(2)

etc.**

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21:49:06

problem 1.3.22. C=f(A) = cost for A sq ft. What do f(10k) and f^-1(20k) represent?

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f(10,000) is the cost of 10,000 sq ft.

f^-1(20,000) is the number of square feet you can cover for $20,000.

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21:49:10

** f(10,000) is the cost of 10,000 sq ft.

f^-1(20,000) is the number of square feet you can cover for $20,000. **

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21:49:37

problem 1.3.38. vert stretch y = x^2 by factor 2 then vert shift 1.

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Vertically stretching y = x^2 we get y = 2 x^2.

The vertical shift adds 1 to all y values, giving us the function y = 2 x^2 + 1.

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21:49:43

** Vertically stretching y = x^2 we get y = 2 x^2.

The vertical shift adds 1 to all y values, giving us the function y = 2 x^2 + 1. **

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21:50:04

Give the equation of the function.Describe your sketch in detail.

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The function would be y = f(x) = 2 x^2 + 1. The factor 2 stretches the y = x^2 parabola verticall and +1 shifts every point of the stretched parabola 1 unit higher.

The result is a parabola which is concave up with vertex at point (0,1). The parabola has been stretched by a factor of 2 as compared to a x^2 parabola.

If the transformations are reversed the the graph is shifted downward 1 unit then stretched vertically by factor 2. The vertex, for example, shifts to (0, -1) then when stretched shifts to (0, -2). The points (-1, 1) and (1, 1) shift to (-l, 1) and (1, 0) and the stretch leaves them there.

The shift would transform y = x^2 to y = x^2 - 1. The subsequent stretch would then transform this function to y = 2 ( x^2 - 1) = 2 x^2 - 2.

The reversed pair of transformations results in a parabola with its vertex at (0, -2), as opposed to (0, -1) for the original pair of transformations.

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21:50:08

** The function would be y = f(x) = 2 x^2 + 1. The factor 2 stretches the y = x^2 parabola verticall and +1 shifts every point of the stretched parabola 1 unit higher.

The result is a parabola which is concave up with vertex at point (0,1). The parabola has been stretched by a factor of 2 as compared to a x^2 parabola.

The shift would transform y = x^2 to y = x^2 - 1. The subsequent stretch would then transform this function to y = 2 ( x^2 - 1) = 2 x^2 - 2.

The reversed pair of transformations results in a parabola with its vertex at (0, -2), as opposed to (0, -1) for the original pair of transformations. **

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21:52:02

problem 1.3.43 (was 1.8.30) estimate f(g(1))what is your estimate of f(g(1))?Explain how you look at the graphs of f and g to get this result

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You have to first find g(1), then apply f to that value.

To find g(1), you note that this is g(x) for x = 1. So you look on the x-axis where x = 1. Then you move up or down to find the point on the graph where x = 1 and determine the corresponding y value. On this graph, the x = 1 point lies at about y = 2.

Then you look at the graph of f(x). You are trying to find f(g(1)), which we now see is f(2). So we look at the x = 2 point on the x-axis and then look up or down until we find the graph, which for x = 2 lies just a little bit above the x axis. Looking over to the y-axis we see that at this point y is about .1.

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21:52:11

*&*& right problem? *&*&

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21:52:16

** You have to first find g(1), then apply f to that value.

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21:53:08

problem 1.5.12 graph, decide if inverse, approximate inverse at x = 20 for f(x) = x^2+e^x and g(x) = x^3 + 3^x

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The inverse of a function at a certain value is the x that would give you that value when plugged into the function. At x = 20 for g(x) = x^2 + e^x is the x value for which x^3 + 3^x = 20. The double use of x is confusing and way the problem is stated in the text isn't as clear as we might wish, but what you have to do is estimate the required value of x.

It would be helpful to sketch the graph of the inverse function by reflecting the graph of the original function through the line y = x, or alternatively and equivalently by making an extensive table for the function, then reversing the columns.

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21:53:14

** The inverse of a function at a certain value is the x that would give you that value when plugged into the function. At x = 20 for g(x) = x^2 + e^x is the x value for which x^3 + 3^x = 20. The double use of x is confusing and way the problem is stated in the text isn't as clear as we might wish, but what you have to do is estimate the required value of x.

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21:57:31

query text problem 1.3 #13 temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts

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the horizontal intercept is the time when the object became the same temperature as the outside

*&*& H is the temperature, t is the clock time. H(30) is the temperature at clock time t = 30, so H(30) = 10 tells us that a clock time t = 30 the temperature was 10 degrees.

The vertical coordinate is the temperature, and the vertical intercept of the graph occurs when t = 0 so the vertical intercept gives us the temperature at clock time 0.

The vertical coordinate is the clock time, and the horizontal intercept occurs when H = 0, so the horizontal intercept gives us the clock time when temperature is 0.

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21:57:38

the verticle ** vertical ** intercept is the temperature of the object when it is placed outside

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21:57:44

the horizontal intercept is the time when the object became the same temperature as the outside

*&*& H is the temperature, t is the clock time. H(30) is the temperature at clock time t = 30, so H(30) = 10 tells us that a clock time t = 30 the temperature was 10 degrees.

The vertical coordinate is the temperature, and the vertical intercept of the graph occurs when t = 0 so the vertical intercept gives us the temperature at clock time 0.

The vertical coordinate is the clock time, and the horizontal intercept occurs when H = 0, so the horizontal intercept gives us the clock time when temperature is 0. *&*&

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üÚ¸«›ÙÙTÃËÁå’½Éþ¶šiÎz|Ý¢Ãó×˜zx

assignment #008

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-05-2008

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22:06:38

What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?

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g(t) = 3t - 5, f(z) = 2^z.

The z is a 'dummy' variable; when we find f(g(t)), the g(t) is substituted for z and we get f(g(t)) = 2^(g(t)) = 2^(3t-5).

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22:06:43

** g(t) = 3t - 5, f(z) = 2^z.

The z is a 'dummy' variable; when we find f(g(t)), the g(t) is substituted for z and we get f(g(t)) = 2^(g(t)) = 2^(3t-5). **

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22:07:31

describe in some detail how we can numerically solve a differential equation dy /dt = f(x), given a point (x0, y0) on its solution curve, an interval (x0, xf) and an increment `dx

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evaluate the function y' for x=x0 and y=x0, which gives you a slope for your y vs. x graph.

Using the chosen increment `dx you then multiply `dx by the slope to determine how much change there will be in y, and you use this information to obtain a new approximate point on your y vs. x graph by adding the change in y to y0 and the change in x to x0.

You then repeat the process starting with the new point.

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22:07:36

** You start with a point (x0, y0) on the y vs. x graph.

You evaluate the function y' for x=x0 and y=x0, which gives you a slope for your y vs. x graph.

You then repeat the process starting with the new point. **

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22:08:06

explain why a numerical solution to differential equation is only an approximate solution in most cases

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assume the slope at the initial point, but that slope generally changes at least a bit by the time you get to the second point. So you are assuming a constant slope when the slope actually changes.

If your interval is small enough the change in slope will have a small effect.

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22:08:11

** You assume the slope at the initial point, but that slope generally changes at least a bit by the time you get to the second point. So you are assuming a constant slope when the slope actually changes.

If your interval is small enough the change in slope will have a small effect. **

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22:08:44

query Problem 1.4.10 Solve 2 * 5^x = 11 * 7^x

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RESPONSE -->

Taking logs of both sides and applying the laws of logarithms we get

log 2 + x log 5 = log 11 + x log 7. Rearranging we obtain

x log 5 - x log 7 = log 11 - log 2 so that

x ( log 5 - log 7) = log 11 - log 2 and

x = (log 11 - log 2) / (log 5 - log 7).

This can be approximated as -5.07.

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22:08:53

** Taking logs of both sides and applying the laws of logarithms we get

log 2 + x log 5 = log 11 + x log 7. Rearranging we obtain

x log 5 - x log 7 = log 11 - log 2 so that

x ( log 5 - log 7) = log 11 - log 2 and

x = (log 11 - log 2) / (log 5 - log 7).

This can be approximated as -5.07. ** DER

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22:09:19

Problem 1.4.6 simplify 2 ln(e^A) + 3 ln(B^e)

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Starting with 2 ln (e^A) + 3 ln (e^B) we use the fact that the natural log and exponential functions are inverses, expressed by the law of logarithms ln(e^x) = x, to get

2 * A + 3 * B or just

2A + 3B.

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22:09:23

** Starting with 2 ln (e^A) + 3 ln (e^B) we use the fact that the natural log and exponential functions are inverses, expressed by the law of logarithms ln(e^x) = x, to get

2 * A + 3 * B or just

2A + 3B. **

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RESPONSE -->

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22:09:44

query Problem 1.4.31 (was 1.7.26) P=174 * .9^twhat is the function when converted to exponential form P = P0 e^(kt)?

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RESPONSE -->

174 * .9^t = 174 * e^(kt) if

e^(kt) = .9^t, which is the case if

e^k = .9. Taking the natural log of both sides we get

ln(e^k) = ln(.9) so that

k = ln(.9) = -.105 approx.

So the function is

P = 174 e^(-.105 t).

.................................................

......!!!!!!!!...................................

22:09:49

** 174 * .9^t = 174 * e^(kt) if

e^(kt) = .9^t, which is the case if

e^k = .9. Taking the natural log of both sides we get

ln(e^k) = ln(.9) so that

k = ln(.9) = -.105 approx.

So the function is

P = 174 e^(-.105 t). **

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RESPONSE -->

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22:10:20

Query problem 1.4.44 (was 1.6.24) population function for exponential growth if 40 meg in 1980 and 56 meg in 1990; doubling timegive a population function and the doubling time

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RESPONSE -->

The population function is exponential and has form P = P0 * e^(kt).

Let t be the time since 1980 and population be in millions. Then we have

40 = P0 e^(k * 0) and

56 = P0 e^(k * 10).

From the first equation we get

40 = P0 so the second equation becomes

56 = 40 * e^(10 k) or

e^(10 k) = 56 / 40 = 1.4. Taking logs we get

10 k = ln(1.4) so that

k = ln(1.4) / 10 = .0336, approx.

Thus our equation is

P = 40 e^(.0336 t).

This doubles when

e^(.0336 t) = 2. Taking the ln of both sides we have

.0336 t = ln(2) so that

t = ln(2) / .0336 = 20.6, approx.

Doubling time is about 20.6 years.

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......!!!!!!!!...................................

22:10:26

** The population function is exponential and has form P = P0 * e^(kt).

Let t be the time since 1980 and population be in millions. Then we have

40 = P0 e^(k * 0) and

56 = P0 e^(k * 10).

From the first equation we get

40 = P0 so the second equation becomes

56 = 40 * e^(10 k) or

e^(10 k) = 56 / 40 = 1.4. Taking logs we get

10 k = ln(1.4) so that

k = ln(1.4) / 10 = .0336, approx.

Thus our equation is

P = 40 e^(.0336 t).

This doubles when

e^(.0336 t) = 2. Taking the ln of both sides we have

.0336 t = ln(2) so that

t = ln(2) / .0336 = 20.6, approx.

Doubling time is about 20.6 years. **

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RESPONSE -->

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22:12:30

query Problem 1.4.45 (was 1.7.42 but changed) time to get to 10% of strontium 90 if half-life 29 yrs

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RESPONSE -->

e^(k * 29) = 1/2. Taking ln of both sides

29 k = ln(1/2) so that

k = ln(1/2) / 29 = -.0239.

So the model is

Q = Q0 * e^(-.0239 t).

Now if we want to get 10% of the original quantity we have Q = Q0 / 10 so that

Q0 / 10 = Q0 e^(-.0239 t) and

e^(-.0239 t) = 1/10.

Taking logs of both sides and solving for t we get

t = ln(1/10) / -.0239 = 96.3 approx.

.................................................

......!!!!!!!!...................................

22:12:37

** If the model is Q = Q0 e^(kt) then since half-life is 29 years we know that

e^(k * 29) = 1/2. Taking ln of both sides

29 k = ln(1/2) so that

k = ln(1/2) / 29 = -.0239.

So the model is

Q = Q0 * e^(-.0239 t).

Now if we want to get 10% of the original quantity we have Q = Q0 / 10 so that

Q0 / 10 = Q0 e^(-.0239 t) and

e^(-.0239 t) = 1/10.

Taking logs of both sides and solving for t we get

t = ln(1/10) / -.0239 = 96.3 approx. **

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RESPONSE -->

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22:13:38

Problem 1.4.26 P=174 * .9^t

What is the function when converted to exponential form P = P0 e^(kt)?

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RESPONSE -->

If P=174(.9)^t is put in the form P = P0 e^(kt) then P0 = 174 and e^(k t) = .9.t so that e^k = .9.

It follows that

e^k = .9 so that

ln(e^k) = ln(.9) or

k = ln(.9) = .105.

The function is therefore

P=174 e^-(.105 t).

.................................................

......!!!!!!!!...................................

22:13:43

If P=174(.9)^t is put in the form P = P0 e^(kt) then P0 = 174 and e^(k t) = .9.t so that e^k = .9.

It follows that

e^k = .9 so that

ln(e^k) = ln(.9) or

k = ln(.9) = .105.

The function is therefore

P=174 e^-(.105 t).

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RESPONSE -->

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22:14:11

problem 1.4.32 population function for exponential growth.

If 40 meg in 1980 and 56 meg in 1990 give the equation and find the doubling time

......!!!!!!!!...................................

RESPONSE -->

P=Po b^t is the form of the function. Initial quantity is 40 * 10^6 so Po = 40 * 10^6. Substituting Po = 40 * 10^6:

P=40*10^6 b^t. At t = 10 we have P = 56 * 10^6 so we substitute for P and t:

56*10^6=40*10^6 b^10. We solve for b:

1.4=b^10

b=1.03

P=40*10^6(1.03)^t is our function.

doubling time occurs when the 40^10^6 grows to 80*10^6:

80*10^6=40*10^6(1.03)^t

2=1.03^t

log2=tlog1.03

t=23.4498

.................................................

......!!!!!!!!...................................

22:14:16

P=Po b^t is the form of the function. Initial quantity is 40 * 10^6 so Po = 40 * 10^6. Substituting Po = 40 * 10^6:

P=40*10^6 b^t. At t = 10 we have P = 56 * 10^6 so we substitute for P and t:

56*10^6=40*10^6 b^10. We solve for b:

1.4=b^10

b=1.03

P=40*10^6(1.03)^t is our function.

doubling time occurs when the 40^10^6 grows to 80*10^6:

80*10^6=40*10^6(1.03)^t

2=1.03^t

log2=tlog1.03

t=23.4498

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RESPONSE -->

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22:17:10

Problem 1.4.45 percent of original strontium -- 90 after century; 2.47% annual decay rate.

What percent of the original strontium -- 90 would remain after a century?

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RESPONSE -->

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22:17:50

Q=Qoe^(-kt)

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RESPONSE -->

Q=Qoe^-.0247t

.................................................

......!!!!!!!!...................................

22:17:57

Q=Qoe^-.0247t

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RESPONSE -->

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22:18:10

That`s all that I can do with that problem at this point

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RESPONSE -->

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22:18:23

** The model is Q(t) = Qo * e^(kt).

You know that you lost .0247 of the quantity in a year. Thus

Q(1) = Qo e^(k* 1) = (1 - .0247) Qo.

So Qo e^(k* 1) = (1 - .0247) Qo.

This equation is easily solved for k.

Then you substitute t = 100 back into the function, using your newly found k. **

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RESPONSE -->

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’®“¿OÖùâÃÝm«³Š³døñÌÙiÖ˜©ÔÍëJ||²

assignment #009

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-05-2008

......!!!!!!!!...................................

22:28:21

Query class notes #10 Describe in your own words how the predictor-corrector method of solving a differential equation works

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RESPONSE -->

We start with the rate dT/dt calculated at the beginning of the t interval for some given initial T. We then use this dT/dt to calculate the amount of change in T over the interval. Then T is calculated for the end of the interval.

We then calculate a dT/dt for this T.

The two values of dT / dt then averaged to obtain a corrected value.

This is then used to calculate a new change in T. This change is added to the original T.

The process is then continued for another interval, then another, until we reach the desired t value. **

.................................................

......!!!!!!!!...................................

22:28:27

** We start with the rate dT/dt calculated at the beginning of the t interval for some given initial T. We then use this dT/dt to calculate the amount of change in T over the interval. Then T is calculated for the end of the interval.

We then calculate a dT/dt for this T.

The two values of dT / dt then averaged to obtain a corrected value.

This is then used to calculate a new change in T. This change is added to the original T.

The process is then continued for another interval, then another, until we reach the desired t value. **

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RESPONSE -->

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22:28:51

Problem 1.5.13. amplitude, period of 5 + cos(3x)

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RESPONSE -->

The function goes through a complete cycle as 3x changes from 0 to 2 pi. This corresponds to a change in x from 0 to 2 pi / 3. So the period of this function is 2 pi /3.

The cosine function is multiplied by 1 so the amplitude is 1.

The function is then vertically shifted 5 units.

.................................................

......!!!!!!!!...................................

22:28:56

*&*& The function goes through a complete cycle as 3x changes from 0 to 2 pi. This corresponds to a change in x from 0 to 2 pi / 3. So the period of this function is 2 pi /3.

The cosine function is multiplied by 1 so the amplitude is 1.

The function is then vertically shifted 5 units. *&*&

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RESPONSE -->

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22:30:06

Explain how you determine the amplitude and period of a given sine or cosine function.

......!!!!!!!!...................................

RESPONSE -->

Amplitude is the multiplication factor of the cosine function, such as the absolute value of a in y = a cos(x).

the period is 2`pi divided by the coefficient of x.

.................................................

......!!!!!!!!...................................

22:30:12

*&*& GOOD ANSWER FROM STUDENT:

Amplitude is the multiplication factor of the cosine function, such as the absolute value of a in y = a cos(x).

the period is 2`pi divided by the coefficient of x. *&*&

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RESPONSE -->

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22:31:17

query Problem 1.5.28. trig fn graph given, defined by 3 pts (0,3), (2,6), (4,3), (6,0), (8,3)

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RESPONSE -->

The complete cycle goes from y = 3 to a max of y = 6, back to y = 3 then to a min of y = 0, then back to y = 3. The cycle has to go thru both the max and the min. So the period is 8.

The mean y value is 3, and the difference between the mean y value and the max y value is the amplitude 3.

The function starts at it mean when x = 0 and moves toward its maximum, which is the behavior of the sine function (the cosine, by contrast, starts at its max and moves toward its mean value).

So we have the function 3 sin( 2 `pi / 8 * x), which would have mean value 0, shifted vertically 3 units so the mean value is 3. The function is therefore

y = 3 + 3 sin( `pi / 4 * x).

.................................................

......!!!!!!!!...................................

22:31:43

What is a possible formula for the graph?

......!!!!!!!!...................................

RESPONSE -->

y = 3 + 3 sin( `pi / 4 * x).

.................................................

......!!!!!!!!...................................

22:31:49

** The complete cycle goes from y = 3 to a max of y = 6, back to y = 3 then to a min of y = 0, then back to y = 3. The cycle has to go thru both the max and the min. So the period is 8.

The mean y value is 3, and the difference between the mean y value and the max y value is the amplitude 3.

The function starts at it mean when x = 0 and moves toward its maximum, which is the behavior of the sine function (the cosine, by contrast, starts at its max and moves toward its mean value).

So we have the function 3 sin( 2 `pi / 8 * x), which would have mean value 0, shifted vertically 3 units so the mean value is 3. The function is therefore

y = 3 + 3 sin( `pi / 4 * x). **

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RESPONSE -->

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22:32:35

problem 1.5.30. Solve 1 = 8 cos(2x+1) - 3 for x.

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RESPONSE -->

2x + 1 = `pi/3, or `pi / 3 + 2 `pi, or `pi / 3 + 4 `pi, ... or `pi / 3 + 2n `pi, ... with n = an integer.

So x = (`pi / 3 - 1) / 2.

Or x = (`pi / 3 + 2 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + `pi.

Or x = (`pi / 3 + 4 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + 2 `pi.

…

Or x = (`pi / 3 + 2 n `pi - 1) / 2 = (`pi / 3 - 1) / 2 + n `pi

...

x = (`pi / 3 - 1) / 2 + n `pi, n an integer, is the general solution.

Since 0 (`pi / 3 - 1) / 2 < `pi he first two solutions are between 0 and 2 `pi, and are the only solutions between 0 and 2 `pi.

We generally want at least the solutions between 0 and 2 `pi.

.................................................

......!!!!!!!!...................................

22:32:40

** 2x + 1 = `pi/3, or `pi / 3 + 2 `pi, or `pi / 3 + 4 `pi, ... or `pi / 3 + 2n `pi, ... with n = an integer.

So x = (`pi / 3 - 1) / 2.

Or x = (`pi / 3 + 2 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + `pi.

Or x = (`pi / 3 + 4 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + 2 `pi.

…

Or x = (`pi / 3 + 2 n `pi - 1) / 2 = (`pi / 3 - 1) / 2 + n `pi

...

x = (`pi / 3 - 1) / 2 + n `pi, n an integer, is the general solution.

Since 0 (`pi / 3 - 1) / 2 < `pi he first two solutions are between 0 and 2 `pi, and are the only solutions between 0 and 2 `pi.

We generally want at least the solutions between 0 and 2 `pi. **

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RESPONSE -->

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22:33:04

problem 1.5.47 (was 1.9.34) arccos fndescribe the graph of the arccos or cos^-1 function, including a statement of its domain and range, increasing/decreasing behavior, asymptotes if any, and concavityexplain why the domain and range are as you describe

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RESPONSE -->

To get the graph of arccos(x) we take the inverse of the graph of cos(x), restricting ourselves to 0 < = x < = `pi, where y = cos(x) takes each possible y value exactly once. The graph of this portion of y = cos(x) starts at (0, 1), decreases at an increasing rate (i.e., concave down) to (`pi/2, 0) the decreases at a decreasing rate (i.e., concave up) to (`pi, -1).

The inverse function reverses the x and y values. Since y values of the cosine function range from -1 to 1, the x values of the inverse function will run from -1 to 1. The graph of the inverse function therefore runs from x = -1 to x = 1.

The graph of the inverse function starts at (-1, `pi) and runs to (0, `pi/2), reversing the values of y = cos(x) between (`pi/2, 0) and (`pi, -1). Since the graph of y = cos(x) is decreasing at a decreasing rate (i.e., concave up) between these points, the graph of the inverse function will be decreasing at a decreasing rate (i.e., concave up) over the corresponding region.

The graph of the inverse function continues from (0, `pi/2) to (1, 0), where it decreases at an increasing rate.

The values of the inverse cosine function range from 0 to `pi; the domain of this function is [-1, 1].

.................................................

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22:33:09

** To get the graph of arccos(x) we take the inverse of the graph of cos(x), restricting ourselves to 0 < = x < = `pi, where y = cos(x) takes each possible y value exactly once. The graph of this portion of y = cos(x) starts at (0, 1), decreases at an increasing rate (i.e., concave down) to (`pi/2, 0) the decreases at a decreasing rate (i.e., concave up) to (`pi, -1).

The graph of the inverse function continues from (0, `pi/2) to (1, 0), where it decreases at an increasing rate.

The values of the inverse cosine function range from 0 to `pi; the domain of this function is [-1, 1].**

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RESPONSE -->

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ûP†Œ©–àKÙS†ˆMÁFÕšþÆ° éîØx

assignment #010

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-05-2008

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22:46:29

query problem 1.6.9 (was 1.10.16)cubic polynomial representing graph. What cubic polynomial did you use to represent the graph?

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RESPONSE -->

The given function has zeros at x = -2, 1, 5, so y = k (x+2)(x-1)(x-5).

At x = 0 the function has value 2, so 2 = k (0+2)(0-1)(0-5), or 2 = 10 k.

Thus k = .2 and the function is

y = .2 ( x+2)(x-1)(x-5).

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22:46:37

*&*& The given function has zeros at x = -2, 1, 5, so y = k (x+2)(x-1)(x-5).

At x = 0 the function has value 2, so 2 = k (0+2)(0-1)(0-5), or 2 = 10 k.

Thus k = .2 and the function is

y = .2 ( x+2)(x-1)(x-5). **

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RESPONSE -->

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22:47:05

Query problem 1.6.15 (formerly 1.4.19) s = .01 w^.25 h^.75what is the surface area of a 65 kg person 160 cm tall?

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RESPONSE -->

s = .01 *65^.25 *160^.75 = 1.277meters^2

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......!!!!!!!!...................................

22:47:10

** Substituting we get

s = .01 *65^.25 *160^.75 = 1.277meters^2 **

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RESPONSE -->

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22:47:36

What is the weight of a person 180 cm tall whose surface area is 1.5 m^2?

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RESPONSE -->

1.5 = .01 w^.25*180^.75 . Dividing both sides by 180:

1.5/180^.75

.01w^.25. Dividing both sides by .01:

3.05237 = w^.25 Taking the fourth power of both sides:

w = 3.052^4 = 86.806

.................................................

......!!!!!!!!...................................

22:47:41

** Substituting the values we get

1.5 = .01 w^.25*180^.75 . Dividing both sides by 180:

1.5/180^.75

.01w^.25. Dividing both sides by .01:

3.05237 = w^.25 Taking the fourth power of both sides:

w = 3.052^4 = 86.806 **

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RESPONSE -->

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22:48:04

For 70 kg persons what is h as a function of s?

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RESPONSE -->

s = .01 *70^.25 h^.75

s = .02893 h^.75

s/.02893 = h^.75.

Taking the 1/.75 = 1.333... power of both sides:

(s/.02893)^1.333 = h

h = 110.7s^1.333... = 110.7 s^(4/3).

.................................................

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22:48:10

** Substituting 70 for the weight we get

s = .01 *70^.25 h^.75

s = .02893 h^.75

s/.02893 = h^.75.

Taking the 1/.75 = 1.333... power of both sides:

(s/.02893)^1.333 = h

h = 110.7s^1.333... = 110.7 s^(4/3). **

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RESPONSE -->

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¿°‰¸ˆKü¹GçùwÉ÷ØqõK¤cØø©ÐÈ

assignment #012

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-05-2008

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23:09:58

What is the seventh power of (x + `dx) (use the Binomial Theorem)?

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RESPONSE -->

x^7+7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7

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23:10:03

** Using the binomial Theorem:

x^7+7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7

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RESPONSE -->

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23:10:46

What is ( (x + `dx)^7 - x^7 ) / `dx and what does the answer have to do with the derivative of x^7?

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RESPONSE -->

(x + `dx)^7 - x^7

= x^7+7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7 - x^7

= 7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7,

[ (x + `dx)^7 - x^7 ] / `dx

= (7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7)/`dx

= 7x^6+21x^6'dx+35x^5'dx^2+35x^3'dx^3+21x^2'dx^4+7x'dx^5+'dx^6.

As `dx -> 0,

every term with factor `dx approaches 0 and the quotient approaches 7 x^6. **

.................................................

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23:10:50

** (x + `dx)^7 - x^7

= x^7+7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7 - x^7

= 7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7,

[ (x + `dx)^7 - x^7 ] / `dx

= (7x^6'dx+21x^6'dx^2+35x^4'dx^3+35x^3'dx^4+21x^2'dx^5+7x'dx^6+'dx^7)/`dx

= 7x^6+21x^6'dx+35x^5'dx^2+35x^3'dx^3+21x^2'dx^4+7x'dx^5+'dx^6.

As `dx -> 0,

every term with factor `dx approaches 0 and the quotient approaches 7 x^6. **

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RESPONSE -->

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23:11:11

Query problem 2.1.16 (prev edition 2.1.19 (was 2.1.8)) sketch position fn s=f(t) is vAve between t=2 and t=6 is same as vel at t = 5

Describe your graph and explain how you are sure that the velocity at t = 5 is the same as the average velocity between t=2 and t= 6.

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RESPONSE -->

The slope of the tangent line at t = 5 is the instantaneous velocity, and the average slope (rise / run between t = 2 and t = 6 points) is the average velocity.

The slope of the tangent line at t = 5 should be the same as the slope between the t = 2 and t = 6 points of the graph.

If the function has constant curvature then if the function is increasing it must be increasing at a decreasing rate (i.e., increasing and concave down), and if the function is decreasing it must be decreasing at a decreasing rate (i.e., decreasing and concave up). **

.................................................

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23:11:24

** The slope of the tangent line at t = 5 is the instantaneous velocity, and the average slope (rise / run between t = 2 and t = 6 points) is the average velocity.

The slope of the tangent line at t = 5 should be the same as the slope between the t = 2 and t = 6 points of the graph.

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RESPONSE -->

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23:12:08

What aspect of the graph represents the average velocity?

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RESPONSE -->

The straight line through two points has a rise representing the change in position and a run representing the change in clock time, so that the slope represents change in position / change in clock time = average rate of change of position = average velocity **

.................................................

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23:12:13

** The straight line through two points has a rise representing the change in position and a run representing the change in clock time, so that the slope represents change in position / change in clock time = average rate of change of position = average velocity **

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RESPONSE -->

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23:12:35

What aspect of the graph represents the instantaneous veocity at t = 5?

......!!!!!!!!...................................

RESPONSE -->

The slope of the tangent line at the t = 5 graph point represents the instantaneous velocity at t = 5.

According to the conditions of the problem this slope must equal the slope between the t = 2 and t = 6 points **

.................................................

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23:12:39

** The slope of the tangent line at the t = 5 graph point represents the instantaneous velocity at t = 5.

According to the conditions of the problem this slope must equal the slope between the t = 2 and t = 6 points **

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RESPONSE -->

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23:13:00

Query problem 2.1.14 (3d edition 2.1.16) graph increasing concave down thru origin, A, B, C in order left to right; origin to B on line y = x; put in order slopes at A, B, C, slope of AB, 0 and 1.What is the order of your slopes.

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RESPONSE -->

The graph is increasing so every slope is positive. The downward concavity means that the slopes are decreasing.

0 will be the first of the ordered quantities since all slopes are positive.

C is the rightmost point and since the graph is concave down will have the next-smallest slope.

The slope of the line from the origin to B is 1. The slope of the tangent line at B is less than the slope of AB and the slope of the tangent line at A is greater than the slope of AB.

So slope at A is the greatest of the quantities, 1 is next, followed by slope at B, then slope of AB, then slope at A and finall 0 (in descending order).

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23:13:04

** The graph is increasing so every slope is positive. The downward concavity means that the slopes are decreasing.

0 will be the first of the ordered quantities since all slopes are positive.

C is the rightmost point and since the graph is concave down will have the next-smallest slope.

The slope of the line from the origin to B is 1. The slope of the tangent line at B is less than the slope of AB and the slope of the tangent line at A is greater than the slope of AB.

So slope at A is the greatest of the quantities, 1 is next, followed by slope at B, then slope of AB, then slope at A and finall 0 (in descending order). **

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RESPONSE -->

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23:13:53

Query problem 2.2.8 (was 2.1.16) f(x) = sin(3x)/x.

Give your f(x) values at x = -.1, -.01, -.001, -.0001 and at .1, .01, .001, .0001 and tell what you think the desired limit should be.

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RESPONSE -->

-.1, 2.9552

-.01, 2.9996

-.001, 3

-.0001, 3

.1, 2.9552

.01, 2.9996

.001, 3

.0001, 3 .

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23:14:00

COMMON ERROR: Here are my values for f(x):

-.1, 2.9552

-.01, 2.9996

-.001, 3

-.0001, 3

.1, 2.9552

.01, 2.9996

.001, 3

.0001, 3 .

So the limiting value is 3.

INSTRUCTOR COMMENT: Good results and your answer is correct. However none the values you quote should be exactly 3. You need to give enough significant figures that you can see the changes in the expressions.

The values for .1, .01, .001 and .0001 are 2.955202066, 2.999550020, 2.999995500, 2.999999954. Of course your calculator might not give you that much precision, but you can see the pattern to these values.

The limit in any case is indeed 3.

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RESPONSE -->

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23:14:30

Describe your graph.

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RESPONSE -->

the graph passes horizontally through the y axes at (0,3), then as x increases it decreases an increasing rate -- i.e., it is concave downward--for a time, but gradually straightens out then decreases at a decreasing rate as it passes through the x axis, etc..

However the important behavior for this graph is near x = 0, where the graph reaches a maximum of 3 at x = 0, and approaches this value as a limit.

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23:14:45

** the graph passes horizontally through the y axes at (0,3), then as x increases it decreases an increasing rate -- i.e., it is concave downward--for a time, but gradually straightens out then decreases at a decreasing rate as it passes through the x axis, etc..

However the important behavior for this graph is near x = 0, where the graph reaches a maximum of 3 at x = 0, and approaches this value as a limit. **

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RESPONSE -->

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23:15:18

Find an interval such that the difference between f(x) and your limit is less than .01.

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RESPONSE -->

As the numbers quoted earlier show, f(x) is within .01 of the limit 3 for -.01 < x < .01. This interval is a good answer to the question.

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23:15:25

** As the numbers quoted earlier show, f(x) is within .01 of the limit 3 for -.01 < x < .01. This interval is a good answer to the question.

Note that you could find the largest possible interval over which f(x) is within .01 of 3. If you solve f(x) = 3 - .01, i.e., f(x) = 2.99, for x you obtain solutions x = -.047 and x = .047 (approx). The maximum interval is therefore approximately -.047 < x < .047.

However in such a situation we usually aren't interested in the maximum interval. We just want to find an interval to show that the function value can indeed be confined to within .01 of the limit.

In general we wish to find an interval to show that the function value can be confined to within a number usually symbolized by `delta (Greek lower-case letter) of the limit. **

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RESPONSE -->

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23:15:45

Query problem 2.2.17 (3d edition 2.3.26 was 2.2.10) f(x) is cost so f(x) / x is cost per unit. Describe the line whose slope is f(4) / 4

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RESPONSE -->

A line from (0, 0) to (4, f(4) ) has rise f(4) and run 4 so the slope of this line is rise / run = f(4) / 4.

Similarly the slope of the line from the origin to the x=3 point has slope f(3) / 3.

If the graph is concave down then the line from the origin to the x = 4 point is less than that of the line to the x = 3 point and we conclude that f(4) / 4 is smaller than f(3) / 3; average cost goes down as the number of units rises. **

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23:15:49

** A line from (0, 0) to (4, f(4) ) has rise f(4) and run 4 so the slope of this line is rise / run = f(4) / 4.

Similarly the slope of the line from the origin to the x=3 point has slope f(3) / 3.

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RESPONSE -->

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Ö¯|¸í—šéËÜ™€±‘¼À°ÐVÝàÍ„v

assignment #013

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-05-2008

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23:28:36

query problem 2.4.37 . Which graph matches the graph of the bus and why?

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RESPONSE -->

The bus only makes periodic stops, whereas the graph for III only comes to a stop once. I would matche the bus with II.

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23:28:41

** The bus only makes periodic stops, whereas the graph for III only comes to a stop once. I would matche the bus with II. **

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RESPONSE -->

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23:29:19

describe the graph of the car with no traffic and no lights

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RESPONSE -->

The car matches up with (I), which is a continuous, straight horizontal line representing the constant velocity of a car with no traffic and no lights.

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23:29:23

** The car matches up with (I), which is a continuous, straight horizontal line representing the constant velocity of a car with no traffic and no lights. *&*&

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RESPONSE -->

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23:29:42

describe the graph of the car with heavy traffic

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RESPONSE -->

The car in heavy traffic would do a lot of speeding up and slowing down at irregular intervals, which would match the graph in III with its frequent increases and decreases.

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23:29:48

** The car in heavy traffic would do a lot of speeding up and slowing down at irregular intervals, which would match the graph in III with its frequent increases and decreases. **

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RESPONSE -->

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23:30:57

query 2.5.10 (was 2.4.8) q = f(p) (price and quantity sold)what is the meaning of f(150) = 2000?

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RESPONSE -->

q = 2000 when p = 150, meaning that when the price is set at $150 we expect to sell 2000 units.

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23:31:01

*&*& q = 2000 when p = 150, meaning that when the price is set at $150 we expect to sell 2000 units. *&*&

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RESPONSE -->

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23:31:29

what is the meaning of f'(150) = -25?

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RESPONSE -->

f' is the derivative, the limiting value of `df / `dp, giving the rate at which the quantity q changes with respect to price p.

If f'(150) = -25, this means that when the price is $150 the price will be changing at a rate of -25 units per dollar of price increase.

Roughly speaking, a one dollar price increase would result in a loss of 25 in the number sold. **

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23:31:35

** f' is the derivative, the limiting value of `df / `dp, giving the rate at which the quantity q changes with respect to price p.

If f'(150) = -25, this means that when the price is $150 the price will be changing at a rate of -25 units per dollar of price increase.

Roughly speaking, a one dollar price increase would result in a loss of 25 in the number sold. **

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RESPONSE -->

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23:32:10

query problem 2.4.7 graph of v vs. t for no parachute.

Describe your graph, including all intercepts, asymptotes, intervals of increasing behavior, behavior for large |t| and concavity, and tell why the graph's concavity is as you indicate.

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RESPONSE -->

When you fall without a parachute v will inrease, most rapidly at first, then less and less rapidly as air resistance increases.

When t = 0 we presume that v = 0.

The graph of v vs. t is therefore characterized as an increasing graph beginning out at the origin, starting out nearly linear (the initial slope is equal to the acceleration of gravity) but with a decreasing slope. The graph is therefore concave downward.

At a certain velocity the force of air resistance is equal and opposite to that of gravity and you stop accelerating; velocity will approach that 'terminal velocity' as a horizontal asymptote.

The reason for the concavity is that velocity increases less and less quickly as air resistance increases; the approach of the velocity to terminal velocity is more and more gradual

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23:32:14

** When you fall without a parachute v will inrease, most rapidly at first, then less and less rapidly as air resistance increases.

When t = 0 we presume that v = 0.

At a certain velocity the force of air resistance is equal and opposite to that of gravity and you stop accelerating; velocity will approach that 'terminal velocity' as a horizontal asymptote.

The reason for the concavity is that velocity increases less and less quickly as air resistance increases; the approach of the velocity to terminal velocity is more and more gradual **

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RESPONSE -->

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23:32:44

What does the t = 0 acceleration indicate?

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RESPONSE -->

t = 0 acceleration is acceleration under the force of gravity, before you build velocity and start encountering significant air resistance.

Acceleration is rate of velocity change, indicated by the slope of the v vs. t graph. **

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23:32:49

** t = 0 acceleration is acceleration under the force of gravity, before you build velocity and start encountering significant air resistance.

Acceleration is rate of velocity change, indicated by the slope of the v vs. t graph. **

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RESPONSE -->

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ª~Æüå›æóÌŒÅ”ÒMÎÛ’ýÿò

assignment #014

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-05-2008

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23:48:11

Query problem 2.5.13. s(t) = 5 t^2 + 3

What are the functions for velocity and acceleration as functions of t?

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RESPONSE -->

The velocity function is s ' (t) = 10 t and the acceleration function is s '' (t) = 10.

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23:48:17

** The velocity function is s ' (t) = 10 t and the acceleration function is s '' (t) = 10. *&*&

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RESPONSE -->

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23:48:43

Query problem 2.5.11. Function negative, increasing at decreasing rate. What are the signs of the first and second derivatives of the function?

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RESPONSE -->

The function is increasing so its derivative is positive.

The slopes are decreasing, meaning that the rate of increase is decreasing. This means that the first derivative, which is represented by the slope, is decreasing.

The second derivative is the rate of change of the derivative. Since the derivative is decreasing its rate of change is negative. Thus the second derivative is negative.

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23:48:48

*&*& The function is increasing so its derivative is positive.

The slopes are decreasing, meaning that the rate of increase is decreasing. This means that the first derivative, which is represented by the slope, is decreasing.

The second derivative is the rate of change of the derivative. Since the derivative is decreasing its rate of change is negative. Thus the second derivative is negative. *&*&

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RESPONSE -->

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23:49:57

Query problem 2.5.23 continuous fn increasing, concave down. f(5) = 2, f '(5) = 1/2. Describe your graph.How many zeros does your function have and what are their locations?

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RESPONSE -->

the curve never actually reaches 2 but comes infinitessimally close

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23:50:08

** The average slope between (0,0) and (5, 2) is .4. If the graph goes thru (0,0) then its slope at (5, 2), which is 1/2, is greater than its average slope over the interval, which cannot be the case if the graph is concave down.

A constant slope of 1/2, with the graph passing through (5,2), would imply an x-intercept at (1, 0). Since the function is concave down the average slope between the x intercept and (5, 2) must be > 1/2 an x intercept must lie between (1,0) and (2,0).

We can't really say what happens x -> infinity, since you don't know how the concavity behaves. It's possible that the function approaches infinity (a square root function, for example, is concave down but still exceeds all bounds as x -> infinity). It can approach an asymptote and 3 or 4 is a perfectly reasonable estimate--anything greater than 2 is possible.

However the question asks about the limit at -infinity.

As x -> -infinity we move to the left. The slope increases as we move to the left, so the function approaches -infinity as x -> -infinity.

f'(1) implies slope 1, which implies that the graph makes an angle of 45 deg with the x axis; it's not horizontal.

Because of the downward concavity the ave slope between x = 1 and x = 5 is greater than the slope at x = 5--i.e., greater than 1/2. This is the only restriction. A slope of 1 is therefore indeed possible.

A slope of 1/4, or any slope less than 1/2, would be impossible. **

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RESPONSE -->

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23:50:30

What is the limiting value of the function as x -> -infinity and why must this be the limiting value?

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RESPONSE -->

The limiting value is 2

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23:50:41

STUDENT RESPONSE AND INSTRUCTOR COMMENT:

The limiting value is 2, the curve never actually reaches 2 but comes infinitessimally close.

INSTRUCTOR COMMENT:

The value of the function actually reaches 2 when x = 5, and the function is still increasing at that point. If there is a horizontal asymptote, which might indeed be the case, it would have to be to a value greater than 2, since the function is increasing. **

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RESPONSE -->

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23:51:16

Is it possible that f ' (1) = 1? Is it possible that f ' (1) = 1/4?

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RESPONSE -->

f'(1) implies slope 1, which implies that if x and y scales are equal the graph makes an angle of 45 deg with the x axis; it's not horizontal.

A slope of 1/4, or any slope less than 1/2, would be impossible.

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23:51:22

** f'(1) implies slope 1, which implies that if x and y scales are equal the graph makes an angle of 45 deg with the x axis; it's not horizontal.

A slope of 1/4, or any slope less than 1/2, would be impossible. **

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RESPONSE -->

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23:51:57

Query problem 5.1.3 speed at 15-min intervals is 12, 11, 10, 10, 8, 7, 0 mph

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RESPONSE -->

upper estimate would assume that the speed on each time interval remained at the maximum value given for that interval. For example on the first interval from 0 to 15 minutes, speeds of 12 mph and 11 mph are given. The upper estimate would assume the higher value, 12 mph, for the 15 minute interval. The estimate would be 12 mi/hr * .25 hr = 3 miles.

For the second interval the upper estimate would be 11 mi/hr * .25 hr = 2.75 mi.

For the third interval the upper estimate would be 10 mi/hr * .25 hr = 2.5 mi.

For the fourth interval the upper estimate would be 10 mi/hr * .25 hr = 2.5 mi.

For the fifth interval the upper estimate would be 8 mi/hr * .25 hr = 2.0 mi.

For the sixth interval the upper estimate would be 7 mi/hr * .25 hr = 1.75 mi.

The upper estimate would therefore be the sum 14.5 mi of these distances.

Lower estimates would use speeds 11, 10, 10, 8, 7 and 0 mph in the same calculation, obtaining a lower estimate of the estimate 11.5 mph. **

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23:52:03

** your upper estimate would assume that the speed on each time interval remained at the maximum value given for that interval. For example on the first interval from 0 to 15 minutes, speeds of 12 mph and 11 mph are given. The upper estimate would assume the higher value, 12 mph, for the 15 minute interval. The estimate would be 12 mi/hr * .25 hr = 3 miles.

For the second interval the upper estimate would be 11 mi/hr * .25 hr = 2.75 mi.

For the third interval the upper estimate would be 10 mi/hr * .25 hr = 2.5 mi.

For the fourth interval the upper estimate would be 10 mi/hr * .25 hr = 2.5 mi.

For the fifth interval the upper estimate would be 8 mi/hr * .25 hr = 2.0 mi.

For the sixth interval the upper estimate would be 7 mi/hr * .25 hr = 1.75 mi.

The upper estimate would therefore be the sum 14.5 mi of these distances.

Lower estimates would use speeds 11, 10, 10, 8, 7 and 0 mph in the same calculation, obtaining a lower estimate of the estimate 11.5 mph. **

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RESPONSE -->

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23:52:32

What time interval would result in upper and lower estimates within .1 mile of the distance?

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RESPONSE -->

The right- and left-hand limits have to differ over an interval (a, b) by at most ( f(b) - f(a) ) * `dx. We want to find `dx that will make this expression at most .1 mile. Thus we have the equation ( f(b) - f(a) ) * `dx <= .1 mile.

Since f(b) - f(a) = 0 - 12 = -12, we have | -12 mph | * `dx <= .1 mile so `dx <= .1 mile / 12 mph = 1/120 hour = 30 sec.

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23:52:37

** The right- and left-hand limits have to differ over an interval (a, b) by at most ( f(b) - f(a) ) * `dx. We want to find `dx that will make this expression at most .1 mile. Thus we have the equation ( f(b) - f(a) ) * `dx <= .1 mile.

Since f(b) - f(a) = 0 - 12 = -12, we have | -12 mph | * `dx <= .1 mile so `dx <= .1 mile / 12 mph = 1/120 hour = 30 sec. **

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RESPONSE -->

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23:57:04

Query problem 5.1.13. Acceleration table for vel, estimate vel (at 1-s intervals 9.81, 8.03, 6.53, 5.38, 4.41, 3.61)

Give your upper and lower estimates of your t = 5 speed and explain how you obtained your estimates.

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RESPONSE -->

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23:57:12

For each interval we multiply the maximum or minimum value by the time interval. For each interval the maximum value given happens to be the left-hand value of the acceleration and the minimum is the right-hand value.

Left-hand values give us the sum

9.81 m/s^2 * 1 sec +8.03 m/s^2 * 1 sec +6.53 m/s^2 * 1 sec +5.38 m/s^2 * 1 sec +4.41 m/s^2 * 1 sec =34.16 m/s.

Right-hand values give us the sum

8.03 m/s^2 * 1 sec +6.53 m/s^2 * 1 sec +5.38 m/s^2 * 1 sec +4.41 m/s^2 * 1 sec +3.61 m/s^2 * 1 sec =27.96 m/s.

So our lower and upper limits for the change in velocity are 27.96 m/s and 34.16 m/s. *&*&

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RESPONSE -->

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23:57:36

What is the average of your estimates, and is the estimate high or low (explain why in terms of concavity)?

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RESPONSE -->

he average of the two estimates is (27.96 m/s + 34.16 m/s) / 2 = 31.06 m/s.

The graph of a(t) is decreasing at a decreasing rate. It is concave upward. On every interval a linear approximation (i.e., a trapezoidal approximation) will therefore remain above the graph. So the graph 'dips below' the linear approximation.

This means that on each interval the average acceleration will be closer to the right-hand estimate, which is less than the left-hand estimate. Thus the actual change in velocity will probably be closer to the lower estimate than to the upper, and will therefore be less than the average of the two estimates.

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23:57:41

** The average of the two estimates is (27.96 m/s + 34.16 m/s) / 2 = 31.06 m/s.

Another way of saying this:

The graph is concave up. That means for each interval the graph dips down between the straight-line approximation of a trapezoidal graph. Your estimate is identical to that of the trapezoidal graph; since the actual graph dips below the trapezoidal approximation the actual velocity will be a bit less than that you have estimated. **

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RESPONSE -->

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¯Òúd„ÓÌÚø‚ÝÐëª£¸ïx½Æ®|]˜€†

assignment #015

015. The differential and the tangent line

08-06-2008

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00:03:24

`q001. Using the differential and the value of the function at x = 3, estimate the value of f(x) = x^5 and x = 3.1. Compare with the actual value.

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RESPONSE -->

The differential of a function y = f(x) is `dy = f ' (x) * `dx. Since for f(x) = x ^ 5 we have f ' (x) = 5 x^4, the differential is `dy = 5 x^4 `dx.

At x = 3 the differential is `dy = 5 * 3^4 * `dx, or

`dy = 405 `dx.

Between x = 3 and x = 3.1 we have `dx = .1 so `dy = 405 * .1 = 40.5.

Since at x = 3 we have y = f(3) = 3^5 = 243, at x = 3.1 we should have y = 243 + 40.5 = 283.5, approx..

confidence assessment: 3

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00:03:34

The differential of a function y = f(x) is `dy = f ' (x) * `dx. Since for f(x) = x ^ 5 we have f ' (x) = 5 x^4, the differential is `dy = 5 x^4 `dx.

At x = 3 the differential is `dy = 5 * 3^4 * `dx, or

`dy = 405 `dx.

Between x = 3 and x = 3.1 we have `dx = .1 so `dy = 405 * .1 = 40.5.

Since at x = 3 we have y = f(3) = 3^5 = 243, at x = 3.1 we should have y = 243 + 40.5 = 283.5, approx..

Note that the actual value of 3.1 ^ 5 is a bit greater than 286; the inaccuracy in the differential is due to the changing value of the derivative between x = 3 and x = 3.1. Our approximation was based on the rate of change, i.e. the value of the derivative, at x = 3.

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RESPONSE -->

self critique assessment: 3

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00:04:42

`q002. Using the differential and the value of the function at x = e, estimate the value of ln(2.8). Compare with the actual value.

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RESPONSE -->

The differential of the function y = f(x) = ln(x) is `dy = f ' (x) `dx or

`dy = 1/x `dx.

If x = e we have

`dy = 1/e * `dx.

Between e and 2.8, `dx = 2.8 - e = 2.8 - 2.718 = .082, approx.. Thus `dy = 1/e * .082 = .030, approx..

Since ln(e) = 1, we see that ln(2.8) = 1 + .030 = 1.030, approx..

confidence assessment: 3

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00:04:48

The differential of the function y = f(x) = ln(x) is `dy = f ' (x) `dx or

`dy = 1/x `dx.

If x = e we have

`dy = 1/e * `dx.

Between e and 2.8, `dx = 2.8 - e = 2.8 - 2.718 = .082, approx.. Thus `dy = 1/e * .082 = .030, approx..

Since ln(e) = 1, we see that ln(2.8) = 1 + .030 = 1.030, approx..

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RESPONSE -->

self critique assessment: 3

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00:07:21

`q003. Using the differential verify that the square root of a number close to 1 is twice as close to 1 as the number. Hint: Find the differential approximation for the function f(x) = `sqrt(x) at an appropriate point.

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RESPONSE -->

confidence assessment: 3

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00:07:31

The differential for this function is easily seen to be `dy = 1 / (2 `sqrt(x) ) * `dx. At x = 1 the differential is therefore

`dy = 1 / 2 * `dx.

This shows that as we move away from x = 1, the change in y is half the change in x. Since y = f(x) = 1 when x = 1, as x 'moves away' from 1 we see that y also 'moves away' from 1, but only half as much.

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RESPONSE -->

self critique assessment: 3

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00:07:49

`q004. Using the differential approximation verify that the square of a number close to 1 is twice as far from 1 as the number.

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RESPONSE -->

The differential for this function is easily seen to be `dy = 2 x * `dx. At x = 1 the differential is therefore

`dy = 2 * `dx.

This shows that as we move away from x = 1, the change in y is double the change in x. Since y = f(x) = 1 when x = 1, as x 'moves away' from 1 we see that y also 'moves away' from 1, but by twice as much.

confidence assessment: 3

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00:07:57

The differential for this function is easily seen to be `dy = 2 x * `dx. At x = 1 the differential is therefore

`dy = 2 * `dx.

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RESPONSE -->

self critique assessment: 3

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00:08:18

`q005. The lifting strength of an athlete in training changes according to the function L(t) = 400 - 250 e^(-.02 t), where t is the time in weeks since training began. What is the differential of this function? At t = 50, what approximate change in strength would be predicted by the differential for the next two weeks?

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RESPONSE -->

The differential is

`dL = L ' (t) * `dt =

-.02 ( -250 e^(-.02 t) ) `dt, so

`dL = 5 e^(-.02 t) `dt.

At t = 50 we thus have

`dL = 5 e^(-.02 * 50) `dt, or

`dL = 1.84 `dt.

The change over the next `dt = 2 weeks would therefore be approximately

`dL = 1.84 * 2 = 3.68.

confidence assessment: 3

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00:08:24

The differential is

`dL = L ' (t) * `dt =

-.02 ( -250 e^(-.02 t) ) `dt, so

`dL = 5 e^(-.02 t) `dt.

At t = 50 we thus have

`dL = 5 e^(-.02 * 50) `dt, or

`dL = 1.84 `dt.

The change over the next `dt = 2 weeks would therefore be approximately

`dL = 1.84 * 2 = 3.68.

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RESPONSE -->

self critique assessment: 3

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00:08:46

`q006. As you move away from a fairly typical source of light, the illumination you experience from that light is given by I(r) = k / r^2, where k is a constant number and r is your distance in meters from the light. Using the differential estimate the change in illumination as you move from r = 10 meters to r = 10.3 meters.

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RESPONSE -->

The differential is

`dI = I ' (r) * `dr,

where I ' is the derivative of I with respect to r.

Since I ' (r) = - 2 k / r^3, we therefore have

`dI = -2 k / r^3 * `dr.

For the present example we have r = 10 m and `dr = .3 m, so

`dI = -2 k / (10^3) * .3 = -.0006 k.

This is the approximate change in illumination.

confidence assessment: 3

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00:08:53

The differential is

`dI = I ' (r) * `dr,

where I ' is the derivative of I with respect to r.

Since I ' (r) = - 2 k / r^3, we therefore have

`dI = -2 k / r^3 * `dr.

For the present example we have r = 10 m and `dr = .3 m, so

`dI = -2 k / (10^3) * .3 = -.0006 k.

This is the approximate change in illumination.

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RESPONSE -->

self critique assessment: 3

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00:47:46

`q007. A certain crystal grows between two glass plates by adding layers at its edges. The crystal maintains a rectangular shape with its length double its width. Its width changes by .1 cm every hour. At a certain instant its width is 5 cm. Use a differential approximation to determine its approximate area 1 hour later.

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RESPONSE -->

If the width of the crystal is x then its length is 2x and its area is 2x * x = 2x^2. So we wish to approximate f(x) = 2x^2 near x = 5.

f ' (x) = 4 x, so when x = 5 we have y = 2 * 5^2 = 50 and y ' = 4 * 5 = 20.

The rate at which area changes with respect to width is therefore close to y ' = 20 units of area per unit of width. A change of .1 cm in width therefore implies a change of approximately 20 * .1 = 2 in area. Thus the approximate area should be 50 + 2 = 52. This can easily be compared with the accurate value of the area which is 52.02.

confidence assessment: 3

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00:47:53

If the width of the crystal is x then its length is 2x and its area is 2x * x = 2x^2. So we wish to approximate f(x) = 2x^2 near x = 5.

f ' (x) = 4 x, so when x = 5 we have y = 2 * 5^2 = 50 and y ' = 4 * 5 = 20.

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RESPONSE -->

self critique assessment: 3

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00:48:21

`q008. The radius of a sphere is increasing at the rate of .3 cm per day. Use the differential to determine the approximate rate at which its volume is increasing on a day when the radius is in the neighborhood of 20 cm.

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RESPONSE -->

The volume of a sphere is V(r) = 4/3 * `pi * r^3. The rate at which the volume changes with respect to the radius is dV / dr = 4 * `pi * r^2.

Thus when r = 20 the volume is changing at a rate of 4 * `pi * 20^2 = 1600 `pi cm^3 volume per cm of radius. It follows that if the radius is changing by .3 cm / day, the volume must be changing at 1600 `pi cm^3 / (cm of radius) * .3 cm of radius / day = 480 `pi cm^3 / day.

confidence assessment: 3

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00:48:31

The volume of a sphere is V(r) = 4/3 * `pi * r^3. The rate at which the volume changes with respect to the radius is dV / dr = 4 * `pi * r^2.

Note that this is the instantaneous rate at the instant r = 20. This rate will increase as r increases.

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RESPONSE -->

self critique assessment: 3

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¦òÐQ‘”øäÈÏœÇ–¼üÅÑÞÙÌÞ

assignment #016

016. Implicit differentiation.

08-06-2008

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00:55:00

`q001. Suppose that y is a function of x. Then the derivative of y with respect to x is y '. The derivative of x itself, with respect to x, is by the power function rule 1, x being x^1. What therefore would be the derivative of the expression x^2 y?

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RESPONSE -->

By the product rule we have (x^2 y) ' = (x^2) ' y + x^2 * y ' = 2 x y + x^2 y ' .

confidence assessment: 3

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00:55:15

By the product rule we have (x^2 y) ' = (x^2) ' y + x^2 * y ' = 2 x y + x^2 y ' .

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RESPONSE -->

self critique assessment: 3

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00:55:54

`q002. Note that if y is a function of x, then y^3 is a composite function, evaluated from a given value of x by the chain of calculations that starts with x, then follows the rule for y to get a value, then cubes this value to get y^3. To find the derivative of y^3 we would then need to apply the chain rule. If we let y' stand for the derivative of y with respect to x, what would be the derivative of the expression y^3 with respect to x?

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RESPONSE -->

The derivative of y^3 with respect to x can be expressed more explicitly by writing y^3 as y(x)^3. This reinforces the idea that y is a function of x and that we have a composite f(y(x)) of the function f(z) = z^3 with the function y(x). The chain rule tells us that the derivative of this function must be

(f ( y(x) ) )' = y ' (x) * f ' (y(x)),

in this case with f ' (z) = (z^3) ' = 3 z^2.

The chain rule thus gives us (y(x)^3) ' = y ' (x) * f ' (y(x) ) = y ' (x) * 3 * ( y(x) ) ^2.

In shorthand notation, (y^3) ' = y ' * 3 y^2.

This shows how the y ' comes about in implicit differentiation.

confidence assessment: 3

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00:56:02

(f ( y(x) ) )' = y ' (x) * f ' (y(x)),

in this case with f ' (z) = (z^3) ' = 3 z^2.

The chain rule thus gives us (y(x)^3) ' = y ' (x) * f ' (y(x) ) = y ' (x) * 3 * ( y(x) ) ^2.

In shorthand notation, (y^3) ' = y ' * 3 y^2.

This shows how the y ' comes about in implicit differentiation.

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RESPONSE -->

self critique assessment: 3

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00:56:22

`q003. If y is a function of x, then what is the derivative of the expression x^2 y^3?

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RESPONSE -->

The derivative of x^2 y^3, with respect to x, is

(x^2 y^3) ' = (x^2)' * y^3 + x^2 * (y^3) ' = 2x * y^3 + x^2 * [ y ' * 3 y^2 ] = 2 x y^3 + 3 x^2 y^2 y '.

confidence assessment: 3

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00:56:33

The derivative of x^2 y^3, with respect to x, is

(x^2 y^3) ' = (x^2)' * y^3 + x^2 * (y^3) ' = 2x * y^3 + x^2 * [ y ' * 3 y^2 ] = 2 x y^3 + 3 x^2 y^2 y '.

Note that when the expression is simplified, the y ' is conventionally placed at the end of any term of which is a factor.

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RESPONSE -->

self critique assessment: 3

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00:57:00

`q004. The equation 2x^2 y + 7 x = 9 can easily be solved for y. What is the result?

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RESPONSE -->

Starting with 2 x^2 y + 7 x = 9 we first subtract seven x from both sides to obtain

2 x^2 y = 9 - 7 x. We then divide both sides by 2 x^2 to obtain

y = (9 - 7 x ) / (2 x^2), or if we prefer

y = 9 / (2 x^2 ) - 7 / ( 2 x ).

confidence assessment: 3

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00:57:21

Starting with 2 x^2 y + 7 x = 9 we first subtract seven x from both sides to obtain

2 x^2 y = 9 - 7 x. We then divide both sides by 2 x^2 to obtain

y = (9 - 7 x ) / (2 x^2), or if we prefer

y = 9 / (2 x^2 ) - 7 / ( 2 x ).

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RESPONSE -->

self critique assessment: 3

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00:57:45

`q005. Using the form y = 9 / (2 x^2 ) - 7 / ( 2 x ), what is y and what is y ' when x = 1?

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RESPONSE -->

y ' = 9/2 * ( 1/x^2) ' - 7/2 ( 1 / x) ' = 9/2 * (-2 / x^3) - 7/2 * (-1 / x^2) = -9 / x^3 + 7 / (2 x^2).

So when x = 1 we have

y = 9 / (2 * 1^2 ) - 7 / ( 2 * 1 ) = 9/2 - 7/2 = 1 and

y ' = -9 / 1^3 + 7 / (2 * 1^2) = -9 + 7/2 = -11 / 2.

confidence assessment: 3

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00:57:54

y ' = 9/2 * ( 1/x^2) ' - 7/2 ( 1 / x) ' = 9/2 * (-2 / x^3) - 7/2 * (-1 / x^2) = -9 / x^3 + 7 / (2 x^2).

So when x = 1 we have

y = 9 / (2 * 1^2 ) - 7 / ( 2 * 1 ) = 9/2 - 7/2 = 1 and

y ' = -9 / 1^3 + 7 / (2 * 1^2) = -9 + 7/2 = -11 / 2.

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RESPONSE -->

self critique assessment: 3

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00:58:16

`q006. The preceding calculation could have been obtained without solving the original equation 2x^2 y + 7 x = 9 for y. We could have taken the derivative of both sides of the equation to get 2 ( (x^2) ' y + x^2 y ' ) + 7 = 0, or 2 ( 2x y + x^2 y ' ) + 7 = 0.

Complete the simplification of this equation, then solve for y ' .

Note that when x = 1 we have y = 1. Substitute x = 1, y = 1 into the equation for y ' and see what you get for y '.

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RESPONSE -->

Starting with 2 ( 2x y + x^2 y ' ) + 7 = 0 we divide both sides by 2 to obtain

2x y + x^2 y ' + 7/2 = 0. Subtracting 2 x y + 7 / 2 from both sides we obtain

x^2 y' = - 2 x y - 7 / 2.

Dividing both sides by x^2 we end up with

y ' = (- 2 x y - 7 / 2) / x^2 = -2 y / x - 7 / ( 2 x^2).

Substituting x = 1, y = 1 we obtain

y ' = -2 * 1 / 1 - 7 / ( 2 * 1^2) = - 11 / 2.

confidence assessment: 3

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00:58:23

Starting with 2 ( 2x y + x^2 y ' ) + 7 = 0 we divide both sides by 2 to obtain

2x y + x^2 y ' + 7/2 = 0. Subtracting 2 x y + 7 / 2 from both sides we obtain

x^2 y' = - 2 x y - 7 / 2.

Dividing both sides by x^2 we end up with

y ' = (- 2 x y - 7 / 2) / x^2 = -2 y / x - 7 / ( 2 x^2).

Substituting x = 1, y = 1 we obtain

y ' = -2 * 1 / 1 - 7 / ( 2 * 1^2) = - 11 / 2.

Note that this agrees with the result y ' = -11/2 obtained when we solved the original equation and took its derivative. This shows that it is not always necessary to explicitly solve the original equation for y. This is a good thing because it is not always easy, in fact not always possible, to solve a given equation for y.

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RESPONSE -->

self critique assessment: 3

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00:58:49

`q007. Follow the procedure of the preceding problem to determine the value of y ' when x = 1 and y = 2, for the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0. Also validate the fact that x = 1, y = 2 is a solution to the equation, because if this isn't a solution it makes no sense to ask a question about the equation for these values of x and y.

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RESPONSE -->

qNote that this time we do not need to solve the equation explicitly for y. This is a good thing because this equation is cubic in y, and while there is a formula (rather a set of formulas) to do this it is a lengthy and messy process.

The derivative of the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0 is (2 x^2 y^3) ' - (3 x y^2) ' - (4)' = 0. Noting that 4 is a constant, we see that (4)' = 0. The derivative of the equation therefore becomes

(2 x^2) ' * y^3 + 2 x^2 * ( y^3) ' - (3 x ) ' * y^2 - 3 x * (y ^2 ) ' = 0,

confidence assessment: 3

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00:58:56

`qNote that this time we do not need to solve the equation explicitly for y. This is a good thing because this equation is cubic in y, and while there is a formula (rather a set of formulas) to do this it is a lengthy and messy process.

(2 x^2) ' * y^3 + 2 x^2 * ( y^3) ' - (3 x ) ' * y^2 - 3 x * (y ^2 ) ' = 0, or

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RESPONSE -->

confidence assessment: 3

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01:02:17

4 x y^3 + 6 x^2 y^2 y' - 3 y^2 - 6xy y' = 0. Subtracting from both sides the terms which do not contain y ' we get

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RESPONSE -->

6 x^2 y^2 y' - 6xy y' =4 x y^3 -3 y^2

confidence assessment: 3

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01:04:05

6 x^2 y^2 y ' - 6 x y y ' = - 4 x * y^3 + 3 y^2. Factoring out the y ' on the left-hand side we have

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RESPONSE -->

y '(6 x^2 y^2 - 6 x y ) = - 4 x * y^3 + 3 y^2

confidence assessment: 3

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01:07:11

y ' ( 6 x^2 y^2 - 6 x y) = -4 x y^3 + 3 y^2. Dividing both sides by the coefficient of y ':

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RESPONSE -->

y ' = (- 4 x * y^3 + 3 y^2) / ( 6 x^2 y^2 - 6 x y )

confidence assessment: 3

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01:08:22

y ' = (- 4 x * y^3 + 3 y^2) / ( 6 x^2 y^2 - 6 x y ) . The numerator and denominator have common factor y so we end up with

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RESPONSE -->

if x = 1 and y = 2 the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0 gives us

2 * 1^2 * 2^3 - 3 * 1 * 2^2 - 4 = 0, or

16 - 12 - 4 = 0, which is true.

Substituting x = 1 and y = 2 into the equation y' = (- 4 x * y^2 + 3 y) / ( 6 x^2 y - 6 x ) we get

y ' = (- 4 * 1 * 2^2 + 3 * 2) / ( 6 * 1^2 * 2 - 6 * 1 ) =

(-16 + 6) / (12 - 6) = -10 / 6 = -5/3 or -1.66

confidence assessment: 3

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01:09:16

y' = (- 4 x * y^2 + 3 y) / ( 6 x^2 y - 6 x ).

Now we see that if x = 1 and y = 2 the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0 gives us

2 * 1^2 * 2^3 - 3 * 1 * 2^2 - 4 = 0, or

16 - 12 - 4 = 0, which is true.

Substituting x = 1 and y = 2 into the equation y' = (- 4 x * y^2 + 3 y) / ( 6 x^2 y - 6 x ) we get

y ' = (- 4 * 1 * 2^2 + 3 * 2) / ( 6 * 1^2 * 2 - 6 * 1 ) =

(-16 + 6) / (12 - 6) = -10 / 6 = -5/3 or -1.66... .

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RESPONSE -->

confidence assessment: 3

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01:10:03

`q008. (Mth 173 only; Mth 271 doesn't require the use of sine and cosine functions). Follow the procedure of the preceding problem to determine the value of y ' when x = 3 and y = `pi, for the equation x^2 sin (y) - sin(xy) = 0.

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RESPONSE -->

Taking the derivative of both sides of the equation we obtain

(x^2) ' sin(y) + x^2 (sin(y) ) ' - ( sin (xy) ) '.

By the Chain Rule

(sin(y)) ' = y ' cos(y) and

(sin(xy)) ' = (xy) ' cos(xy) = ( x ' y + x y ' ) cos(xy) = (y + x y ' ) cos(xy).

So the derivative of the equation becomes

2 x sin(y) + x^2 ( y ' cos(y)) - ( y + x y ' ) cos(xy) = 0. Expanding all terms we get

2 x sin(y) + x^2 cos(y) y ' - y cos(xy) - x cos(xy) y ' = 0. Leaving the y ' terms on the left and factoring ou y ' gives us

[ x^2 cos(y) - x cos(xy) ] y ' = y cos(xy) - 2x sin(y), so that

y ' = [ y cos(xy) - 2x sin(y) ] / [ x^2 cos(y) - x cos(xy) ].

Now we can substitute x = 3 and y = `pi to get

y ' = [ `pi cos( 3 * `pi) - 2 * 3 sin(`pi) ] / [ 3^2 cos(`pi) - 3 cos(3 * `pi) ] = [ `pi * -1 - 2 * 3 * 0 ] / [ 9 * -1 - 3 * -1 ] = -`pi / 6.

confidence assessment: 3

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01:10:10

Taking the derivative of both sides of the equation we obtain

(x^2) ' sin(y) + x^2 (sin(y) ) ' - ( sin (xy) ) '.

By the Chain Rule

(sin(y)) ' = y ' cos(y) and

(sin(xy)) ' = (xy) ' cos(xy) = ( x ' y + x y ' ) cos(xy) = (y + x y ' ) cos(xy).

So the derivative of the equation becomes

2 x sin(y) + x^2 ( y ' cos(y)) - ( y + x y ' ) cos(xy) = 0. Expanding all terms we get

2 x sin(y) + x^2 cos(y) y ' - y cos(xy) - x cos(xy) y ' = 0. Leaving the y ' terms on the left and factoring ou y ' gives us

[ x^2 cos(y) - x cos(xy) ] y ' = y cos(xy) - 2x sin(y), so that

y ' = [ y cos(xy) - 2x sin(y) ] / [ x^2 cos(y) - x cos(xy) ].

Now we can substitute x = 3 and y = `pi to get

y ' = [ `pi cos( 3 * `pi) - 2 * 3 sin(`pi) ] / [ 3^2 cos(`pi) - 3 cos(3 * `pi) ] = [ `pi * -1 - 2 * 3 * 0 ] / [ 9 * -1 - 3 * -1 ] = -`pi / 6.

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RESPONSE -->

self critique assessment: 3

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assignment #015

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-06-2008

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00:49:58

query problem 5.2.23 integral of x^2+1 from 0 to 6. What is the value of your integral, and what were your left-hand and right-hand estimates?

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RESPONSE -->

An antiderivative of x^2 + 1 is x^3 / 3 + x. The change in the antiderivative from x = 0 to x = 6 is equal to the integral:

Integral = 6^3 / 3 + 6 - (0^3 / 3 + 0) = 216/3 + 6 = 234/3 = 78.

The values of the function at 0, 2, 4, 6 are 1, 5, 17 and 37.

The left-hand sum would therefore be 2 * 1 + 2 * 5 + 2 * 17 = 46.

The right-hand sum would be 2 * 5 + 2 * 17 + 2 * 37 = 118.

The sums differ by ( f(b) - f(a) ) * `dx = (37 - 1) * 2 = 72.

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00:50:12

** An antiderivative of x^2 + 1 is x^3 / 3 + x. The change in the antiderivative from x = 0 to x = 6 is equal to the integral:

Integral = 6^3 / 3 + 6 - (0^3 / 3 + 0) = 216/3 + 6 = 234/3 = 78.

The values of the function at 0, 2, 4, 6 are 1, 5, 17 and 37.

The left-hand sum would therefore be 2 * 1 + 2 * 5 + 2 * 17 = 46.

The right-hand sum would be 2 * 5 + 2 * 17 + 2 * 37 = 118.

The sums differ by ( f(b) - f(a) ) * `dx = (37 - 1) * 2 = 72.**

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RESPONSE -->

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00:50:41

From the shape of the graph which estimate would you expect to be low and which high, and what property of the graph makes you think so?

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RESPONSE -->

The graph is increasing so the left-hand sum should be the lesser.

the graph is concave upward and increasing, and for each interval the average value of the function will therefore be closer to the left-hand estimate than the right. This is corroborated by the fact that 46 is closer to 78 than is 118.

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00:50:48

** The graph is increasing so the left-hand sum should be the lesser.

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RESPONSE -->

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00:51:22

query problem 5.2.32 f(x) piecewise linear from (-4,1) to (-2,-1) to (1,1) to (3,0) then like parabola with vertex at (4,-1) to (5,0). What is your estimate of the integral of the function from -3 to 4 and how did you get this result?

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RESPONSE -->

We use the area interpretation of the integral. This graph can be broken into a trapezoidal graph and since all lines are straight the areas can be calculated exactly.

From x=-3 to x=0 the area between the graph and the x axis is 2 (divide the region into two triangles and a rectangle and find total area, or treat it as a trapezoid). The area is below the x axis so the integral from x = -3 to x = 0 is -2.

From x = 0 to x = 3 the area is 2, above the x axis, so the integral on this interval is 2.

If the graph from x=3 to x=4 was a straight line from (3,0) to (4,-1) the area over this interval would be .5 and the integral would be -.5. Since the graph curves down below this straight line the area will be more like .6 and the integral closer to -.6.

The total integral from x=-3 to x=4 would therefore be about -2 + 2 - .6 = -.6.

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00:51:27

** We use the area interpretation of the integral. This graph can be broken into a trapezoidal graph and since all lines are straight the areas can be calculated exactly.

From x = 0 to x = 3 the area is 2, above the x axis, so the integral on this interval is 2.

The total integral from x=-3 to x=4 would therefore be about -2 + 2 - .6 = -.6. **

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RESPONSE -->

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00:51:54

query problem 5.3.20 ave value of e^t over [0,10]. What is the average value of the function over the given interval?

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RESPONSE -->

An antiderivative of f(t) = e^t is F(t) = e^t. Using the Fundamental Theorem you get the definite integral by calculating F(10) - F(0). This gives you e^10 - e^0 = 22025, approx.

The average value of the function is the integral divided by the length of the interval. The length of the interval is 10 and 22025 / 10 is 2202.5.

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00:52:05

** An antiderivative of f(t) = e^t is F(t) = e^t. Using the Fundamental Theorem you get the definite integral by calculating F(10) - F(0). This gives you e^10 - e^0 = 22025, approx.

The average value of the function is the integral divided by the length of the interval. The length of the interval is 10 and 22025 / 10 is 2202.5. **

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RESPONSE -->

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00:52:32

What would be the height of a rectangle sitting on the t axis from t = 0 to t = 10 whose area is the same as that under the y = e^t graph from t=0 to t=10?

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RESPONSE -->

The height of the rectangle would be the average value of the function. That would have to be the case so that height * length would equal the integral.

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00:52:39

** The height of the rectangle would be the average value of the function. That would have to be the case so that height * length would equal the integral. **

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RESPONSE -->

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00:53:04

query problem 5.3.26 v vs. t polynomial (0,0) to (1/6,-10) thru (2/6,0) to (.7,30) to (1,0). When is the cyclist furthest from her starting point?

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RESPONSE -->

The change in position of the cyclist from the start to any instant will be the integral of the v vs. t graph from the t = 0 to that instant.

From t = 0 to t = 1/3 we see that v is negative, so the cyclist is getting further from her starting point. The average velocity over this interval appears to be about -6 mph, so the integral over this segment of the graph is about -6 * 1/3 = -2, indicating a displacement of -2 miles. At t = 1/3 she is therefore 3 miles from the lake.

For the remaining 2/3 of an hour the cyclist is moving in the positive direction, away from the lake. Her average velocity appears to be a bit greater than 15 mph--say about 18 mph--so the integral over this segment is about 18 * 2/3 = 12. She gets about 12 miles further from the lake, ending up about 15 miles away.

The key to interpretation of the integral of a graphed function is the meaning of a rectangle of the Riemann Sum. In this case any rectangle in the Riemann sum has dimensions of velocity (altitude) vs. time interval (width), so each tiny rectangle represents a product of velocity and time interval, giving you a displacement. The integral, represented by the total area under the curve, therefore represents the displacement.

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00:53:09

** The change in position of the cyclist from the start to any instant will be the integral of the v vs. t graph from the t = 0 to that instant.

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RESPONSE -->

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‹|xóÝkÂæ¶º’¹‚•Ô²’Ñ‹z‘—£ëšå{

assignment #016

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-06-2008

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01:12:47

5.4.2 fn piecewise linear from (0,1)-(1,1)-(3,-1)-(5,-1)-(6,1). F ' = f, F(0) = 0. Find F(b) for b = 1, 2, 3, 4, 5, 6.What are your values of F(b) for b = 1, 2, 3, 4, 5, 6 and how did you obtain them.

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RESPONSE -->

f(x) is the 'rate function'. The change in the 'amount function' F(x) is represented as the accumulated area under the graph of f(x). Since F(0) = 0, F(x) will just the equal to the accumulated area.

We can represent the graph of f(x) as a trapezoidal approximation graph, with uniform width 1.

The first trapezoid extends from x = 0 to x = 1 and has altitude y = 1 at both ends. Therefore its average altitude is 1 and its what is 1, giving area 1. So the accumulated area through the first trapezoid is 1. Thus F(1) = 1.

The second trapezoid has area .5, since its altitudes 1 and 0 average to .5 and imply area .5. The cumulative area through the second trapezoid is therefore still 1 + .5 = 1.5. Thus F(2) = 1.5.

The third trapezoid has altitudes 0 and -1 so its average altitude is -.5 and width 1, giving it 'area' -.5. The accumulated area through the third trapezoid is therefore 1 + .5 - .5 = 1. Thus F(3) = 1.

The fourth trapezoid has uniform altitude -1 and width 1, giving it 'area' -1. The accumulated area through the fourth trapezoid is therefore 1 + .5 - .5 - 1 = 0. Thus F(4) = 0.

The fifth trapezoid has uniform altitude -1 and width 1, giving it 'area' -1. The accumulated area through the fifth trapezoid is therefore 1 + .5 - .5 - 1 - 1 = -1. Thus F(5) = -1.

The sixth trapezoid has altitudes 1 and -1 so its average altitude, and therefore it's area, is 0. Accumulated area through the sixth trapezoid is therefore 1 + .5 - .5 - 1 - 1 + 0 = -1. Thus F(6) = -1.

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01:12:53

** f(x) is the 'rate function'. The change in the 'amount function' F(x) is represented as the accumulated area under the graph of f(x). Since F(0) = 0, F(x) will just the equal to the accumulated area.

We can represent the graph of f(x) as a trapezoidal approximation graph, with uniform width 1.

The second trapezoid has area .5, since its altitudes 1 and 0 average to .5 and imply area .5. The cumulative area through the second trapezoid is therefore still 1 + .5 = 1.5. Thus F(2) = 1.5.

The fourth trapezoid has uniform altitude -1 and width 1, giving it 'area' -1. The accumulated area through the fourth trapezoid is therefore 1 + .5 - .5 - 1 = 0. Thus F(4) = 0.

The fifth trapezoid has uniform altitude -1 and width 1, giving it 'area' -1. The accumulated area through the fifth trapezoid is therefore 1 + .5 - .5 - 1 - 1 = -1. Thus F(5) = -1.

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RESPONSE -->

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01:14:34

If you made a trapezoidal graph of this function what would be the area of teach trapezoid, and what would be the accumulated area from x = 0 to x = b for b = 1, 2, 3, 4, 5, 6? What does your answer have to do with this question?

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01:15:32

Query 5.4.12. integral of e^(x^2) from -1 to 1.

How do you know that the integral of this function from 0 to 1 lies between 0 and 3?

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RESPONSE -->

The max value of the function is e, about 2.71828, which occurs at x = 1. Thus e^(x^2) is always less than 3.

On the interval from 0 to 1 the curve therefore lies above the x axis and below the line y = 3. The area of this interval below the y = 3 line is 3. So the integral is less than 3.

Alternatively e^(x^2) is never as great as 3, so its average value is less than 3. Therefore its integral, which is average value * length of interval, is less that 3 * 1 = 3.

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01:15:38

** The max value of the function is e, about 2.71828, which occurs at x = 1. Thus e^(x^2) is always less than 3.

On the interval from 0 to 1 the curve therefore lies above the x axis and below the line y = 3. The area of this interval below the y = 3 line is 3. So the integral is less than 3.

Alternatively e^(x^2) is never as great as 3, so its average value is less than 3. Therefore its integral, which is average value * length of interval, is less that 3 * 1 = 3. **

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îÐ¯„â™õÊ{wÒª~¤å»ÔàòžÚÈ¡ŠÞù

assignment #017

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-06-2008

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01:18:17

Explain in terms of the contribution to the integral from a small increment `dx why the integral of f(x) - g(x) over an interval [a, b] is equal to the integral of f(x) over the interval minus the integral of g(x) over the same integral.

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RESPONSE -->

In terms of the contribution from a single increment, the rectangle has 'altitude' f(x) - g(x) so its area is [ f(x) - g(x) ] `dx. Adding up over all increments we get

sum { [ f(x) - g(x) ] `dx, which by the distributive law of addition over multiplication is

sum [ f(x) `dx - g(x) `ds ], which is not just a series of additions and can be rearrange to give

sum (f(x) `dx) - sum (g(x) `dx). As the interval `dx shrinks to zero, the sums approach the definite integrals and we get

int( f(x), x, a, b ) - int(g(x), x, a, b),

where int ( function, variable, left limit, right limit) is the definite integral of 'function' with respect to 'variable' from 'left limit' to 'right limit'.

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01:18:23

** In terms of the contribution from a single increment, the rectangle has 'altitude' f(x) - g(x) so its area is [ f(x) - g(x) ] `dx. Adding up over all increments we get

sum { [ f(x) - g(x) ] `dx, which by the distributive law of addition over multiplication is

sum [ f(x) `dx - g(x) `ds ], which is not just a series of additions and can be rearrange to give

sum (f(x) `dx) - sum (g(x) `dx). As the interval `dx shrinks to zero, the sums approach the definite integrals and we get

int( f(x), x, a, b ) - int(g(x), x, a, b),

where int ( function, variable, left limit, right limit) is the definite integral of 'function' with respect to 'variable' from 'left limit' to 'right limit'.**

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RESPONSE -->

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01:18:48

Explain why the if f(x) > m for all x on [a,b], the integral of f(x) over this interval is greater than m (b-a).

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RESPONSE -->

This is also in the text, so look there for an alternative explanation and full rigor.

The idea is that if f(x) > m for all x, then for any interval the contribution to the Riemann sum will be greater than m * `dx. So when all the contributions are added up the result is greater than the product of m and the sum of all `dx's.

The sum of all `dx's is equal to the length b-a of the entire interval. So the Riemann sum must be greater than the product of m and this sum--i.e., greater than m * ( b - a ).

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01:18:53

** This is also in the text, so look there for an alternative explanation and full rigor.

The sum of all `dx's is equal to the length b-a of the entire interval. So the Riemann sum must be greater than the product of m and this sum--i.e., greater than m * ( b - a ). **

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01:19:18

Explain why the integral of f(x) / g(x) is not generally equal to the integral of f(x) divided by the integral of g(x).

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RESPONSE -->

For now assume that f and g are both positive functions. The integral of f(x) represents the area beneath the curve between the two limits, and the integral of g(x) represents the area beneath its curve between the same two limits. So the integral of f(x) divided by the integral of g(x) represents the first area divided by the second.

The function f(x) / g(x) represents the result of dividing the value of f(x) by the value of g(x) at every x value. This gives a different curve, and the area beneath this curve has nothing to do with the quotient of the areas under the original two curves.

It would for example be possible for g(x) to always be less than 1, so that f(x) / g(x) would always be greater than f(x) so that the integral of f(x) / g(x) would be greater than the integral of f(x), while the length of the interval is very long so that the area under the g(x) curve would be greater than 1. In this case the integral of f(x) divided by the integral of g(x) would be less than the integral of f(x), and would hence be less than the integral of f(x) / g(x).

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01:19:23

** For now assume that f and g are both positive functions. The integral of f(x) represents the area beneath the curve between the two limits, and the integral of g(x) represents the area beneath its curve between the same two limits. So the integral of f(x) divided by the integral of g(x) represents the first area divided by the second.

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01:20:18

Given a graph of f(x) and the fact that F(x) = 0, explain how to construct a graph of F(x) such that F'(x) = f(x). Then explain how, if f(x) is the rate at which some quantity changes with respect to x, this construction gives us a function representing how much that quantity has changed since x = 0.

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RESPONSE -->

To construct the graph you could think of finding areas. You could for example subdivide the graph into small trapezoids, and add the area of each trapezoid to the areas of the ones preceding it. This would give you the approximate total area up to that point. You could graph total area up to x vs. x.

This would be equivalent to starting at (0,0) and drawing a graph whose slope is always equal to the value of f(x). The 'higher' the graph of f(x), the steeper the graph of F(x). If f(x) falls below the x axis, F(x) will decrease with a steepness that depends on how far f(x) is below the axis.

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01:20:23

** To construct the graph you could think of finding areas. You could for example subdivide the graph into small trapezoids, and add the area of each trapezoid to the areas of the ones preceding it. This would give you the approximate total area up to that point. You could graph total area up to x vs. x.

If you can see why the two approaches described here are equivalent, and why if you could find F(x) these approaches would be equivalent to what you suggest, you will have excellent insight into the First Fundamental Theorem. **

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01:26:47

Query problem 3.1.16 (3d edition 3.1.12 ) (formerly 4.1.13) derivative of fourth root of x.

What is the derivative of the given function?

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RESPONSE -->

1/(4*x^(3/4))

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01:26:59

The derivative y=x^(1/4), which is of the form y = x^n with n = 1/4, is

y ' = n x^(n-1) = 1/4 x^(1/4 - 1) = 1/4 x^(-3/4). *&*&

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01:29:28

Query problem 3.1.27 (formerly 4.1.24) derivative of (`theta-1)/`sqrt(`theta)

What is the derivative of the given function?

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RESPONSE -->

15*t^4-5/(2sqrt(t))-7/(t^2)

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01:29:34

** (`theta-1) / `sqrt(`theta) =

`theta / `sqrt(`theta) - 1 / `sqrt(`theta) =

`sqrt(`theta) - 1 / `sqrt(`theta) =

`theta^(1/2) - `theta^(-1/2).

The derivative is therefore found as derivative of the sum of two power functions: you get

1/2 `theta^(-1/2) - (-1/2)`theta^(-3/2), which simplifies to

1/2 [ `theta^(-1/2) + `theta^(-3/2) ]. . **

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01:33:17

Query problem 3.1.60 (3d edition 3.1.48) (formerly 4.1.40) function x^7 + 5x^5 - 4x^3 - 7

What is the eighth derivative of the given function?

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RESPONSE -->

The first derivative is 7 x^6 + 25 x^4 - 12 x^3. The second derivative is the derivative of this expression. We get 42 x^5 + 200 x^3 - 36 x^2.

It isn't necessary to keep taking derivatives if we notice the pattern that's emerging here.

If we keep going the highest power will keep shrinking but its coefficient will keep increasing until we have just 5040 x^0 = 5050 for the seventh derivative. The next derivative, the eighth, is the derivative of a constant and is therefore zero.

The main idea here is that the highest power is 7, and since the power of the derivative is always 1 less than the power of the function, the 7th derivative of the 7th power must be a multiple of the 0th power, which is constant. Then the 8th derivative is the derivative of a constant and hence zero.

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01:33:22

** The first derivative is 7 x^6 + 25 x^4 - 12 x^3. The second derivative is the derivative of this expression. We get 42 x^5 + 200 x^3 - 36 x^2.

It isn't necessary to keep taking derivatives if we notice the pattern that's emerging here.

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RESPONSE -->

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01:33:44

Query problem 3.2.4 (3d edition 3.2.6) (formerly 4.2.6) derivative of 12 e^x + 11^x.

What is the derivative of the given function?

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RESPONSE -->

The derivative of a^x is ln(a) * a^x. So the derivative of 11^x is ln(11) * 11^x.

The derivative of the given function is therefore 12 e^x + ln(11) * 11^x.

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01:33:49

** The derivative of a^x is ln(a) * a^x. So the derivative of 11^x is ln(11) * 11^x.

The derivative of the given function is therefore 12 e^x + ln(11) * 11^x. **

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01:34:12

Query problem 3.2.10 (3d edition 3.2.19) (formerly 4.2.20) derivative of `pi^2+`pi^x.

What is the derivative of the given function?

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RESPONSE -->

`pi is a constant and so therefore is `pi^2. Its derivative is therefore zero.

`pi^x is of the form a^x, which has derivative ln(a) * a^x. The derivative of `pi^x is thus ln(`pi) * `pi^x.

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01:34:17

** `pi is a constant and so therefore is `pi^2. Its derivative is therefore zero.

`pi^x is of the form a^x, which has derivative ln(a) * a^x. The derivative of `pi^x is thus ln(`pi) * `pi^x. **

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01:34:41

Query problem 3.2.40 (3d edition 3.2.30) (formerly 4.2.34)

value V(t) = 25(.85)^6, in $1000, t in years since purchase. What are the value and meaning of V(4) and ov V ' (4)?

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RESPONSE -->

V(4) is the value of the automobile when it is 4 years old.

V'(t) = 25 * ln(.85) * .85^t. You can easily calculate V'(4). This value represents the rate at which the value of the automobile is changing, in dollars per year, at the end of the 4th year.

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01:34:46

** V(4) is the value of the automobile when it is 4 years old.

V'(t) = 25 * ln(.85) * .85^t. You can easily calculate V'(4). This value represents the rate at which the value of the automobile is changing, in dollars per year, at the end of the 4th year. **

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Ò~Û‹íÙÀàwÑ¶UÙnŠý…•¹ï³’fõBù‹§ÍÓ

assignment #018

Ì§ëµÖ¥~‡åÃ ¿w]ÉÅðp¤…šåëöz

Calculus I

08-06-2008

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01:38:20

The figure is a pair of nested rectangles, one whose dimensions are f(x) by g(x), the other with dimensions f(x + `dx) by g(x+`dx).

The product of the two functions is initially f(x) * g(x), represented in the figure by the area of the smaller rectangle.

The product after the change `dx is f(x + `dx) * g(x + `dx), represented by the area of the larger rectangle.

The additional area is represented by three regions, one whose dimensions are f(x) * [ g(x + `dx) - g(x) ], another with dimensions g(x) * [ f(x + `dx) - f(x) ] and the third with dimensions [ f(x + `dx) - f(x) ] * [ g(x + `dx) - g(x) ].

We can approximate g(x + `dx) as g ' (x) * `dx, and we can approximate f(x + `dx) - f(x) by f ' (x) `dx, so the areas of the three regions are

f(x) * g ' (x) `dx, g(x) * f ' (x) `dx and f ' (x) `dx * g ' (x) `dx, giving us total additional area f(x) g ' (x) `dx + g(x) f ' (x) `dx + f ' (x) g ' (x) `dx^2.

This area, recall, represents the change in the product f(x) * g(x) as x changes by `dx. The average rate of change of the product is therefore

ave rate = change in product / `dx = [ f(x) g ' (x) `dx + g(x) f ' (x) `dx + f ' (x) g ' (x) `dx^2 ] / `dx = f(x) g ' (x) + g(x) f ' (x) + `dx.

As `dx -> 0 the average rate of change f(x) g ' (x) + g(x) f ' (x) + `dx approaches the instantaneous rate of change, which is f(x) g ' (x) + g(x) f ' (x).

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01:38:44

Explain in your own words why the derivative of the product of two functions cannot be expected to be equal to the product of the derivatives.

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RESPONSE -->

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01:38:54

The product of the derivatives would involve just the limiting behavior of the small rectangle in the upper-right-hand corner and would not involve the behavior of the elongated rectangles along the sides of the figure.

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01:39:21

Query problem 3.3.16 (3d edition 3.3.14) (formerly 4.3.14) derivative of (t - 4) / (t + 4).

What is the derivative of the given function?

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RESPONSE -->

By the Quotient Rule the derivative of f / g is (f ' g - g ' f) / g^2. For this problem f(t) is the numerator t - 4 and g(t) is the denominator t + 4. We get f ' (t) = 1 and g ' (t) = 1 so that the derivative is

[(1(t + 4)) - (1(t - 4))] / [(t + 4)^2] = 8 / ((t + 4)^2) = g'(t)

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01:39:28

*&*& By the Quotient Rule the derivative of f / g is (f ' g - g ' f) / g^2. For this problem f(t) is the numerator t - 4 and g(t) is the denominator t + 4. We get f ' (t) = 1 and g ' (t) = 1 so that the derivative is

[(1(t + 4)) - (1(t - 4))] / [(t + 4)^2] = 8 / ((t + 4)^2) = g'(t) **

DER

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01:41:10

Query problem 3.3.56 (3d edition 3.3.47) (was 4.3.40) f(v) is gas consumption in liters/km of a car at velocity v (km/hr); f(80) = .05 and f ' (80) = -.0005.

What is the function g(v) which represents the distance this car goes on one liter at velocity v?

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RESPONSE -->

This is an interpretation problem. If you graph f(v) vs. v you are graphing the number of liters/km (just like gallon used per mile but measured in metric units). f(80) is the number of liters/km used at a speed of 80 km/hour--for this particular car .05 liters / km. If the car uses .05 liters every kilometer then it takes it 1/ (.05 liters / km) = 20 kilometers /liter: it takes 20 km to use a liter.

Generalizing we see that liters / km and km / liter have a reciprocal relationship, so g(v) = 1 / f(v).

g(80) represents the number of km traveled on a liter when traveling at 80 km/hour. f(80) = .05 means that at 80 km/hour the car uses .05 liters per kilometer. It therefore travels 20 km on a liter, so g(80) = 20.

Using the reciprocal function relationship g(v) = 1 / f(v) we obtain the same result.

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01:41:17

** This is an interpretation problem. If you graph f(v) vs. v you are graphing the number of liters/km (just like gallon used per mile but measured in metric units). f(80) is the number of liters/km used at a speed of 80 km/hour--for this particular car .05 liters / km. If the car uses .05 liters every kilometer then it takes it 1/ (.05 liters / km) = 20 kilometers /liter: it takes 20 km to use a liter.

Generalizing we see that liters / km and km / liter have a reciprocal relationship, so g(v) = 1 / f(v).

Using the reciprocal function relationship g(v) = 1 / f(v) we obtain the same result. **

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01:41:39

What are the meanings of f ' (80) and f(80)?

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RESPONSE -->

f ' (80) is the rate at which the consumption rate per km is changing with respecting velocity. It would be the slope of the f(v) vs. v graph at v = 80. The rise on a graph of f(v) vs. v would represent the change in the number of liter / km -- in this case going faster increases the amount of fuel used per kilometer. The run would repesent the change in the velocity. So the slope represents change in liter/km / (change in speed in km / hr).

f'(80) = .0005 means that for an increase of 1 km / hr in speed, fuel consumption per km goes up by .0005 liter/km.

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