course mth 271
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12:11:09 Note 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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12:11:17 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.
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12:14:43 `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 --> t=10=71 t=40=7 t=90=-9
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12:15:42 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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12:18:14 `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 --> 64/30=2.133 16/50=0.32
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12:19:26 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.
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RESPONSE --> the original question had a t=40 not t=20
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12:24:00 `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 --> avg t= 10 to11= 1.79 avg t= 10 -10.1 =1.799
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12:24:36 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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12:25:31 `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?
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RESPONSE --> .02(10)-2= -1.8
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12:25:34 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.
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12:26:11 `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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12:26:53 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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12:26:58 `q007. What is the change in depth between these clock times?
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12:27:09 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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12:27:14 `q008. What is the average rate at which depth changes between these clock time?
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12:27:23 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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12:28:25 `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 --> -1.8
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12:28:29 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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~|h{ễΪ assignment #005 ݺӯD Applied Calculus I 06-20-2006
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12:24:56 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 top of a trapezoid is sloped similar to the curve of the line. the narrow the trapezoid the more accurate it will be.
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12:25:04 ** the specific idea is that ave rate of depth change is [change in depth / change in time] ; rise represents change in depth and run represents change in time so slope = rise/run represents ave rate of depth change. **
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12:27:32 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 --> if the top of the trapezoid and the curve of the line are similar and their x points or run points are the same, the area below the trapezoid top line and the area below a curve line will be exact except for the variances in the top lines.
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12:27:40 ** The average altitude represents the avg. velocity. The area of a trapezoid involves the altitude, which represents the avg. velocity, and the width, which represents the change in clock time. When you multiply ave altitude by width you are representing ave vel * change in clock time, which gives change in position. This reasoning isn't confined to velocities. For any rate vs. clock time graph, average altitude represents approximate average rate, which multiplied by the change in time (not by the time itself) gives you the change in quantity **
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12:32:23 text problem 0.5 #10 add x/(2-x) + 2/(x-2)
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RESPONSE --> x^2-2x+4-2x/-x^2-4+4x -1-1-1 -3
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12:32:43 ** common denominator could be [ (2-x)(x-2) ]. In this case we have x / (2-x) + 2 / (x-2) = [ (x-2) / (x-2) ] * [ x / (2-x) ] + [ (2-x) / (2-x) ] * [ 2 / (x-2) ] = x(x-2) / [ (2-x)(x-2) ] + 2 (2-x) / [ (2-x)(x-2) ] = [x(x-2) + 2(2-x) ] / [ (2-x)(x-2) ] = [ x^2 - 2x + 4 - 2x ] / [ (2-x)(x-2) ] = (x^2-4x+4) / [ -x^2+4x-4 ] = (x-2)^2 / [-(x-2)^2] = -1. NOTE however that there is a SIMPLER SOLUTION: We can note that x-2 = -(2-x) so that the original problem is -x/(x-2) + 2 /(x-2) = (-x + 2) / (x-2) = -(x-2)/(x-2) = -1. **
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12:37:51 text problem 0.5 #50 cost = 6 x + 900,000 / x, write as single fraction and determine cost to store 240 units
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RESPONSE --> (6x^2+900,000)/x 5190
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12:38:02 ** express with common denominator x: [x / x] * 6x + 900,000 / x = 6x^2 / x + 900,000 / x = (6x^2 + 900,000) / x so cost = (6x^2+900,000)/x Evaluating at x = 240 we get cost = (6 * 240^2 + 900000) / 240 = 5190. **
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ƘrC~m Student Name: assignment #006 006. goin' the other way
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12:41:44 Note 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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12:42:14 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.
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RESPONSE --> *****review*****8
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12:44:16 `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 --> 80cm -(30s-20s)4cm= 40cm less accurate. the flow will decrease as time progresses.
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12:44:22 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.
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12:46:43 `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 --> at t=20 flow 1s 4cm/s at t=30 flow is 3cm/s the depth will be greater since the rate of flow is slowing as time progresses.
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12:46:55 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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12:53:42 `q004. What is your specific estimate of the depth at t = 30 seconds?
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RESPONSE --> t=20 @ 4cm/s, t=30@ 3cm/s @ t=20 the depth was 80cm @t=30 the depth would be approx. 3-4/10=-.1 -.1x80 =-8 80-8=72cm/s
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12:54:18 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.
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RESPONSE --> ***** review**********8
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12:57:52 `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 --> t=20----- y'=.1(20)-6= -4cm/s t=30-----y'=.1(30)-6= -3cm/s
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12:57:56 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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12:58:51 `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 --> 0=.1t-6 6=.1t t=60sec
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12:58:54 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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12:59:49 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.
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`ύ{Szv؛ assignment #006 ݺӯD Applied Calculus I 06-21-2006
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12:30:02 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 --> y=ax^3 the pile is in a conical shape
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12:30:32 ** 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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12:35:02 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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12:37:15 ** 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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RESPONSE --> ***** for a surface (2dimensional) use y=k x^2 for 3D use y=k x^3
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12:40:49 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 --> for y=k x^3 y'=k 3x^2 substituting 5 into the equation to find the inst rate. multiply the inst rate by the delta x =.01. since the run change is so small the volume will not be off very much.
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12:41:29 ** 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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RESPONSE --> ***********review
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12:43:41 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 --> y=.02(30)^2 -3(30) + 6 y= -66
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12:45:03 ** 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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RESPONSE --> for rate of depth change take the derivative.
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12:50:55 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 --> y=kx^3 y=2(1.5)^3
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12:51:24 ** 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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RESPONSE --> ***************review
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12:52:11 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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12:52:51 ** 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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RESPONSE --> ^^^^^^^^^^review
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12:55:59 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 --> y=kx^3 1.5=30x^3 0.3684
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12:56:29 ** 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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RESPONSE --> *************** review
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ŒGѥٚY}z assignment #006 ݺӯD Applied Calculus I 06-21-2006
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13:00:52
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13:00:54 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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13:00:56 ** 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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13:00:58 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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13:01:00 ** 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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13:01:05 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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13:01:07 ** 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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13:01:10 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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13:01:13 ** 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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13:01:15 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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13:01:18 ** 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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13:01:21 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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13:01:26 ** 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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13:01:29 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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13:01:31 ** 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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RESPONSE --> ...M~˽DLj assignment #006 ݺӯD Applied Calculus I 06-23-2006
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18:20:24 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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18:20:26 ** 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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18:20:28 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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18:20:30 ** 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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18:23:22 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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18:23:26 ** 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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18:23:30 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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18:23:33 ** 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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18:23:37 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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18:23:40 ** 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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18:23:41 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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18:23:45 ** 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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18:23:53 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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18:23:57 ** 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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18:33:05 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 --> y=(.002x^3)/2
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18:33:43 ** 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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RESPONSE --> i divided instead of multiplying
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18:36:07 query problem 4. number of swings vs. length data. Which function fits best?
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RESPONSE --> y=ax^p
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18:37:14 ** 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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18:38:11 problem 7. time per swing model. For your data what expression represents the number of swings per minute?
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RESPONSE --> y
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18:38:56 ** 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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18:39:16 If the time per swing in seconds is y, then what expression represents the number of swings per minute?
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RESPONSE --> x
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18:39:41 ** 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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18:42:00 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 --> T=60/ ax^5
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18:42:14 ** 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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18:42:59 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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18:43:23 ** 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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18:43:36 query problem 9. length ratio x2 / x1.
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18:43:45 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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18:43:49 ** 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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18:43:53 What expression, in terms of x1 and x2, represents the ratio of the frequencies of the two pendulums?
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18:43:56 ** 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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18:44:03 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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18:44:05 ** 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. 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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18:50:42 Query problem 1.1.18 are (0,4), (7,-6) and (-5,11) collinear?
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RESPONSE --> (7,-6) (0,4) (-5,11) 0-7/4- -6= -7/10 -5-0/11-4= -5/7 no
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18:51:13 ** The distance between (-5,11) and (7,-6) is approximately 20.81: d3 = sqr rt [(7+5)^2 + -(6 +11)^2] d3 = sqr rt 433 d3 = 20.81 Using the distance formula the distances between (-5,11) and (0,4) is 8.6 and the distance between (0,4) and (7, -6) is 12.2. 'Collinear' means 'lying along the same straight line'. If three points are collinear then the sum of the distances between the two closer pairs of points will equal the distance between the furthest two. Since 8.6 + 12.2 = 20.8 the points are on the same straight line. **
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18:56:10 Query problem 1.1.24 find x | dist (2,-1) to (x,2) is 5What value of x makes the distance 5?
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RESPONSE --> 5= (x-2)^2 + (2- -1)^2 (x-2)^2= -4 0
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18:56:37 ** The expression for the distance from (2, -1) to (x, 2) is = sqrt((x-2)^2 + (2+1)^2). This distance is to be 5, which gives us the equation5 = sqrt((x-2)^2 + (2+1)^2) Starting with the equation 5 = sqrt((x-2)^2 + (2+1)^2) we first square both sides to get 25 = (x-2)^2 + 9 or (x-2)^2 = 16. Solutions are found by taking the square root of both sides, keeping in mind that (x-2)^2 doesn't distinguish between positive and negative values of x - 2. We find that (x - 2) = +_ sqrt(16) = +- 4. (x-2) = 4 gives us the solution x = 6 and (x-2) = -4 gives us the solution x = -2. **
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18:57:46 Query problem 1.1.34 percent increase in Dow What are the requested percent increases?
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18:58:00 ** Aug 94 to Nov 94 Change in value = 3900 - 3800 = 100 At a percent of the initial value we have 100/3800 = 2.6% increase Dec 95 to June 96: change in value = 5600-5300 = 300 As percent of initial value: 300/5300 = 5.5% approx.. **
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шL߽Ţ̭躌 assignment #007 ݺӯD Applied Calculus I 06-23-2006
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20:30:03 **** Query class notes #07 **** Explain 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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RESPONSE --> finding the derivative. the rate is instantaneous so a small change in x can be used to express a small change in y
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20:30:33 ** 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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20:32:35 Query class notes #08 **** What 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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RESPONSE --> dT/dt=k(T-20)
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20:32:48 ** y proportional to x means that for some k we have y = k x. The rate of change of the temperature is the derivative dT/dt. The difference between temp and room temp is T 20. So the statement says that dT/dt = k (T 20). Whenever the rate dT/dt is proportional to a quantity like T - Troom, which is a linear function of T, the result is that T-Troom is an exponential function of clock time. **
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20:37:22 Query problem 1.2.08 graph matching y = `sqrt(9-x^2) Describe the graph that matches this function and explain how you know this is the graph
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RESPONSE --> inverted parabola
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20:38:12 ** It turns out that this graph is in fact the upper half of a circle of radius 3 centered at the origin. We can show that this graph is part of a circle: If y = `sqrt(9 - x^2) then y^2 = 9 - x^2 so x^2 + y^2 = 9. This is the Pythagorean Theorem for a right triangle defined by center (0,0) and legs x and y; we see that the square of the hypotenuse is 9 so the hypotenuse is 3. The hypotenuse represents the radius of the circle. **
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RESPONSE --> ***********missed
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20:47:18 query problem 1.2.10 air freshener init 30 grams, evap at 2 g/day or 12%/day What is the formula if 2 grams evaporate per day?
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RESPONSE --> y=30-(2x)
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20:47:47 ** If 12% evaporates per day then 88% remains at the end of each day. That is, the growth rate is -.12 so the growth factor would be 1 + (-.12) = .88 and the function would be Q = 30 gram * .88^t. **
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RESPONSE --> **********review
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20:50:37 query problem 1.2.18 formula for exponential function through left (1,6) and (2,18) what is the formula for the function?
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20:52:17 ** STUDENT ANSWER AND INSTRUCTOR COMMENT: I did not use simultaneous equation to solve. I just tried different number for the given original and the number which would be raised to the x power. I then plugged in the two points and found an equation. }INSTRUCTOR COMMENT: Trial and error might work for this problem but only simultaneous equations will work if the numbers are less obvious, so you need to understand that procedure. Using P = P0 * a^t we plug in the coordinates of the first point to get 6 = P0 * a^1. For the second point we get 18 = P0 * a^2. Dividing the second equation by the first we obtain 18/6 = (P0 a^2) / (P0 a^1) or 3 = a, so we know that a = 3. Substituting this into the first equation we find that 6 = P0 * 3^1, which we easily solve for P0 to obtain P0 = 2. So our model P = P0 a^t becomes P = 2 * 3^t. **
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RESPONSE --> ***********remember
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20:54:28 Query problem 1.2.28 graph of y=`sqrt(x+1) Describe your graph, including coordinates of intercepts, whether increasing or decreasing (if both, where it does each), and concavity.
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20:54:36 ** This graph intercepts the x axis where y = 0, which occurs when x+1 = 0 or x = -1. As x increases the square roots increase, but more and more slowly (just consider the square roots for x = 0, 1, 2, 3 and you'll see how the values increase by less and less each time). So the graph will be increasing at a decreasing rate, which means it is concave downward. The square root of a negative number is not a real number, so this function is undefined when x + 1 < 0, which happens when x < -1. So the function is undefined, and there is no graph at all, for x < -1. **
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21:02:00 Query problem 1.2.52 pts of intersection of x+y=7 and 3x-2y=11 What are the point(s) of intersection?
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RESPONSE --> y=-x+7 2y=3x-11 y=3x/2-11/2 x+7=3x/2-11/2 x-3x/2= -11/2-7 -x/2=-25/2 2x=25 x=12.5 y= -5.5 y= -1.33333
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21:02:26 ** This system could be solved by elimination but that solution is confined to linear equations (for which it is very appropriate) and won't be demonstrated here. The methods used here can be used with nonlinear equations. We can solve both equations for y and then set the two results equal: The first equation is x + y = 7. Subtract x from both sides to get y = 7 - x. The second equation is 3x - 2y = 11. Subtract 3x from both sides: -2y = 11 - 3 x. Divide both sides by -2: y = (11 - 3x) / (-2) so y = -11/2 + 3 x / 2. Now set both expressions for y equal to one another: 7 - x = -11/2 + 3 x / 2. Add x and 11 /2 to both sides: 7 - x + x + 11/2 = -11/2 + 3 x / 2 + x + 11/2. 7 + 11/2 = 3 x / 2 + x. Put each side over common denominator: 14 / 2 + 11 / 2 = 3 x / 2 + 2 x / 2. Add: 25 / 2 = 5 x / 2. Multiply both sides by 2/5: 2/5 * 25 / 2 = 2/5 * 5x / 2. Simplify 5 = x. So at the point of intersection x = 5. Thus, substituting this result into the first equation, y = 7 - x = 7 - 5 = 2. Alternatively we could have substituted into the second equation to get y = -11/2 + 3 x / 2 = -11/ 2 + 3 * 5 / 2 = -11/2 + 16 / 2 = 4 / 2 = 2. We get the same y value either way, which must be the case at a point of intersection. So the intersection point is at x = 5, y = 2, i.e., the point (5, 2). ALTERNATIVE SOLUTION: Solve first equation for y then substitute into the second. You could have solved your first equation for y, obtaining y = 7 - x. Substituting into the second equation you would have obtained 3x - 2(7-x) = 11 so 5x - 14 = 11 and x = 5. Then substituting this value in either equation would have given you y = 2. **
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21:03:32 **** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I still need a lot of practice. Time and apparently age does not allow me to absorb all of the data.
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F埈b༙̟Z Student Name: assignment #007 007. Depth functions and rate functions.
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19:02:58 Note 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 --> y=.05(20)^2-6(20)+100 = 0 y=.05(30)^2-6(30)+100 = -35 0- -35= 35
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19:03:01 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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19:08:29 `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 --> y'=.1(30)-6 =-3 y'= -3x 10 =-30 y=.05(30)^2 -6(30) +100 =-35 y=.05(40)^2 -6(40) +100 =-160 -35- -160 =125
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19:08:58 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.
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RESPONSE --> wrong
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19:11:40 `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 --> y'=.1(30)-6 =-3 y'=.1(40)-6= -2
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19:12:05 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.
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19:13:09 `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 --> due to the 40cm depth variance in c
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19:13:18 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.
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19:15:44 `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 --> a=.05, b= -6, c=100 y'=.1t-6
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19:15:53 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
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19:16:23 `q006. For the function y = .05 t^2 - 6 t + 30, what is the function y ' ?
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RESPONSE --> y'=.1t-6
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19:16:33 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.
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19:18:07 `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 --> y'=.1t-6 y=.5(.1)t^2 -6t
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19:18:23 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.
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19:18:45 `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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19:19:05 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.
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19:19:36 `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 --> their a and b is the same
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19:19:46 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.
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~YͩC}֓ Student Name: assignment #008 008. Approximate depth graph from the rate function
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21:06:54 `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 --> t y' 0 -6 10 -5 30 -3 60 0 100 4
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21:07:00 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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21:15:53 `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 --> y= 1/2(.1)t^2 -6t+100 y=.05t^2-6t+100 y(10)=50 y(30)=-35 y(50)= -75 y(70)= -75 y(90)= -35 y(110)= 45
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21:16:04 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.
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21:20:27 `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 --> 100-y/10-0= -6 10-y/10= -6 -y/10=-16 y=160 100-160=-60
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21:20:52 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).
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RESPONSE --> *********review
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21:22:29 `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 --> .1*20 -6 -4
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21:22:35 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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21:22:43 `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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21:22:45 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).
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ڐGO assignment #008 ݺӯD Applied Calculus I 06-24-2006
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06:47:21 What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?
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RESPONSE --> g(t)=(3t-5)^1 f(z)=2^(3t-5)
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06:47:41 CORRECT STUDENT RESPONSE: f(z)=2^z and g(t)=3t-5, so that f(g(t)) = 2^g(t) = 2^(3t-5).
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06:47:49 19:11:56
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06:48:28 prob 1.3.66 temperature conversion
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06:49:43 What linear equation relates Celsius to Fahrenheit?
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RESPONSE --> c=5/9f-160/9
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06:49:56 CORRECT STUDENT RESPONSE: degrees Fahrenheit=1.8(degrees Celsius)+32, or F = 1.8 C + 32. INSTRUCTOR COMMENT: Since each Fahrenheit degree is 1.8 Celsius degrees a graph of F vs. C will have slope 1.8. Since F = 32 when C = 0 the graph will have vertical intercept at (0, 32) so the y = m x + b form will be F = 1.8 C + 32.
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06:54:38 How did you use the boiling and freezing point temperatures to get your relationship?
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RESPONSE --> (212-32)/(100-0)=1.8 f=1.8c+b 32= (1.8x0)+b b=32 f=1.8c+32
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06:54:42 ** A graph of Fahrenheit vs. Celsius temperatures gives us the two (x,y) points (o,32) and (100,212). We use these two points to find the slope m=(y2-y1)/(x2-x1) = (212 - 32) / 100 = 1.8. Now we insert the coordinates of the point (0,32) and into the point-slope form y = 1.8 x + b of a line to get 32 = 1.8 * 0 + b. We easily solve to get b = 32. {So the equation is y = 1.8 x + 32, or F = 1.8 C + 32. **
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