course Mth 173 ????????x?assignment #005
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11:35:53 `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 --> this curve shows that it is constantley changing. and the rate of depth change is the slope of the line confidence assessment: 2
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11:36:03 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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RESPONSE --> ok self critique assessment: 2
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11:39:50 `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 --> first the derivative must be taken of the model which it is .02 t -2 and when the t is plugged in for all of the times the depths are -1.8, -1.2, -.2 confidence assessment: 2
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11:40:49 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 --> i see that the derivative is not even needed to find the depth of all the times. self critique assessment: 3
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11:47:08 `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 --> there is a difference of -17 cm between clock times 10 20. so the equation set up is change in depth/ change in clock time so its -17 cm/ 10sec = -1.7cm/s. and the same thing applies for the change in clock time between 40 and 90 sec and it yields -1.26cm/s confidence assessment: 3
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11:47:38 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 --> ok self critique assessment: 3
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11:54:34 `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 --> first i found the depths at t 10 and t 11 y(10)= .01(10^2)-2(10)+90 and the depth is 71 and the depth at t =11 is 69.21 cm. then i set the change in depths over the change in times so the equation was (69.21 - 71) cm/ [ (11 - 10) sec ] = -1.79 cm/s. and the avg. rate between clock time t=10 and t=10.1 is -1.79 cm/s confidence assessment: 3
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11:55:02 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=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 --> ok self critique assessment: 3
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11:57:15 `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?
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RESPONSE --> i really dont know but i do know that the intervals starting from time slowly decrease. confidence assessment: 1
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11:57:56 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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RESPONSE --> ok i see that the two coresspond with each other. self critique assessment: 2
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12:01:22 `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 equation is y(t1) = .01(t1)^2-(t1)+90 amd for t=t1+'dt the equation is y(t1+`dt) =.01(t1+`dt)^2-2(t1+`dt)+90. confidence assessment: 3
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12:01:49 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 --> ok self critique assessment: 3
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12:03:31 `q007. What is the change in depth between these clock times?
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RESPONSE --> the change in depth is not clear to me confidence assessment: 0
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12:04:10 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 --> i still dont understand where bothe times were plugged in at self critique assessment: 1
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12:06:29 `q008. What is the average rate at which depth changes between these clock time?
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RESPONSE --> the average rate is ( .02t1`dt+-2`dt+.01`dt^2 ) / `dt =.02t1-2+.01`dt. confidence assessment: 1
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12:07:19 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 --> is the change in clock time the derivative of the equation? self critique assessment: 1
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12:09:55 `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 --> the value of t1= 10 is -1.8 when plugged in as .02 * 10 -2= -1.8 confidence assessment: 3
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12:10:00 `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 --> confidence assessment: 2
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12:10:24 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 --> ok self critique assessment: 3
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