course Mth 173 ?Vw??????????W??assignment #005005.
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09:32:52 `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 --> Since the graph of a water depth vs. clock time system is a smooth curve, this means that the rate of depth change is constantly changing. We use the derivative to find the rate of depth change at a point. confidence assessment: 2
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09:33:05 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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09:35:14 `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 --> Here we would simply plug the time into the equation since the result is the depth at that particular time: t=10: 71cm t=40: 26cm t=90: -9cm. confidence assessment: 2
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09:36:45 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 --> Ok. self critique assessment: 3
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09:38:52 `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 --> Here we use rate = rise / run, or rate = depth change / time change. t=10 to t=40: (26-71) / (40 -10) = -1.5cm/s. t=40 to t=90: (-9-26) / (90-40) = -0.7cm/s. confidence assessment: 3
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09:43:53 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 clock times here are not the same as in the preceding question, nor are the depths the same. In the question posed to me, I had the points (10, 71), (40, 26), and (90, -9). Still, the idea is the same, and I think my results are pretty close. self critique assessment: 2
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09:46:35 `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 --> Depth at t=10: 71cm. Depth at t=11: 69.21cm. Rate between t=10 and t=11: (69.21-71) / (11-10) = -1.79cm/s. Depth at t=10.1: 70.8201. Rate between t=10 and t=10.1: (70.8201-71) / (10.1-10) = -1.799cm/s. confidence assessment: 2
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09:47:25 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 --> Yep: don't round early, round late. Especially in tiny numbers. self critique assessment: 3
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09:48:08 `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?
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RESPONSE --> It seems like the rate is approaching -1.8cm/s, since each smaller measurement has been getting closer and closer to -1.8cm/s. confidence assessment: 2
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09:48:23 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. self critique assessment: 3
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10:18:17 `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 --> Depth at clock time t: y = a(t1)^2 + b(t1) + c. Depth at clock time t1+'dt: y = a(t1+'dt)^2 + b(t1+'dt) + c. confidence assessment: 2
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10:19:09 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 --> I thought it was asking for the standard form, not the specific one for this equation. I see what I missed. self critique assessment: 3
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10:20:31 `q007. What is the change in depth between these clock times?
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RESPONSE --> Depth change is y(t+dt) - y(t). To find the specific form, we have to use algebra: .01(t+dt)^2 - 2(t+dt) + 90 - (.01t^2 - 2t + 90) .01(t^2 + 2t(dt) + dt^2) - 2t - 2dt + 90 - .01t^2 + 2t - 90 .01t^2 + .02t(dt) + .01dt^2 - 2t - 2dt + 90 - .01t^2 + 2t - 90 .01dt^2 + .02t(dt) - 2dt confidence assessment: 2
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11:54:35 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 --> Ok. self critique assessment: 3
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11:56:16 `q008. What is the average rate at which depth changes between these clock time?
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RESPONSE --> The rate would be rise / run with rise being the change in depth and run being the change in clock time. We found that the change in clock time is 'dt, and the change in depth is y(t+'dt) - y(t). Therefore: average rate = (y(t+'dt) - y(t)) / 'dt. We found that y(t+'dt) - y(t) = .01dt^2 + .02t(dt) - 2dt, so we can substitute and find a more exact answer: (.01dt^2 + .02t(dt) - 2dt) / dt = .01dt + .02t - 2. As 'dt shrinks toward 0, we can safely ignore it, giving us the equation: rate = .02t - 2. confidence assessment: 2
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11:58:49 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 --> Ok. self critique assessment: 3
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12:00:16 `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 t=10, rate = -1.8cm/s. This is the same number as we previously estimated by shrinking the change in time down to smaller and smaller numbers. confidence assessment: 3
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12:00:24 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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