course Mth173 assignment #005005.
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21:06:11 `qNote that there are 9 questions in this assignment. `q001. We see that the water depth vs. clock time system likely behaves in a much more predictable detailed manner than the stock market. So we will focus for a while on this system. An accurate graph of the water depth vs. clock time will be a smooth curve. Does this curve suggest a constantly changing rate of depth change or a constant rate of depth change? What is in about the curve at a point that tells you the rate of depth change?
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RESPONSE --> The curve represents a constantly changing rate of change, a constant rate of change is a straight line. The changing slope of the curve is what gives us the hint of different rates. confidence assessment: 2
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21:06:30 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 --> right self critique assessment: 3
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21:16:57 `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 --> The line made with the model will fit the three points exactly and predeict the points within fairly accuratly, however the farther outwards of the points you go your results will not be as accurate and eventually will be downright wrong. by plugging in the times to the equation we get the points. (10, 40), (40, 26), (90, -9) The point (90, -9) is probobly outward of the points used for the model because it is impossible to have a -9cm of fluid in a container. As we see the results are skewed. confidence assessment: 2
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21:19:10 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 used the y=.01(40) - 2(40) +90 I must have misread the input for the model. self critique assessment:
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21:25:12 `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 --> By using the slope of the lines we can find the average rate of change for each segment. (71-54) / (10- 20) = -1.7cm/s (54-(-9)) / (20 - 90) = -.9cm/s confidence assessment: 2
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21:29:23 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 --> for the second point I made a calulator error, switching the (sub1) and (sub2) values. reworking the problems I got the correct -1.26cm/s instead of my original and wrong -.9cm/s self critique assessment: 2
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21:35:54 `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 --> the average rate of change between t=10 and t=11 is (69.21-71) / (11-10) = -1.79 cm/s and the average rate of change between t=10 and t=11 is (70.82 - 71) / (10.1 - 10) = -1.8 cm/s confidence assessment: 2
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21:41:04 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 --> my calculator rounded, had I used derive I probobly would have a more precise answer. self critique assessment: 2
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21:44:52 `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?
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RESPONSE --> using the universal rate equation 2(.01)t+(2) when t=10 the rate of depth change is: 2(.01)(10)+(2)= 2.20 cm/s confidence assessment: 2
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21:51:56 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 --> This is why rounding is bad at such a short point, because the -1.8 is the rounded rate between 10 and 10.1 but it is actually -1.799999 but rounded it becomes -1.8, why wouldn't the rate equation of 2ax+b work in this instance? self critique assessment: 2
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21:56:54 `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 --> The depth for t=t1 is (.01(t1)^2 - 2(t1) +90) - (.01(t1 + (2(.01)(t1)+(2)))^2 -2(t1 +(2(.01)(t1)+(2))) + 90 confidence assessment: 1
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21:58:01 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 --> 'dt would = 2(.01)t+(2), right? self critique assessment: 2
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22:01:00 `q007. What is the change in depth between these clock times?
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RESPONSE --> The change in depth would be the depth change between t1 and t1+'dt confidence assessment: 1
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22:05:34 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 --> You plugged in the (t1+'dt) into the model equation and simplified untill you reached the simplfied equation in the other window, the formula for the depth change between the times. self critique assessment: 2
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22:11:43 `q008. What is the average rate at which depth changes between these clock time?
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RESPONSE --> This would be the past equation of t1 and t1+'dt plugged into the quadratic model /((t1+'dt) - t1) confidence assessment: 1
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22:16:15 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 --> okay self critique assessment: 0
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22:20:20 `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 .02 t1 - 2 when t1=10 .02 (10) -2= -1.8 The rate of change at exactly 10s confidence assessment: 2
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22:22:59 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 --> this whole process made a rate equation for this model, the way that the universal rate equation was made, but it isn't applicable in this situation. self critique assessment: 2
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