course Mth 173 You are probably going to think that I am a complete idiot. But I did not realize to write down all the equations for the next questions. ??????????assignment #005005.
......!!!!!!!!...................................
08:05:17 `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?
......!!!!!!!!...................................
RESPONSE --> A constantly changing rate of depth because the steepness of the curve is constantly changing. confidence assessment: 1
.................................................
......!!!!!!!!...................................
08:05:35 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
08:09:43 `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?
......!!!!!!!!...................................
RESPONSE --> y(t) = .01(t^2) - 2(t) + 90 y = .01(10^2) - 2(10) + 90 y = .01(100) - 2(10) + 90 y = 1. - 20 + 90 y = 71 y(t) = .01(t^2) - 2(t) + 90 y = .01(40^2) - 2(40) + 90 y = .01(1600) - 2(40) + 90 y = 16 - 80 + 90 y = 26 y(t) = .01(t^2) - 2(t) + 90 y = .01(90^2) - 2(90) + 90 y = .01(8100) - 2(90) + 90 y = 81 - 180 + 90 y = -9 confidence assessment: 3
.................................................
......!!!!!!!!...................................
08:10:21 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.
......!!!!!!!!...................................
RESPONSE --> I did not figure for 20 but for 40. I thought that is what the question asked for. self critique assessment: 1
.................................................
......!!!!!!!!...................................
08:15:49 `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?
......!!!!!!!!...................................
RESPONSE --> y = a(10^2) + b(10) + c y = a(100) + b(10) + c y = a(11^2) + b(11) + c y = a(121) + b(11) + c y = a(10.1^2) + b (10.1) + c y = a(102.01) + b(10.1) + c I don't have enough information to finish the question. I need numbers for a, b, and c. confidence assessment: 1
.................................................
......!!!!!!!!...................................
08:16:38 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.
......!!!!!!!!...................................
RESPONSE --> I did not realize to use the numbers for a, b, and c from before. self critique assessment: 2
.................................................
......!!!!!!!!...................................
08:21:00 `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?
......!!!!!!!!...................................
RESPONSE --> y'(10) confidence assessment: 1
.................................................
......!!!!!!!!...................................
08:28:14 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.
......!!!!!!!!...................................
RESPONSE --> I don't know where the progression numbers came from. self critique assessment: 1
.................................................
......!!!!!!!!...................................
08:30:17 `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?
......!!!!!!!!...................................
RESPONSE --> y = a(t1^2) + b(t1) + c y = a(t1+'dt^2) + b(t1+'dt) + c confidence assessment: 0
.................................................
......!!!!!!!!...................................
08:31:09 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.
......!!!!!!!!...................................
RESPONSE --> I am being really dumb here. Again I did not realize to use the previous equation. I really do know how to do this. self critique assessment: 3
.................................................
......!!!!!!!!...................................
08:32:33 `q007. What is the change in depth between these clock times?
......!!!!!!!!...................................
RESPONSE --> You must take the two answers from the previous question and subtract them in order to get the change in depth between these two. confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:33:00 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.
......!!!!!!!!...................................
RESPONSE --> I can not remember the equation from box to the next box. self critique assessment: 3
.................................................
......!!!!!!!!...................................
08:33:46 `q008. What is the average rate at which depth changes between these clock time?
......!!!!!!!!...................................
RESPONSE --> You would use the equation 'dy/'dt to get the average. confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:35:02 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
08:36:32 `q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results?
......!!!!!!!!...................................
RESPONSE --> y = .02(10) - 2 y = .2 - 2 y = -1.8 confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:36:42 At t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
"