course Mth 271 here are about half of what I owe and I will do some more assignments later and tomorrow. CUp~assignment #003
......!!!!!!!!...................................
08:34:23 `q002. Look at your results for the slopes, and look the results for the average rates of change. What do you notice? In what way then does the graph represent the average rate of change?
......!!!!!!!!...................................
RESPONSE --> I wasnt sure how to graph on this program confidence assessment: 0
.................................................
Xf assignment #003 003. 11-04-2008
......!!!!!!!!...................................
08:44:40 `qNote that there are four questions in this assignment. `q001. Sketch a graph similar to that you constructed for the stock values, this time for the depth of the water vs. clock time (depths 80, 40, 20 at clock times 10, 40, 90). Your first point, for example, will be (10, 80). Connect these points with straight lines and determine the slopes of the lines.
......!!!!!!!!...................................
RESPONSE --> The slopeschange between the first two points at m= - 3/4 and between the last two is m=-5/2 confidence assessment: 1
.................................................
......!!!!!!!!...................................
08:45:52 The three points are (10, 80), (40, 40) and (90, 20). From the first point to the second the rise is from 80 to 40, or -40, and the run is from 10 to 40, or 30. So the slope is -40 / 30 = -1.33. From the second point to the third the rise is from 40 to 20, or -20, and the run is from 40 to 90, or 50, so the slope is -20 / 50 = -.4. Click on 'Next Picture' to see graph.
......!!!!!!!!...................................
RESPONSE --> I just got my x's and y's mixed up somehow. self critique assessment: 1
.................................................
......!!!!!!!!...................................
08:47:52 `q002. Look at your results for the slopes, and look the results for the average rates of change. What do you notice? In what way then does the graph represent the average rate of change?
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 0
.................................................
......!!!!!!!!...................................
08:47:56 `q002. Look at your results for the slopes, and look the results for the average rates of change. What do you notice? In what way then does the graph represent the average rate of change?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
08:48:04 The slopes and the rates of change are numerically equal. For example between the second and third points the rise of -20 represents the -20 cm change in depth and the run of 50 represents the 50 seconds required to make this change, so the slope represents the -20 cm / (50 sec) average rate of change over the second time interval. We therefore see that slope represents average rate of change.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 1
.................................................
......!!!!!!!!...................................
08:50:08 `q003. To what extent do you think your graph with three points and straight line segments between them accurately depicts the detailed behavior of the water over the 80-second period of observation? How do you think the actual behavior of the system differs from that of the graph? How do you think the graph of the actual behavior of the system would differ from that of the graph you made?
......!!!!!!!!...................................
RESPONSE --> The three points cant depict it very accurately because the slope is constantly changing, as proven by the difference in slopes between the three points, so to say the slope is accurate would be inaccurate.The velocity of water changes constantly. The graph of the actual behavior would be more radical and would be in less of a line. confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:50:24 The straight line segments would indicate a constant rate of change of depth. It is fairly clear that as depth decreases, the rate of change of depth will decrease, so that the rate of change of depth will not be constant. The graph will therefore never be straight, but will be a curve which is decreasing at a decreasing rate.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
08:52:02 `q004. From the given information, do you think you can accurately infer the detailed behavior of the water depth over the 80-second period? Do you think you can infer the detailed behavior better than you could the values of the stocks? Why or why not?
......!!!!!!!!...................................
RESPONSE --> I believe that i would be a little easier to decide the rate of change for water than it would the stock values but honestly they both have a lot more that just one ting that effects their behavior so that it makes them both almost impossible to accurately depict a detailed behavior of these two things. confidence assessment: 3
.................................................
......!!!!!!!!...................................
08:52:07 It will turn out that three data points will be sufficient to infer the detailed behavior, provided the data are accurate. However you might or might not be aware of that at this point, so you could draw either conclusion. However it should be clear that the behavior of the water depth is much more predictable than the behavior of the stock market. We don't know on a given day whether the market will go up or down, but we do know that if we shoot a hole in the bottom of a full bucket the water level will decrease, and we expect that identical holes in identical buckets should result in the same depth vs. clock time behavior.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
ˉϦOѡJΆ|Yð assignment #004 004. 11-04-2008
......!!!!!!!!...................................
08:53:04 These questions duplicate those of Asst 3 so you may exit.
......!!!!!!!!...................................
RESPONSE --> Same as 3 so this asst is not necesary self critique assessment: 3
.................................................
~zjvSޒjoГ\ assignment #005 005. 11-04-2008
......!!!!!!!!...................................
08:54:36 `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 --> The curve lets us know that there will not be a constant rate of change and that the rate of change will constantly be changing and that we may be able to find the average rate of change, or the velocity. confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:54:45 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: 1
.................................................
......!!!!!!!!...................................
09:01:26 `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 --> at t=10 the depth is 87 cm at t=40 the depth is 102 cm at t=90 the depth is -9 cm confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:01:57 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 --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:02:44 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 my math wrong at t=10 and t=40 somehow but I know what I did wrong. self critique assessment: 1
.................................................
......!!!!!!!!...................................
09:04:09 `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals?
......!!!!!!!!...................................
RESPONSE --> between the first two (10, 80) & (40,40) it is -4/3 and between the (40,40) & 90, 20) it is -2/5 confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:04:13 `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
09:04:51 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.
......!!!!!!!!...................................
RESPONSE --> Oh! I thought we were doing the 10, 80 and 40, 40 and 90,20 points! self critique assessment: 1
.................................................
......!!!!!!!!...................................
09:05:09 `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 --> confidence assessment:
.................................................
......!!!!!!!!...................................
09:05:29 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 --> self critique assessment: 4
.................................................
......!!!!!!!!...................................
09:11:16 `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?
......!!!!!!!!...................................
RESPONSE --> well at t=10 the depth is 71 and that instnatneous rate of change would be figured out by the derivative of the original function. y(t)=.01t^2-2t+90 y'=.002t-2 so when t=10 y'=-1.98 confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:11:25 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 --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:13:45 `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 --> the function is y=.01t2-2t+90 if t =t1 then the answer is y=.01(t1)^2-2(t1)+90 and the derivative is y'=.002t-2 so when t =t1 y'=.002(t1)-2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:13:55 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 --> self critique assessment: 1
.................................................
......!!!!!!!!...................................
09:14:27 `q007. What is the change in depth between these clock times?
......!!!!!!!!...................................
RESPONSE --> Im not sure how to compute the change in depth between the last two points confidence assessment: 0
.................................................
......!!!!!!!!...................................
09:14:31 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 --> self critique assessment: 0
.................................................
......!!!!!!!!...................................
09:14:53 `q008. What is the average rate at which depth changes between these clock time?
......!!!!!!!!...................................
RESPONSE --> Same question as the last! confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:15:11 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 --> I wasnt sure what to do with the `dt self critique assessment: 0
.................................................
......!!!!!!!!...................................
09:16:08 `q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results?
......!!!!!!!!...................................
RESPONSE --> It is consistent with the other becasue the average rate of change is consistent. confidence assessment: 0
.................................................
......!!!!!!!!...................................
09:16:13 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: 1
.................................................
ۍGOȄƷ assignment #006 006. goin' the other way 11-04-2008
......!!!!!!!!...................................
09:18:14 `qNote 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?
......!!!!!!!!...................................
RESPONSE --> The depth would be 84 cm confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:19:17 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.
......!!!!!!!!...................................
RESPONSE --> OK i see what happened I missed the part about the -4cm being on a ONE second interval self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:21:39 `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?
......!!!!!!!!...................................
RESPONSE --> at t=30 the depth would be 40 cm I believe this would be less accurate because as you move further away rfom the original depth and time perception it becomes less likely that the rate of change will be constant. confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:21:45 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 0
.................................................
......!!!!!!!!...................................
09:23:38 `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?
......!!!!!!!!...................................
RESPONSE --> well if it is changing at -3cm per second as oppose to -4cm per second than the difference between t=21 at y=76 and t=30 would have a depth of 49 cm. confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:24:07 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.
......!!!!!!!!...................................
RESPONSE --> I thought that I could start at t=21 with a depth of 76 which is how I got confused. self critique assessment: 0
.................................................
......!!!!!!!!...................................
09:28:53 `q004. What is your specific estimate of the depth at t = 30 seconds?
......!!!!!!!!...................................
RESPONSE --> Well to find the specific depth at t =30 i would need an original position function. And if I use the original point for this (which were t=20 with a depth of 80 cm) and a slope of -4cm y=4x so at t=30 I would say the depth is 120cm confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:29:34 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.
......!!!!!!!!...................................
RESPONSE --> I thought I was using a contant rate of change of -4 cm/sec... i didnt realize that the -3cm/sec carried over. self critique assessment: 0
.................................................
......!!!!!!!!...................................
09:32:39 `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.
......!!!!!!!!...................................
RESPONSE --> y'=.1(20)-6 is y'=-4 cm/sec which proves that the rate of change is -4 and y'=.1(30)-6 is y'=-3 cm/sec which proves that the rate of change at t=30 is -3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:32:43 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:46:50 `q006. For the rate function y ' = .1 t - 6, at what clock time does the rate of depth change first equal zero?
......!!!!!!!!...................................
RESPONSE --> ok so we set the original function to 0 because that function is the position function and that is what needs to be zero. confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:47:45 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.
......!!!!!!!!...................................
RESPONSE --> I was confused a little bit. But now I see how if the position is 0 (which was my response) how the rate of chnage would also be zero. self critique assessment: 1
.................................................
......!!!!!!!!...................................
09:50:13 `q007. How much depth change is there between t = 20 sec and the time at which depth stops changing?
......!!!!!!!!...................................
RESPONSE --> well at t=20 the rate of change is -4 and at t=60 the rate of change is 0. at t=20 the depht is 76 and at t=60 we would be a 0 cm so the difference would be 76 cm. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:50:18 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
"