QA005

#$&*

course Mth 173

6/26 10:05am

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Question: `qNote that there are 9 questions in this assignment.

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Question: `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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Your solution:

The curve represents a constant rate of depth change because the graph is a smooth curve. The second question is a little confusing.

confidence rating #$&*: Can improve

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Given Solution:

`aThe 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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Self-critique (if necessary):

I thought that since the line was considered a smooth curve that that would make the graph constant.

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A smooth curve can change its slope is any way, as long as it remains smooth.

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Self-critique Rating: 2

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Question: `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 = 20 and t = 90?

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Your solution:

y(10) = .01 * 10^2 - 2 * 10 + 90

= 1 - 20 + 90

y(10) = 71cm

y(20) = .01 * 20^2 - 2* 20 + 90

= 4 - 40 +90

y(20) = 54cm

y(90) = .01 * 90^2 - 2 * 90 + 90

= 81 - 180 + 90

y(90) = -9cm which means that the container would be empty.

confidence rating #$&*: Really Confident

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Given Solution:

`aAt 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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Self-critique (if necessary):

Would not the container be empty at y(90)? Because there cannot be a negative amount of a liquid right???

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If you assume the hole is at y = 0, then the water would stop descending at the level of the hole and would not reach -9 cm.

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@&

However this doesn't mean that a model based on three data points wouldn't predict this. Due to uncertainties in the data, such a model cannot be assumed to be accurate in all respects.

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Self-critique Rating: 3

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Question: `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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Your solution:

y(10) = 71, y(20) = 54, y(90) = -9

Interval 1:

71 - 54 = 17 / 10 = 1.7cm/s

Interval 2:

54 - -9 = 63 / 70 = .9cm/s

confidence rating #$&*: Confident

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Given Solution:

`aFrom 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 two 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 = 20 sec and t = 90 sec, so the change in clock time is 80 sec - 20 sec = 70 sec. The average rate of change between these to clock times is therefore

ave rate = change in depth / change in clock time = -63 cm / 70 sec = -.9 cm/s.

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Self-critique (if necessary):

Why are the values negative if you’re going from clock time 1 to clock time 2 and from clock time 2 to clock time 3??? Click time 1 has the most centimeters while clock time 3 has the less??

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Tha's pretty much it.

The depth is measured in the upward vertical direction, so when the depth decreases the change in depth is negative.

If you use 20 - 10 as the 'run' between the first two points, then you have to use 54 - 71 for the 'rise'.

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Self-critique Rating:

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Question: `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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Your solution:

t = 10 => t = 11

y(10) = 71

y(11) = .01 * 11^2 - 2 * 11 + 90

= 1.21 - 22 + 90

y(11) = 69.21

69.21 - 71 = -1.79/1 = -1.79cm/s

t = 10 => t = 10.1

y(10) = 71

y(10.1) = .01 * 10.1^2 - 2 * 10.1 + 90

= 1.0201 - 20.2 + 90

y(10.1) = 70.8201

70.8201 - 71 = -.1799/.1 = -1.799cm/s

confidence rating #$&*: Confident

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Given Solution:

`aAt 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?

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Your solution:

I did not really know how to solve this problem, but after looking at the solution it all makes sense. Since the change in time at .10sec is -1.799cm/s them it can be inferred that at t = 10 the rate of change is -1.8cm/s.

confidence rating #$&*:

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Not too confident. I had to look at the solution to understand the problem better.

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Given Solution:

`aThe 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.

STUDENT COMMENT:

I don't really understand this even after reading the solution

INSTRUCTOR RESPONSE:

You did some rounding in your solutions up to this point (your solutions were otherwise correct), and didn't get all the 9's in some of the numbers.

Done without rounding, the rates are -1.7 cm/s, -1.79 cm/s and -1.799 cm/s.

These represent average rates over shorter and shorter intervals starting at t = 10 sec.

It appears that these average rates are approaching a limit of -1.8 cm/s, which we therefore take to be the instantaneous rate at t = 10 sec.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `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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Your solution:

y(t1) = a t1^2 + b*t1 + c

y(t1+`dt) = (a t1^2 + b*t1 + c) * (a `dt^2 + b*`dt = c)

confidence rating #$&*:

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I was confident until I saw the solution

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Given Solution:

`aAt 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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Self-critique (if necessary):

I see where my mistake is. I did not understand that I was to plug the variables into the equation from the previous problem. I used the regular derivative formula.

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Self-critique Rating: 2

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Question: `q007. What is the change in depth between these clock times?

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Your solution:

y(a) = (t1) - (t1 + `dt)

[(.01 t1^2 - 2t1 + 90) - (.01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90)] / a

confidence rating #$&*:

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A little lost

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Given Solution:

`aThe 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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Self-critique (if necessary):

I understand a little bit more after I looked at the solution. The change in depth on the first line was factored to get the solution on the last line. I can do better on problems like these.

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Self-critique Rating: 1

@&

You just didn't simplify your expression. I believe you could have, and you will the next time.

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Question: `q008. What is the average rate at which depth changes between these clock time?

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Your solution:

.02 t1 `dt + - 2 `dt + .01 `dt^2/ `dt

= .02 t1 - 2 +.01 `dt

confidence rating #$&*:

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A little confident

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Given Solution:

`aThe 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.

STUDENT COMMENT

don’t understand how that the dt in this equation approaches 0 when .02(t1)-2

INSTRUCTOR RESPONSE

If you divide your previous result

.02 (t1 dt) + - 2 (dt) + .01 (dt^2)

by `dt you get .02 t1 - 2 + .01 * `dt.

The shorter the time interval the smaller `dt will be.

As `dt gets shorter and shorter it approaches 0. This doesn't affect the terms .02 t1 and -2, but it does affect .01 * `dt.

As `dt shrinks to zero, .01 * `dt also shrinks to 0.

The limiting value of our expression, as `dt shrinks to 0, is therefore .02 t1 - 2.

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Self-critique (if necessary):

Good explanation.

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Self-critique Rating:

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Question: `q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results?

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Your solution:

y(10) = .02 * 10 - 2

= -1.8

This answer goes with the answer of a previous question. This shows that at t = 10 the solution would be -1.8cm/s.

confidence rating #$&*: Confident

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Given Solution:

`aAt t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10.

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Self-critique (if necessary):

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#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#