QA 5

course MTH 271

09/18,1630

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assignment #005

005.

09-18-2009

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15:33:05

`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 smooth curve of the water depth v. clock time graph indicates a constantly changing rate of depth change. A constant rate of depth change would have a more linear graph. At a certain point on the curve, we are able to determine the rate of depth change by dividing the change in y values by the change in time (x values).

confidence rating: 3

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15:41:01

`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 depths for this system are as follows:

t=10, y=71cm

t=40, y=26cm

t=90, y=-9cm

confidence rating: 3

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15:45:23

`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 -->

Average rates of depth change:

Between points (10,71) and (40,26):

(26-71)/(40-10)=(-45)/30=-3cm/2s=-1.5cm/s

Between points (40,26) and (90,-9):

(-9-26)/(90-40)=(-35)/50=-7cm/10s=-.7cm/s

confidence rating: 3

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15:52:23

`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 -->

At t=11, y=69.21. The average rate of depth change between (10,71) and (11,69.21) is:

(69.21-71)/(11-10)=-1.79cm/s

At t=10.1, y=70.82. The average rate of depth change between (10,71) and (10.1,70.82):

(70.82-71)/(10.1-10)=(-.1799)/.1=-1.799cm/s

confidence rating: 3

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16:09:36

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

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RESPONSE -->

The precise rate can be found using y=2at + b. Using our original equation, 2a=.01, a=.005 and b=-2.

2(.005)(10)-2=-1.8

confidence rating: 2

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16:18:23

`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 -->

t1=t therefore the depth is y(t)=.01t^2 -2t + 90

The depth at t=t1+ 'dt is y(t1 + 'dt). (t1 + 'dt) is substituted into the original y(t) equation for t.

y(t1 + 'dt)=.01(t + 'dt)^2 -2(t + 'dt) +90

=.01t^2 + .02t'dt + .01'dt^2 - 2t-2'dt + 90

(Since t1=t, t is used here for simplicity)

confidence rating: 3

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16:21:31

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

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RESPONSE -->

depth changes requires subtracting the value for y(t) from the value of y(t + 'dt)

(.01t^2 + .02t'dt + .01'dt^2 -2t -2'dt + 90) - (.01t^2 - 2t + 90)

=.02t'dt + .01'dt^2 -2'dt

confidence rating: 2

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16:25:18

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

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RESPONSE -->

Finding the average rate involves taking the change in y values we found for the preceding problem (.02t'dt + .01'dt^2 -2'dt) and divide it by the change in time: (t + 'dt) - t

(.02t'dt + .01'dt^2 -2'dt)/(t + 'dt -t)

(.02t'dt + .01'dt^2 -2'dt)/('dt)

The quantity 'dt can be factored from the equation.

'dt (.02t + .01'dt -2)/'dt

The 'dt cancels from top and bottom.

.02t + .01'dt -2

confidence rating: 2

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16:26:43

`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 .02t -2 at t=10:

.02(10) -2

=-1.8

This is the same value for t=10 as the previous result.

confidence rating: 2

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