ass 16

course mth 272

Applied Calculus IIAsst # 16

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07-12-2006

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10:52:15

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**** Query problem 6.2.54 (was 6.2.50) solid of revolution y = x e^x x = 0 to 1 about x axis

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11:02:37

area 1

*&*& The volume over an interval `dx will be approximately equal to pi ( x e^x)^2 `dx, so we will be integrating pi x^2 e^(2x) with respect to x from x = 0 to x = 1.

Integrating just x^2 e^(2x) we let u = x^2 and dv = e^(2x) dx so that du = 2x dx and v = 1/2 e^(2x).

This gives us u v - int(v du) = 2x (1/2 e^(2x)) - int(1/2 * 2x e^(2x) ). The remaining integral is calculated by a similar method and we get a result that simplifies to

e^(2x) ( 2 x^2 - 2x + 1) / 4.

Our antiderivative is therefore pi e^(2x) (2 x^2 - 2x + 1) / 4. This antiderivative changes between x = 0 and x = 1 by pi ( e^2 / 4 - 1/4) = 5.02, approx.. *&*&

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11:02:38

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**** What expression do you integrate to obtain the desired result?

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11:04:37

type of present value

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11:04:38

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**** What is your result?

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11:04:44

area as one

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11:04:44

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**** For your first step, what was u and what was dv, what were du and v, and what integral did you obtain?

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11:04:56

i dont know.

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11:04:56

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**** Answer the same questions for your second step.

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11:05:03

same.

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11:05:03

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**** Problem 6.2.62 (was 6.2.58) revenue function 410.5 t^2 e^(-t/30) + 25000

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11:15:30

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11:15:32

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**** What expression did you evaluate to obtain the average daily receipts during the first quarter, and what was your result?

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11:15:35

&& If we integrate the revenue function from t = 0 to t = 90 we will have the first-quarter revenue. If we then divide by 90 we get the average daily revenue.

To get an antiderivative of t^2 e^(-t/30) we first substitute u = t^2 and dv = e^(-t/30), obtaining du = 2t dt and v = -30 e^(-t/30). We proceed through the rest of the steps, which are very similar to steps used in preceding problems, to get antiderivative

- 30•e^(- t/30)•(t^2 + 60•t + 1800).

Our antiderivative of 410.5 t^2 e^(-t/30) + 25000 is therefore 410.5 (- 30•e^(- t/30)•(t^2 + 60•t + 1800) ) + 25000 t.

The change in this antiderivative function between t = 0 and t = 90 is found by substitution to be about 15,000,000, representing $15,000,000 in 90 days for an average daily revenue of $15,000,000 / 90 = $167,000, approx.. &&

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11:15:35

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**** What expression did you evaluate to obtain average daily receipts during the fourth quarter and was your result?

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11:15:37

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11:15:38

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**** What expression did you evaluate to obtain the year's total daily receipts and was your result?

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11:15:41

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11:15:42

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**** Problem 6.2.68 (was 6.2.64) c = 5000 + 25 t e^(-t/10), r=6%, t1=10 yr, find present value

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11:23:21

5252

*&*& The income during a time interval `dt is ( 5000 + 25 t e^(-t/10) ) `dt. To get this amount after t years we would have to invest ( 5000 + 25 t e^(-t/10) ) `dt * e^(-.06 t).

Integrating this expression from t = 0 to t = 10 we obtain

int(( 5000 + 25 t e^(-t/10) ) * e^(-.06 t) dt , t from 0 to 10).

Our result is $38,063.

Note that the entire income stream gives us int(( 5000 + 25 t e^(-t/10) ) dt , t from 0 to 10) = $50,660 over the 10-year period.

The meaning of our solution is that an investment of $38,063 for the 10-year period at 6% continuously compounded interest would yield $50,660 at the end of the period. So $38,063 is the present value of the income stream. *&*&

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11:23:22

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**** What expression do you integrate to obtain the present value of the income function and what is your result?

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11:23:42

i plugged in the values given because rate and t were already given in this case.

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11:23:43

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**** Explain the meaning of the expression you integrated--why does this function give the present value?

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11:32:51

based on the rate and time and the time span gives you the present value

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11:32:52

"

You did not include enough details in your solutions for me to be able to tell exactly what you did. I've inserted solutions; let me know if you have any questions.