Assignment 16

course Mth 272

Ү\|nXassignment #016

016.

Applied Calculus II

03-15-2009

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18:08:49

Query problem 6.2.50 (was 6.2.48) solid of revolution y = x e^x x = 0 to 1 about x axis

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RESPONSE -->

first use the solid of revolution formula

pi from 0 to 1~ [xe^x]^2 dx

pi~ (x^2e^2x) dx

u= x^2

du= 2xdx

dv=e^2xdx

v= 2e^2x

v would be 1/2 e^(2x)

x^2*2e^2x- 2~ e^2x*2xdx then integrate the second part

x^2*2e^2x -2 [e^2x*x^2]

= x^2*2e^2x- 2e^2x*-2x^2 from 0 to 1

(x2e^2x)(x+2x)

pi[((1)2e^2(1))((1)+2(1)) - (0)] = 33.334pi

confidence assessment: 3

Good, except for a couple of easily-corrected errors when integrating e^(2x).

*&*& 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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18:09:06

What expression do you integrate to obtain the desired result?

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RESPONSE -->

answered

confidence assessment: 3

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18:09:12

What is your result?

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answered

confidence assessment: 3

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18:09:18

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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answered

confidence assessment: 3

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18:09:31

Answer the same questions for your second step.

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RESPONSE -->

answered

confidence assessment: 3

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18:28:02

Problem 6.2.58 (was 6.2.56) revenue function 410.5 t^2 e^(-t/30) + 25000

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RESPONSE -->

u= t^2

du= 2tdx

dv=e^(-t/30)dx

v= -1/3e^(-t/30)

v would be -30 e^(-t/30).

t^2*-1/3e^(-t/30) + 1/3~ e^(-t/30)2tdx

t^2*-1/3e^(-t/30) + 1/3[e^(-t/30)t^2]

t^2*-1/3e^(-t/30) + 1/3e^(-t/30)1/3t^2 + 25000t

((t^2+1/3e^(-t/30)) (-2/3 + 25000t)

when x=0 u=0

when x= 90 u=

confidence assessment:

*&*& 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

- 30e^(- t/30)(t^2 + 60t + 1800).

Our antiderivative of 410.5 t^2 e^(-t/30) + 25000 is therefore 410.5 (- 30e^(- t/30)(t^2 + 60t + 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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18:28:04

Problem 6.2.58 (was 6.2.56) revenue function 410.5 t^2 e^(-t/30) + 25000

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RESPONSE -->

confidence assessment: 3

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18:29:14

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

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RESPONSE -->

I tried this one, but i really couldn't find in the book where how to do this problem was. I found the exact problem, but not how to solve it.

confidence assessment:

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18:29:18

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

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18:29:21

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

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confidence assessment:

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20:08:35

Problem 6.2.64 (was 6.2.62) c = 5000 + 25 t e^(-t/10), r=6%, t1=10 yr, find present value

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RESPONSE -->

~0 to 10

5000 + ~25te^(-t/10)

u= 25t

du= 25dt

dv=e^(-.1t)dt

v=-10e^(-.1t)

= -250te^(-.1t) + 10 ~ 25e^(-.1t)dt

= -250te^(-.1t) + 250te^(-.1t)

at this point i got lost. i don't really understand where the 6% fits in?

confidence assessment:

At 6% continuous interest the factor by which you multiply your principal to find its value after t years is e^(.06 t). So to find the principal you need now to end up with a certain amount after t years, you divide that amount by e^(.06 t), which is the same as multiplying it by e^(-.06 t).

*&*& 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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20:08:42

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

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confidence assessment:

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20:08:48

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

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20:09:59

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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The last couple, the revenue and the present value problems were pretty difficult for me to understand. The book wasn't very much help with these problems maybe you can give me a little insight, thanks.

confidence assessment: 3

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The revenue function is pretty straightforward. The solution I appended should clarify that for you, but if not let me know.

Present and future value is more challenging. I recommend Assignment 10 in the qa program

http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/query_cl1_qa_format_thru_16_080121.exe,

which takes you step-by-step through the reasoning behind problems of this type. You can just run it for your own information, or if you wish you are welcome to submit it for my further review and comment. Either way I think it will clarify the idea.