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course mth 272

ĺbѰwassignment #018

018.

Applied Calculus II

07-20-2008

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15:28:47

Query problem 6.3.54 time for disease to spread to x individuals is 5010 integral(1/[ (x+1)(500-x) ]; x = 1 at t = 0.

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RESPONSE -->

alright, for this problem. I'm not sure what to do. Do we integrate this according to a logistical growth function or just integrate it regularly?

confidence assessment: 0

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15:31:02

How long does it take for 75 percent of the population to become infected?

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RESPONSE -->

I think once you find your function from previous equation, you plug in the equivalent of 75% of 500 as x and solve for t.

But I was hoping the answer would be shown to me from the last one, because then i'm pretty confident I could figure out how long it takes for 75 percent to become infected.

confidence assessment: 1

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15:38:31

What integral did you evaluate to obtain your result?

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RESPONSE -->

Again, im not sure. BUT, I will make an attempt here.

=5010 integral of ((ln(x+1)) +((ln(500-y))

=5010 * ln((x+1) / (500-y))

confidence assessment: 1

Good work on the integral. Compare with the technique used in the following, which also uses the result to answer the original question:

1 / ( (x+1)(500-x) ) = A / (x+1) + B / (500-x) so

A(500-x) + B(x+1) = 1 so

A = 1 / 501 and B = 1/501.

The integrand becomes 5010 [ 1 / (501 (x+1) ) + 1 / (501 (500 - x)) ] = 10 [ 1/(x+1) + 1 / (500 - x) ].

Thus we have

t = INT(5010 ( 1 / (x+1) + 1 / (500 - x) ) , x).

Since an antiderivative of 1/(x+1) is ln | x + 1 | and an antiderivative of 1 / (500 - x) is - ln | 500 - x | we obtain

t = 10 [ ln (x+1) - ln (500-x) + c].

(for example INT (1 / (500 - x) dx) is found by first substituting u = 1 - x, giving du = -dx. The integral then becomes INT(-1/u du), giving us - ln | u | = - ln | 500 - x | ).

We are given that x = 1 yields t = 0. Substituting x = 1 into the expression t = 10 [ ln (x+1) - ln (500-x) + c], and setting the result equal to 0, we get c = 5.52, appxox.

So t = 10 [ ln (x+1) - ln (500-x) + 5.52] = 10 ln( (x+1) / (500 - x) ) + 55.2.

75% of the population of 500 is 375. Setting x = 375 we get

t = 10 ln( (375+1) / (500 - 375) ) + 55.2 = 66.2.

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15:42:19

How many people are infected after 100 hours?

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RESPONSE -->

now from looking in the book, I'm pretty sure you DO want a logistic growth model function

possiblly like y = 5010 / ( x ) ^( x )t

and then you would simply plug in 100 for t and get your answer.

confidence assessment: 1

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15:43:56

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

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RESPONSE -->

i cannot solve for the right equation here. just wondering-- where are the critiques that usually follow the student response. Those are usually a good way for me to learn.

confidence assessment:

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Email me tomorrow on the status of the Query program. I'm in the middle of Monday crunch time (when most people submit their work) and need to stay on task with responses, but I need to look at that and I'll be glad to let you know.

In the meantime you had a good start here and I think you'll understand the given solution. You're always welcome to submit additional questions about any of my responses.