Assignment 18

course Mth 272

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018.

Applied Calculus II

10-26-2008

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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, in a population of 500.

22:52:04

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

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RESPONSE -->

25,948.368 hours

confidence assessment: 1

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23:01:44

What integral did you evaluate to obtain your result?

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RESPONSE -->

(0.75)(5010) integral ln(x+1) ln(500-x)

= 3757.5 integral [ln(x) + ln(1)]*[ln(500)-ln(x)]

= 3757.5 integral 6.21*[ln (x)]^2

confidence assessment: 2

** Right setup, but you can't integrate that way. The integral of f * g is not equal to the integral of f multiplied by the integral of g.

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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23:06:42

How many people are infected after 100 hours?

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RESPONSE -->

100= 6.215-[ln(x)^2]

93.785= ln(x^2)

1.97= x^2

x= 1.4 people affected after 100 hours

confidence assessment: 0

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23:06:53

What equation did you solve to obtain your result?

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RESPONSE -->

see previous answer

confidence assessment: 3

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23:07:32

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

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RESPONSE -->

I think I may have the principles of the problem correct, but where I mess up is my algebra. If I went wrong, please attempt to tell me where. Thanks.

confidence assessment: 3

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Assignment 18

You didn't perform the integration correctly. Be sure you understand the use of partial fractions on integrals of this nature.

&#Let me know if you have questions. &#