Assignment 18

course Mth 272

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16:55:15

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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Im not sure if this is the right way to go about this problem. I dont understand what to do with the x = 1 and t = 0.

5010 int. 1/(x+1)(500-x) = 5010(A/(500-x)) + b/(x+1))

1/(x+1)(500-x) = A(x+1) + b(500-x)

when x = -1 B = 1/501

when x = 500, A = 1/501

t = 5010(1/501 ln(500-x) + 1/501 ln(x+1)

t = 10 ln(500-x) + 10 ln(x+1)

What do you do next if this is the right way?

confidence assessment: 0

Good so far. Now substitute the known information. The entire solution:

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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16:55:22

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

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16:55:24

What integral did you evaluate to obtain your result?

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16:55:26

How many people are infected after 100 hours?

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16:55:28

What equation did you solve to obtain your result?

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16:55:31

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

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confidence assessment:

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All you need to do is substitute the given conditions to evaluate your integration constants. See my notes. I think you'll be fine.

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Let me know if you have questions. &#