Assignment 18

course Mth 272

ᗳSПV|Ѓtassignment #018

018.

Applied Calculus II

03-15-2009

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01:01:19

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 -->

partial fractions can be used to find this

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

A(500-x) + B(x+1) =1 substitute 500 and -1 for x

B(500+1) = 1 = 1/501 = B

A(500 + 1) = 1/501= A

1/501~ (x+1) +(500-x)

t = 5010*(1/501[ln(x+1) + ln(500-x)]) now x= 375 (75% of pop)

t= 5010 * (1/501[[ln(375+1) + ln(500-375)])

t= 35.907

when t= 100

100= 5010*(1/501[ln(x+1) + ln(500-x)])

1045= x

confidence assessment: 3

Nearly right but I believe you neglected to solve for your integration constant c, which must fit the given condition x = 1 when t = 0.

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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01:01:24

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

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RESPONSE -->

answered

confidence assessment: 3

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01:01:30

What integral did you evaluate to obtain your result?

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RESPONSE -->

answered

confidence assessment: 3

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01:01:35

How many people are infected after 100 hours?

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RESPONSE -->

answered

confidence assessment: 3

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01:02:16

What equation did you solve to obtain your result?

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RESPONSE -->

I divided 375/35.907 = 10.44 then * 100 = 1045

confidence assessment: 3

t = 10 ln( (x+1) / (500 - x) ) + 55.2. To find x we solve this equation for x:

10 ln( (x+1) / (500 - x) ) = t - 55.2 so

ln( (x+1) / (500 - x) ) = (t - 55.2 ) / 10. Exponentiating both sides we have

(x+1) / (500 - x) = e^( (t-55.2) / 10 ). Multiplying both sides by 500 - x we have

x + 1 = e^( (t-55.2) / 10 ) ( 500 - x) so

x + 1 = 500 e^( (t-55.2) / 10 ) - x e^( (t-55.2) / 10 ). Rearranging we have

x + x e^( (t-55.2) / 10 ) = 500 e^( (t-55.2) / 10 ) - 1. Factoring the left-hand side

x ( 1 + e^( (t-55.2) / 10 ) ) = 500 e^( (t-55.2) / 10 ) - 1 so that

x = (500 e^( (t-55.2) / 10 ) - 1) / ( 1 + e^( (t-55.2) / 10 ) ).

Plugging t = 100 into this expression we actually get x = 494.4, approx.

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01:02:57

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

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RESPONSE -->

The book did not have any examples like this one, so it was difficult to figure out.

confidence assessment: 3

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&#Good responses. See my notes and let me know if you have questions. &#