Assignment 18

course Mth 272

7/28 10:30 am

018.

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Question: `qQuery 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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Your solution: To solve this using partial fractions and find the time, we can first break the denominator of the fraction down into its components, which are (x+1) and (500-x). So we can rewrite our expression as 1 = A/(x+1) + B/(500-x) and then 1 = (500-x)A + (x+1)B. Using algebra, we find that both A and B equal 1/501. Our new expression is 1/501 (500-x) + 1/501(x+1). When we take the integral of this according to our derivative rules, we end up with 1/501 ln(abs(500-x)) + 1/501ln(abs(x+1)) + C. When we multiply this by our original coefficient of 5010, our final answer becomes -10ln(abs(500-x)) + 10ln(abs(x+1)) + C. Finally to find the time from [0,1], we use can plug in 1 and 0 to our expression. So (-10ln(abs(500-1)) + 10ln(abs(1+1)) + C = 0, and we find that C = 5.52. So the final equation becomes t = -10ln(abs(500-x) + 10(ln(abs(1 +x)) + 5.52.

confidence rating: 2

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Given Solution:

`a 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.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qHow long does it take for 75 percent of the population to become infected?

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Your solution: To find out how long it would take 75 % of the population to become infected, we need to use the formula from the earlier question, t = 10 ln( (x+1) / (500 - x) ) + 55.2. Since the total population is 500, 75% of that is 375. Now we can simply plug 375 into the formula. So t= 10 ln( (375+1) / (500 -375) ) + 55.2 = 66.213.

confidence rating: 3

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Given Solution:

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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat integral did you evaluate to obtain your result?

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Self-critique (if necessary): To obtain the result, we needed to evaluate the integral of the original equation, is 5010 integral(1/[ (x+1)(500-x) ].

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Self-critique Rating:

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Question: `qHow many people are infected after 100 hours?

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Your solution: Since we are given a number for t, we can simply plug in 100 for t in our expression and solve. So 100 = 10 [ ln (x+1) - ln (500-x) + 5.52]. This becomes 10 = ln(x+1) – ln(500-x) +5.52, which simplifies to 4.48 = ln(x+1) –ln (500 – x). When we apply e to both sides, we end up with e ^4.48 = (x+1) – (500-x), and then e^4.48 = x -401. Using algebra, we find that x = 494.

confidence rating: 3

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Given Solution:

`a 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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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.

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