Assignment 18

course MTH272

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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: A/x+1 + b/500-x

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

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

B=1/501

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

A = 1/501

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

INT 5010[1/(501(x+1))+ 1 /(501(500-x))]

INT 10[1/x+1 + 1/500-x]

10[ln(x+1)-ln(500-x)+c

0=10[ln2-ln499]+c

c= 5.519

t=10[ln(x+1)-ln(500-x)+5.519]

confidence rating #$&*3

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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):ok

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Self-critique rating #$&*3

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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: 500*.75= 375

t=10[ln376/125 +55.2 = 66.2

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):ok

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Self-critique rating #$&*3

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

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Self-critique (if necessary):is it the one we used at the top ?

right

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Self-critique rating #$&*3

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

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Your solution: 100 = 10ln ((x+1)/(500-x) + 55.2

100-55.2= 10ln((x+1)/500-x)

44.8=10ln((x+1)/(500-x)

4.48=ln((x+1)/(500-x)

e^4.48=x+1/500-x

(500-x)(e^4.48)=x+1

(500-x)(e^4.48) - 1 = x

500(e^4.48)- e^4.48x - 1 = x

500(88.2346)-88.2346x - 1

88.2346x-x=1-44117.33

x(88.2346-1)=44116.33

x=44116.33/88.2345

x=500

confidence rating #$&*2

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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.

Self critique (if necessary):hey pretty close

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Self-critique rating #$&*2

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

&#Very good responses. Let me know if you have questions. &#