Assignment 18

course mth 272

If your solution to a stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

018.

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Question: `qQuery problem 6.2.54 (7th edition 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(500-x) + B(x+1) = 1

500A – Ax +Bx +B = 1

500A + B + (B – A)x = 1

A = 1 / 501 and B = 1/501.

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

10 [ 1/(x+1) + 1 / (500 - x) ].

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

Plug in 0 and 1 and you get a c of 5.52

t = 10 (ln(x+1) - ln (500-x) + 5.52)

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

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Your solution:

500 times .75 = 375

t = 10 (ln(375+1) - ln (500-375) + 5.52)

66.213

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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):

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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:

100 = 10 (ln(x+1) - ln (500-x) + 5.52)

= 494.386

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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):

I just plug this into the calculator and solved if I would have taken the time I would have came up with the equation that you got.

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Self-critique rating #$&*3

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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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&#Your work looks good. Let me know if you have any questions. &#

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