course Mth 272 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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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My final answer I did not multiply ever thing by 10 ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qHow long does it take for 75 percent of the population to become infected? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: let x = 375 for this is 75 percent of 500 people t = 10[ln|375+1| - ln|500-375| + 5.5194] t = 10[ln|376| - ln|125| + 5.5194] = 10[5.9269 – 4.828 + 5.5194] = 10[6.6180] t = 66.18 confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qWhat integral did you evaluate to obtain your result? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qHow many people are infected after 100 hours? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t = 10[ln|x+1| - ln|500-X| + 5.5194] t – 55.194 = 10lnx+1 / 500 – X (t – 55.194) / 10 = ln((x+1) /(500 – x)) e^((t – 55.194)/10) = (x+1) / (500 – x) multipy both sides by 500-x [e^((t – 55.194)/10)]*(500 – x) = x +1 confidence rating #$&*1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a t = 10 ln( (x+1) / (500 - x) ) + 55.2. To find x we solve this equation for x: 10 ln( (x+1) / (500 - x) ) = t - 55.2 so ln( (x+1) / (500 - x) ) = (t - 55.2 ) / 10. Exponentiating both sides we have (x+1) / (500 - x) = e^( (t-55.2) / 10 ). Multiplying both sides by 500 - x we have x + 1 = e^( (t-55.2) / 10 ) ( 500 - x) so x + 1 = 500 e^( (t-55.2) / 10 ) - x e^( (t-55.2) / 10 ). Rearranging we have x + x e^( (t-55.2) / 10 ) = 500 e^( (t-55.2) / 10 ) - 1. Factoring the left-hand side x ( 1 + e^( (t-55.2) / 10 ) ) = 500 e^( (t-55.2) / 10 ) - 1 so that x = (500 e^( (t-55.2) / 10 ) - 1) / ( 1 + e^( (t-55.2) / 10 ) ). Plugging t = 100 into this expression we actually get x = 494.4, approx. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Had a mental blank once I multiplied both side by (500 – x) and I looked at the answer to see how it was done. I understand the problem once I seen that. ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. "