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course 272 11/19 8:30pm 018..............................................
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): ok ------------------------------------------------ Self-critique Rating: 3
********************************************* Question: `qHow long does it take for 75 percent of the population to become infected? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 75% of 500 people = 375 so x = 375 t = 10 (ln |375+1| / |500-375| + 55.2 = 66.2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: 75% of the population of 500 is 375. Setting x = 375 we get t = 10 ln( (375+1) / (500 - 375) ) + 55.2 = 66.2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3
********************************************* Question: `qWhat integral did you evaluate to obtain your result? 5010 int(1/(x+1) + int(1/(500-x) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow many people are infected after 100 hours? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t = 10 (ln |x+1| / |500-x| + 55.2 t - 55.2 = 10 (ln |x+1| / |500-x|) (t-55.2 / 10) = (ln |x+1| / |500-x|) e^(t-55.2 / 10) = (x+1) / (500-x) 500 e^(t-55.2 / 10) - x e^(t-55.2 / 10) = x + 1 or x + x e^(t-55.2 / 10)= 500 e^(t-55.2 / 10) - 1 x (1+ e^(t-55.2 / 10) = 500 e^(t-55.2 / 10) - 1 x = 500 e^(t-55.2 / 10) - 1/ (1+ e^(t-55.2 / 10) now we can plug in 100 for t x = 500 e^(100-55.2 / 10) - 1/ (1+ e^(100-55.2 / 10) x = 494.4 approx confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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): ok ------------------------------------------------ Self-critique Rating: 3
********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. Nothing too surprising, feeling comfortable with this. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. Nothing too surprising, feeling comfortable with this.#$&* course 272 11/19 8:30pm 018.
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Given Solution: `a 1 / ( (x+1)(500-x) ) = A / (x+1) + B / (500-x) so A(500-x) + B(x+1) = 1 so A = 1 / 501 and B = 1/501. The integrand becomes 5010 [ 1 / (501 (x+1) ) + 1 / (501 (500 - x)) ] = 10 [ 1/(x+1) + 1 / (500 - x) ]. Thus we have t = INT(5010 ( 1 / (x+1) + 1 / (500 - x) ) , x). Since an antiderivative of 1/(x+1) is ln | x + 1 | and an antiderivative of 1 / (500 - x) is - ln | 500 - x | we obtain t = 10 [ ln (x+1) - ln (500-x) + c]. (for example INT (1 / (500 - x) dx) is found by first substituting u = 1 - x, giving du = -dx. The integral then becomes INT(-1/u du), giving us - ln | u | = - ln | 500 - x | ). We are given that x = 1 yields t = 0. Substituting x = 1 into the expression t = 10 [ ln (x+1) - ln (500-x) + c], and setting the result equal to 0, we get c = 5.52, appxox. So t = 10 [ ln (x+1) - ln (500-x) + 5.52] = 10 ln( (x+1) / (500 - x) ) + 55.2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow long does it take for 75 percent of the population to become infected? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 75% of 500 people = 375 so x = 375 t = 10 (ln |375+1| / |500-375| + 55.2 = 66.2 confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat integral did you evaluate to obtain your result? 5010 int(1/(x+1) + int(1/(500-x) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow many people are infected after 100 hours? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t = 10 (ln |x+1| / |500-x| + 55.2 t - 55.2 = 10 (ln |x+1| / |500-x|) (t-55.2 / 10) = (ln |x+1| / |500-x|) e^(t-55.2 / 10) = (x+1) / (500-x) 500 e^(t-55.2 / 10) - x e^(t-55.2 / 10) = x + 1 or x + x e^(t-55.2 / 10)= 500 e^(t-55.2 / 10) - 1 x (1+ e^(t-55.2 / 10) = 500 e^(t-55.2 / 10) - 1 x = 500 e^(t-55.2 / 10) - 1/ (1+ e^(t-55.2 / 10) now we can plug in 100 for t x = 500 e^(100-55.2 / 10) - 1/ (1+ e^(100-55.2 / 10) x = 494.4 approx 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. Nothing too surprising, feeling comfortable with this. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. Nothing too surprising, feeling comfortable with this.