#$&* course MTH272 There was one I just couldn’t see working out could you give more details please. 007. `query 7If you had trouble with the concept and/or procedures for substitution, consider submitting all or part of the document
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Given Solution: `a If we let u = 1 - x^3 then u ' = - 3 x^2 and the x^2 in our integrand is - u ' / 3. (1-x^3)^2 is u^2, so the integrand is - u ' / 3 * u^2 = -1/3 u^3 u ' . So the integral is you have -1/3 u^2 du. The integral of u^2 u ' is 1/3 u^3. Thus the integral of -1/3 u^2 u ' is -1/3 of 1/3 u^3, or -1/9 u^3. So your integral should be -1/9 u^3 = -1/9 (1-x^3)^3. The general antiderivative is -1/9 ( 1 - x^3)^3 + c. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ************************************************************************* ********************************************* Question: `qWhat is the derivative of your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: g(x)=(1-x^3)^3/9 (-1/9*-3x^2)(3(1-x^3) ^2) (-1/9*-9x^2)(1-x^3)^2 g(x’)=x^2(1-x^3)^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The derivative of -1/9 (1-x^3)^3, using the Chain Rule, is the product of -1/9, 3(1-x^3)^2, and the derivative -3x^2 of the 'inner function' (1-x^3). Multiplying these factors we get -1/9 (-3x^2) * 3(1-x^3)^2 = x^2 (1-x^3)^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): mine looks a little different but I believe it is the same ------------------------------------------------ Self-critique Rating:3 ********************************************************************* ********************************************* Question: `q 5.2.54 (was 5.2.4) (previously 5.2.54 (was 5.2.52)) find x | dx/dp = -400/(.02p-1)^3, x=10000 when p=100 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -400/(.02p-1)^3 -400(.02p-1)^-3 rearrange the equation u=.02p-1 10000= confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The equation rearranges to dx = -400 * dp * (.02 p - 1)^-3. An antiderivative of the left-hand side could be just x. An antiderivative of dp * (.02 p - 1)^-3 is found using u = .02 p - 1, so du = .02 dp and dp = du / .02 = 50 du. Thus the right-hand side becomes -400 * 50 u^-3 du = -20000 u^-3 du, with antiderivative 20000 / 2 * u^-2 + c = 10,000 u^-2 + c. So we have x = 10,000 * u^-2 + c = 10,000 * (.02 p - 1)^-2 + c. Note that dx / dp is therefore 10,000 * -2 * .02 (p-1)^-3 = -400 (.02 p - 1)^-3, consistent with the original equation. Since x = 10,000 * (.02 p - 1)^-2 + c and x = 10,000 when p = 100 we have 10,000 = 10,000 * (.02 * 100 - 1)^2 + c 10,000 = 10,000 / 1^2 + c 10,000 = 10,000 + c so c = 0. The solution is therefore x = 10,000 * (.02 p - 1)^-2 + 0 or just x = 10,000 * (.02 p - 1)^-2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have to be honest that even after looking through your notes this does not make any sense what so ever. Could you explain with any more details than this please? ------------------------------------------------ Self-critique Rating:
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Given Solution: `a Simple substitution u = -.25 x gives us du/dx = -.25 so that du = -.25 dx and dx = du / (-.25) = -4 du. Our original integrand e^(-.25 x) dx therefore becomes e^u * (-4 du) = -4 e^u du. Our general antiderivative will be -4 e^u + c, meaning -4 e^(-.25 x) + c. The derivative of -4 e^(-.25 x) + c is -4 ( -.25 e^-.25 x) = e^-.25 x, verifying our result. The General Exponential Rule is equivalent to this: u = -.25 x so du/dx = -.25. Thus the integral is of e^u / (du/dx) = e^(-.25 x) / (-1/4) = -4 e^(-.25 x). *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ******************************************************************************* ********************************************* Question: `q 5.3.10 (was 5.3.2) (previously 5.3.10 (was 5.3.10)) integral of 3(x-4)e^(x^2-8x) by Exponential Rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3(x-4)e^(x^2-8x) u=2-8x du=x-4 Multiplying (1/2)(2-8x)=1-4x taking the antiderivative of this leaves you with 2-8x 3e^(x^2-8x)/2 placing u over 2 multiplying it 3 gives you the antiderivative confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a if u=x^2 - 8x then du / dx = 2x - 8 x-4 = 1/2(2x-8) so 3(x-4) = 3/2 du/dx. Thus 3(x-4)e^(x^2-8x) is 3/2 e^u du/dx. The general antiderivative of e^u du/dx is e^u + c, so the integral of 3/2 e^u du/dx is 3/2 e^u. Substituting x^2 - 8x for u we have 3/2 e^(x^2-8x) + c. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ******************************************************************** ********************************************* Question: `qproblem 5.3.16 (was 5.3.3) (previously 5.3.16) integral of 1/(6x-5) by Log Rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u= (6x-5) du=6 1/u dx=(1/du) ln(1/u) dx 1/(6x-5)dx=(1/6)ln1/(6x-5)dx multiplying and dividing by du (1/6)ln1/(6x-5)+c add in the constant confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a du/dx is the derivative of 6x-5, so du/dx = 6 If we let u = 6x - 5 then du = 6 dx so dx = 1/6 du and the integral becomes that of ln(u) * du/6 = 1/6 ln(u) du The integral of 1/6 ln(u) du is 1/6 * 1 / u + c Substituting u = 6x - 5 we get the final result int(1 / (6x - 5) = 1/6 * 1 / (6x-5) = 1 / [ 6(6x-5) ]. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): although it doesn’t look like yours it does works out to the derivative and it is from the information covered in the book ------------------------------------------------ Self-critique Rating:
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Given Solution: `a If we let u = x^2 + 4 we get du/dx = 2x so that the x in the numerator is 1/2 du/dx. The integral of x / (x^2 + 4) is the integral of 1/2 * ( 2x / (x^2+4) ) = 1/2 (1/u du/dx). The general antiderivative is therefore 1/2 ln(u) + c = 1/2 ln |x^2+4| + c. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: ********************************************************************************* ********************************************* Question: `qWhat is the derivative of your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ½ ln (x^2 + 4) ½*2x*1 / (x^2 + 4)= confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The derivative of ln(x^2+4) * (1/2) is 1/2 * 2x * 1 / (x^2 + 4) or x / (x^2 + 4). This confirms that ln(x^2+4) * (1/2) is a solution to the equation. The general antiderivative is of course ln(x^2+4) * (1/2) + c. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ******************************************************************************* ********************************************* Question: `q 5.3.28 (was 5.3.7) (previously 5.3.28 (was 5.3.24) (was 5.3.24) ) integral of e^x/(1+e^x) by Log Rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: e^x/(1+e^x) 1/x dx = ln|x| + c 1/1+e^x*e^x = ln|1+e^ + c using the simple log rule confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a let u = 1 + e^x. Then du/dx = e^x. We are therefore integrating 1 / (1 + e^x) * e^x, which is 1/u du/dx. The antiderivative is ln |u| + c = ln | 1 + e^x | + c. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I worked like the book instructed but I seem to get the proper antiderivative without using the absolute value. Could you explain? ------------------------------------------------ Self-critique Rating:
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Given Solution: `a Here are two detailed solutions: (6x + e^x) `sqrt( 3x^2 + e^x) = `sqrt(u) * du/dx = u^(1/2) du/dx. The antiderivative is thus 2/3 u^(3/2) = 2/3 (3x^2 + e^x)^(3/2). Alternatively If u = 3x^2 + e^x then du = 6x + e^x and we have the integral of `sqrt(u) du, which is just 2/3 u^(3/2) + c = 2/3 (3x^2 + e^x)^(3/2) + c. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):NA ------------------------------------------------ Self-critique Rating: *************************************************************** ********************************************* Question: `q 5.3.58 (previously 5.3.11) (previously 5.3.58 (was 5.3.54) (was 5.3.52) ) dP/dt = -125 e^(-t/20), t=0, P=2500 and interpretation. Give your complete solution. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -125 e^(-t/20), t=0, P=2500 and interpretation {(t) = 0} 2500e^0/20 + c = 2500 c=0 using t = 0 c=0 {(t) = 15} 2500e^-15/20 + c = 1180.92 c=1180.92 using t = -15 c=1180.92 P=2500/2= 1250 2500e^t/20+c=0 .99/2500 = e^t/20 divide both sides by 2500 ln.99/2500= lne^t/20 multiply both sides by the ln (20/1)-7.83 = (20/1)t/20 multiply both sides by 20 -156.68=t 2500e^-156.68/20=.99 156.68 days until all fish are assumed dead for the exception of the .99 fish confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a If dP/dt = -125 e^(-t/20) then dp = -125 e^(-t/20) dt. Integrating both sides we get p = 2500 e^(-t/20) + c ( to integrate the right-hand side start with u = -t / 20, etc. If p = 2500 when t = 0 we have 2500 = 2500 e^(-0/20) + c so 2500 = 2500 + c and c = 0. The final solution is thus p = 2500 e^(-t/20) After 15 days the population is p(15) = 2500 e^(-15/20) = 1000, give or take a couple hundred (you can evaluate the expression). All the trout are considered dead when the population is below 1/2. So you need to solve 1/2 = 2500 e^(-t/20) for t. Dividing both sides of this equation by 2500 then taking the natural log of both sides you get -t/20 = ln( 1/2500 ) so t = -20 * ln (1/2500) = -11 or -12 or so. Thus t is about 200 days, give or take a little. Alternative reasoning of the particular solution: If u = -t/20 then e^u du/dt = e^(-t/20) * -1/20. -125 e^(-t/20) is 2500 * ( -1/20 e^(-t/20) ) = 2500 e^u du/dx. The integral is 2500 e^u + c = 2500 e^(-t/20) + c. If t = 0, P=2500 then 2500 = 2500 e^0 + c = 2500 + c, so c = 0. Thus the particular solution is P = 2500 e^(-t/20). Alternative solution for the time when all trout are dead: 2500 e^(-t/20) < .5 means e^(-t/20) < .0002 so -t/20 < ln(.0002) so -t < ln(.0002) * 20 so -t < -170.34 and t > 170.34. The probability is that all trout are dead by day 171. STUDENT QUESTION: I couldn't figure out the time for all the trout to die because the ln 0 is undefined ** When the population falls below 1/2 of a fish it rounds off to 0 and you assume that all the trout are dead. You can think of this in terms of probability. The function doesn't really tell us the precise number but the probable number. When the probability is againt that last fish being alive we figure that it's most likely dead. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I calculated .99 because I figured anything less than one was good. ------------------------------------------------ Self-critique Rating: