course MTh 272 4/11/10 @ 10:00 am 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. 016. ********************************************* ********************************************* Question: `qQuery problem 6.2.50 (was 6.2.48) solid of revolution y = x e^x x = 0 to 1 about x axis YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: integral xe^x dx from 0 to 1 let: dv = e^x dx v = e^x u = x du = dx uv – fvdu = xe^x – intergral e^x dx = xe^x – e^x + c [xe^x – e^x] 0 to 1 = [1e^1 – e^1 - 0] = [2.718 – 2.718 - 0] = 0 confidence rating #$&*1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The volume over an interval `dx will be approximately equal to pi ( x e^x)^2 `dx, so we will be integrating pi x^2 e^(2x) with respect to x from x = 0 to x = 1. Integrating just x^2 e^(2x) we let u = x^2 and dv = e^(2x) dx so that du = 2x dx and v = 1/2 e^(2x). This gives us u v - int(v du) = 2x (1/2 e^(2x)) - int(1/2 * 2x e^(2x) ). The remaining integral is calculated by a similar method and we get a result that simplifies to e^(2x) ( 2 x^2 - 2x + 1) / 4. Our antiderivative is therefore pi e^(2x) (2 x^2 - 2x + 1) / 4. This antiderivative changes between x = 0 and x = 1 by pi ( e^2 / 4 - 1/4) = 5.02, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??????????do not rember doing a problem like this in homework ------------------------------------------------ Self-critique rating #$&* ********************************************* ********************************************* Question: `qProblem 6.2.58 (was 6.2.56) revenue function 410.5 t^2 e^(-t/30) + 25000 u = t^2 du = 2t dt dv = e^(-t/30) v = -30 e^(-t/30) uv – intgal v*dv = t^2 * e^(-t/30) – (-30) intgal e^(-t/30) * 2t dt = t^2*e^ * e^(-t/30) – (-30) intgal 2t *e^(-t/30) dt = t^2*e^(-t/30) – (-30) *2t * (-30) e^(-t/30) = t^2*e^(-t/30) – (-60t * (-30) e^(-t/30)
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Given Solution: 0 `a If we integrate the revenue function from t = 0 to t = 90 we will have the first-quarter revenue. If we then divide by 90 we get the average daily revenue. To get an antiderivative of t^2 e^(-t/30) we first substitute u = t^2 and dv = e^(-t/30), obtaining du = 2t dt and v = -30 e^(-t/30). We proceed through the rest of the steps, which are very similar to steps used in preceding problems, to get antiderivative - 30•e^(- t/30)•(t^2 + 60•t + 1800). Our antiderivative of 410.5 t^2 e^(-t/30) + 25000 is therefore 410.5 (- 30•e^(- t/30)•(t^2 + 60•t + 1800) ) + 25000 t. The change in this antiderivative function between t = 0 and t = 90 is found by substitution to be about 15,000,000, representing $15,000,000 in 90 days for an average daily revenue of $15,000,000 / 90 = $167,000, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ?????????? I am confusing my self. I understand the concept of intergration by parts but confusing my self with the math. I understand how to sub the antiderivative back into the equation. my problem is getting to the antiderivative. ------------------------------------------------ Self-critique rating #$&*1 ********************************************* ********************************************* Question: `qProblem 6.2.64 (was 6.2.62) c = 5000 + 25 t e^(-t/10), r=6%, t1=10 yr, find present value YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: e^(0.06t) intgal 5000 + 25te^(-t/10) dt dv = e^(-t/10) v = e^(-t/10) dt = 10 e^(-t/10) u = t du = dt uv – integral v du = 10te^(-t/10) – 10 integral e^(-t/10) dt = 10te^(-t/10) – 100e^(-t/10) Present value = 5000 + 25 integral te^(-t/10) = 5000 + 25 [ 10te^(-t/10) – 100e^(-t/10)] 0 to 10 = 5025 [10 *10e^(-10/10) – 100e^(10/10) – 0] = 5025[36.7879 – 36.7879] = 5025 / e^0.06(10) = 5025 / 1.822 = 2,757.77 confidence rating #$&*0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a At 6% continuous interest the factor by which you multiply your principal to find its value after t years is e^(.06 t). So to find the principal you need now to end up with a certain amount after t years, you divide that amount by e^(.06 t), which is the same as multiplying it by e^(-.06 t). *&*& The income during a time interval `dt is ( 5000 + 25 t e^(-t/10) ) `dt. To get this amount after t years we would have to invest ( 5000 + 25 t e^(-t/10) ) `dt * e^(-.06 t). Integrating this expression from t = 0 to t = 10 we obtain int(( 5000 + 25 t e^(-t/10) ) * e^(-.06 t) dt , t from 0 to 10). Our result is $38,063. Note that the entire income stream gives us int(( 5000 + 25 t e^(-t/10) ) dt , t from 0 to 10) = $50,660 over the 10-year period. The meaning of our solution is that an investment of $38,063 for the 10-year period at 6% continuously compounded interest would yield $50,660 at the end of the period. So $38,063 is the present value of the income stream. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ????I though I was doing it right ------------------------------------------------ Self-critique rating #$&*1 ********************************************* ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. ??????????????Had problem with finding Present value and revenue problems. I believe that the problems in chapter 6 are different in the eight edition.