#$&* course MTH 272 10/6 about 11:45 am 004. `query 4
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Given Solution: `a Substituting the coordinates of the first and second points into the form y = C e^(k t) we obtain the equations .5 = C e^(3*k)and 5 = Ce^(4k) . Dividing the second equation by the first we get 5 / .5 = C e^(4k) / [ C e^(3k) ] or 10 = e^k so k = 2.3, approx. (i.e., k = ln(10) ) Thus .5 = C e^(2.3 * 3) .5 = C e^(6.9) C = .5 / e^(6.9) = .0005, approx. The model is thus close to y =.0005 e^(2.3 t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I had no clue what the question was asking so I had to look at the given solution. I do understand the problem and did not look at the solution step by step while solving on paper. I did look over the example in the book and noted another solution to the problem to better understand. ------------------------------------------------ Self-critique Rating:3
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Given Solution: `a The details of the process: dy/dt = 5.2y. Divide both sides by y to get dy/y = 5.2 dt. This is the same as (1/y)dy = 5.2dt. Integrate the left side with respect to y and the right with respect to t: ln | y | = 5.2t +C. Therefore e^(ln y) = e^(5.2 t + c) so y = e^(5.2 t + c). This is the general function which satisfies dy/dt = 5.2 y. Now e^(a+b) = e^a * e^b so y = e^c e^(5.2 t). e^c can be any positive number so we say e^c = A, A > 0. y = A e^(5.2 t). This is the general function which satisfies dy/dt = 5.2 y. When t=0, y = 18 so 18 = A e^0. e^0 is 1 so A = 18. The function is therefore y = 18 e^(5.2 t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I relied heavily on the given solution to figure this problem out. I did have to study it for awhile to understand the solution. I hope that I will be able to solve a problem similar to this on an exam but I think I can do it. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q4.6.5 (previously 4.6.25 (was 4.5.25)) Init investment $1000, rate 12%. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A = P * e^(rt) A = 1000 * e^(0.12t) 2000 = 1000 * e^(0.12t) divide by 2000 2 = e^(0.12t) take the nat. log of both sides Ln 2 = 0.12 t solve for t using calculator and dividing both side by 0.12 t = 5.78 years to double the initial investment at the given rate At 10 years A = 1000 * e^(0.12 * 10) = $3320.12 At 25 years A = 1000 * e^(0.12 * 25) = $20,085.50 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Rate = .12 and initial amount is $1000 so we have amt = $1000 e^(.12 t). The equation for the doubling time is 1000 e^(.105 t) = 2 * 1000. Dividing both sides by 1000 we get e^(.12 t) = 2. Taking the natural log of both sides .12t = ln(2) so that t = ln(2) / .12 = 5.8 yrs approx. after 10 years we have • amt = 1000e^(.12(10)) = $3 320 after 25 yrs we have • amt = 1000 e^(.12(25)) = $20 087 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I had to look at the given solution to see what we were supposed to do with the given information. I fully understand the problem. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 4.6.8 (previously 4.6.44 (was 4.5.42)) demand fn p = C e^(kx) if when p=$5, x = 300 and when p=$4, x = 400 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Plug in the values then divide the equations. 4/5 = (Ce^(400k)) / (Ce^(300k)) 4/5 = e^(100k) take the nat. log of both sides Ln(4/5) = 100k divide both sides by 100 k = (ln (4/5))/100 = -0.00223 Now plug k into on of the previous equations with one set of the given values to find C. 4 = Ce^(-0.00223 * 400) C = 4/(e^(-0.00223 * 400) = 9.76 so then plug in values C and k into the original equation. p = 9.76 * e^(-0.00223 * x) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a You get 5 = C e^(300 k) and 4 = C e^(400 k). If you divide the first equation by the second you get 5/4 = e^(300 k) / e^(400 k) so 5/4 = e^(-100 k) and k = ln(5/4) / (-100) = -.0022 approx.. Then you can substitute into the first equation: } 5 = C e^(300 k) so C = 5 / e^(300 k) = 5 / [ e^(300 ln(5/4) / -100 ) ] = 5 / [ e^(-3 ln(5/4) ] . This is easily evaluated on your calculator. You get C = 9.8, approx. So the function is p = 9.8 e^(-.0022 t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: