#$&* course MTH272 If your solution to a stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a Substituting the coordinates of the first and second points into the form y = C e^(k t) we obtain the equations .5 = C e^(3*k)and 5 = Ce^(4k) . Dividing the second equation by the first we get 5 / .5 = C e^(4k) / [ C e^(3k) ] or 10 = e^k so k = 2.3, approx. (i.e., k = ln(10) ) Thus .5 = C e^(2.3 * 3) .5 = C e^(6.9) C = .5 / e^(6.9) = .0005, approx. The model is thus close to y =.0005 e^(2.3 t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ************************************************************ ********************************************* Question: `q 4.6.2 (previously 4.6.10 (was 4.5.10)) solve dy/dt = 5.2 y if y=18 when t=0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y’=Cy y’ = 5.2 y if y=18 when t=0 y’ = 5.2y separate variables by dividing by y y’/y = 5.2 y’/y dt = 5.2 dt integrate with respect to t 1/y = 5.2 dt dy = y’dt ln(y) = 5.2t + C find antiderivative for each side (e) ln(y) = (e)5.2t + C solve for y y = Ce^(5.2*0) 18= Ce^(5.2*0) Y=18 18 = C e^(k0) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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/d t = 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 took the information from a text I had from calculus 1. The way our text explains it was a bit confusing. ------------------------------------------------ Self-critique 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): There was a slight difference in the 25 year example, but I believe it was due to rounding through the process. ------------------------------------------------ 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: p = C e^(kx) if when p=$5, x = 300 and when p=$4, x = 400 5 = C e^(300k) 4 = C e^(400k) 5/4 = (e^(300k))/ (e^(400k)) divide the first problem by the second problem 5/4 = e^(-100k) ln5/4 = -100k take a natural log (ln5/4)/-100 = k isolate k k = -.0022 5 = C e^(300*-.0022) plug in k to solve C 5/ (e^(300*-.0022)) = C Isolate C C = 9.67 P = 9.67 e^(-.0022x) missing variables to solve the two demand function confidence rating #$&*: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):N/A ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!