course Mth 272 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. 004. `query 4 ********************************************* ********************************************* Question: `q4.6.1 (previously 4.6.06 (was 4.5.06)) y = C e^(kt) thru (3,.5) and (4,5) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: first equation we let y = .5 and t = 3 second equation we let y = 5 and t = 4 .5 = C*e^kt = .5 = C*e^(3k) 5 = C*e^(4k) in the second equation we substitute C = .5 / e^3k 5 = (.5 / e^(3k)) * 4k we divide both sides by .5 10 = e^(1k) take the ln of both sides ln10 = 1k k = ln10/1 = 2.30258 next we substitute k in to find C .5 = C*e^3k = C = .5 / e^(3k) = .5 / e^(3*2.30258) = .5 / 999.999 C = 0.0005 Since we have found both C and k the growth model is y = 0.0005 *e^2.30258(t) confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Substituting the coordinates of the first and second points into the form y = C e^(k t) we obtain the equations .5 = C e^(3*k)and 5 = Ce^(4k) . Dividing the second equation by the first we get 5 / .5 = C e^(4k) / [ C e^(3k) ] or 10 = e^k so k = 2.3, approx. (i.e., k = ln(10) ) Thus .5 = C e^(2.3 * 3) .5 = C e^(6.9) C = .5 / e^(6.9) = .0005 The model is thus close to y =.0005 e^(2.3 t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* ********************************************* 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 18e^5.2t confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ?????????????Not understanding the thought process with this one
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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)) = $20 087 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*: ********************************************* ********************************************* 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=$4, x = 400 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: first equation 5 = C*e^(300x) second equation 4 = C*e^(400x) C = 5 / e^300x substitute this into the second equ 4 = (5 / e^(300x)) * 400x we divide both sides by 5 0.800 = e^(100x) take the ln of both sides ln0.8 = 100x divide both sides by 100 x = ln0.8 / 100 = -0.00223 substitute x into the first equation to find C C = 5 / e^(300x) = 5 / e(300 * -0.00223) = 5 / 0.512 C = 9.765 confidence rating #$&*: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): did not find the function because the problem only asked to solve for C and K. I do understand where the function comes from. ------------------------------------------------ Self-critique rating #$&*:3