course MTH 272 02/23, 1200 assignment #004004. `query 4 Applied Calculus II 02-14-2010
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18:15:45 4.6.06 (was 4.5.06) y = C e^(kt) thru (3,.5) and (4,5)
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RESPONSE --> y=Ce^(kt) .5=Ce^(3k) 5=Ce^(4k) 5/(e^(4k))=C .5 = [5/(e^(4k))] e^(3k) .5 = 5/(e^k) .5 e^k = 5 e^k=10 ln e^k=ln 10 k=2.30 5/(e^(4*2.3)) =C .0005=C y=.0005e^(2.3t) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:15:57 Substituting the coordinates of the first and second points into the form y = C e^(k t) we obtain the equations .5 = C e^(3*k)and 5 = Ce^(4k) . Dividing the second equation by the first we get 5 / .5 = C e^(4k) / [ C e^(3k) ] or 10 = e^k so k = 2.3, approx. (i.e., k = ln(10) ) Thus .5 = C e^(2.3 * 3) .5 = C e^(6.9) C = .5 / e^(6.9) = .0005, approx. The model is thus close to y =.0005 e^(2.3 t). **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*: 3
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18:17:52 4.6.10 (was 4.5.10) solve dy/dt = 5.2 y if y=18 when t=0
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RESPONSE --> y=Ce^(kt) 18=Ce^(k*0) 18=Ce^0 18=C y=18e^(5.2t) k>0...Exponential Growth confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:18:06 The details of the process: dy/dt = 5.2y. Divide both sides by y to get dy/y = 5.2 dt. This is the same as (1/y)dy = 5.2dt. Integrate the left side with respect to y and the right with respect to t: ln | y | = 5.2t +C. Therefore e^(ln y) = e^(5.2 t + c) so y = e^(5.2 t + c). This is the general function which satisfies dy/dt = 5.2 y. Now e^(a+b) = e^a * e^b so y = e^c e^(5.2 t). e^c can be any positive number so we say e^c = A, A > 0. y = A e^(5.2 t). This is the general function which satisfies dy/dt = 5.2 y. When t=0, y = 18 so 18 = A e^0. e^0 is 1 so A = 18. The function is therefore y = 18 e^(5.2 t). **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*: 3
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18:22:25 4.6.25 (was 4.5.25) init investment $750, rate 10.5%, find doubling time, 10-yr amt, 25-yr amt) New problem is init investment $1000, rate 12%.
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RESPONSE --> A=Pe^(rt) P=1000 r=.12 A.) Doubling Time 2P=Pe^(rt) 2=e^(.12t) ln 2=ln e^(.12t) .693=.12t 5.78yrs.=t B.) A=1000e^(.12*10) A=$3320, after 10 years C.) A=1000e^(.12*25) A=$20,085 after 25 years confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:23:13 When rate = .105 we have amt = 1000 e^(.105 t) and the equation for the doubling time is 750 e^(.105 t) = 2 * 750. Dividing both sides by 750 we get e^(.105 t) = 2. Taking the natural log of both sides .105t = ln(2) so that t = ln(2) / .105 = 6.9 yrs approx. after 10 years amt = 750e^.105(10) = $2,143.24 after 25 yrs amt = 7500 e^.105(25) = $10,353.43 *
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RESPONSE --> Differing values between given and #25 in the book. ------------------------------------------------ Self-critique rating #$&*: 3
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18:34:01 4.6.44 (was 4.5.42) demand fn p = C e^(kx) if when p=$5, x = 300 and when p=$4, x = 400
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RESPONSE --> p=Ce^(kx) (300,5) 5=Ce^(300k) 5/(e^(300k))=C (400,4) 4=Ce^(400k) 4=(5/(e^(300k))*(e^400k) 4=5*(e^100k) (4/5)= e^100k ln (.8) = ln e^(100k) -.223=100k -.002=k 5/(e^(300k))=C 5/(e^(300*-.002)=C 9.11=C y=9.11e^(-.002k) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:34:30 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). **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*: 3
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