course Mth 272 ¬ÔJ”ÆTÅ{N˜åÂxѵþÀàøåÁº¼ƒÜ“assignment #004
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16:26:25 4.6.06 (was 4.5.06) y = C e^(kt) thru (3,.5) and (4,5)
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RESPONSE --> y= 0.5 y= 5 0.5= Ce^(3*k) 5 = Ce^(4*k) 5 / .5 = C e^(4k) / [ C e^(3k) ] =10 = e^k solve for k= 2.3 0.5 = Ce^(2.3 * 3)=Ce^(6.9) C= 0.0005 y=.0005e^(2.3t) confidence assessment: 3
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16:26:34 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 --> ok self critique assessment: 3
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16:35:21 4.6.10 (was 4.5.10) solve dy/dt = 5.2 y if y=18 when t=0
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RESPONSE --> dy/dt = 5.2y= 5.2 dt=(1/y)dy e^(ln y) = e^(5.2t + C) y=e^(5.2t + C) y=e^C*e^(5.2t) y = X*e^(5.2t). 18= X*e^0 X=18 y =18e^(5.2t) confidence assessment: 2
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16:35:31 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 --> ok self critique assessment: 3
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16:55:36 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 --> r= 0.105 initial investment= $750 750e^(.105t) = 2*750 e^(.105t) = 2 0.105t = ln(2) t= 6.9 years doubling time 2nd part t=10 years 750e^0.105*10= $2,143.24 3rd part t=25= $10,353.43 confidence assessment: 3
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16:55:45 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 --> ok self critique assessment: 3
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17:03:05 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 = C*e^(k*x) p=5 x=300 5 = C*e^(300*k) p=4 x=400 4= C*e^(400*k) 5/4= C*e^(300*k)/C*e^(400*k) k = ln(5/4)/(-100) k = -0.002 C = 5 /e^(300*k) solve for C C= 9.8 fn p = 9.8*e^(-.0022*t) confidence assessment: 3
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17:03:12 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 --> ok good self critique assessment: 3
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