course Mth 272 x}w툿assignment #004
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20:41:48 4.6.06 (was 4.5.06) y = C e^(kt) thru (3,.5) and (4,5)
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RESPONSE --> First the values of x and y must be plugged into the equations .5=Ce^(3k) 5=Ce^(4k) then division 5/.5= (Ce^(4k)) / (Ce^(3k)) factor out a Ce^(3k) and an e^k is left over and 5/.5 is 10 so 10=e^k then take the ln of each side k= ln(10) or 2.3 then plug in the values .5=Ce^(ln(10)*3)) .5=Ce^(6.9) then divide C= .5/e^(6.9)= .00054 so the model is y=.00054e^(2.3t) confidence assessment: 3
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20:42:07 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 --> got it self critique assessment: 3
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22:06:30 4.6.10 (was 4.5.10) solve dy/dt = 5.2 y if y=18 when t=0
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RESPONSE --> The problem can be integrated, from the form of y`/y=5.2t` which becomes ln(y)=5.2t+c then its put into terms of e e^(lny) which will = y = e^(5.2t+c) so y= Xe^(5.2t) and when t=0 and y= 18 18= Xe^(5.2*0) X=18 the model is y=18e^(5.2t) confidence assessment: 3
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22:06:41 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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23:22:30 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 --> for this one the numbers must be plugged into the exponential growth formula for doubling so 750e^(.105t)=750*2 then division of 750 e^(.105t)=2 then ln of this .105t=ln(2) then division t=6.6 years when t is 10 years amount= 750e^(.105)(10)= $2143.24 amount= 750e^(.105)(25)= $10353.43 confidence assessment: 3
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23:22:39 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 --> got it self critique assessment: 3
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23:44:08 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 --> this is similar to a previous problem, just different values. the 2 equations are 5=ce^(300k) 4= Ce^(400k) then division 5/4= Ce^(300k)/ Ce^(400k) 5/4= e^(-100k) then ln of both sides ln(5/4)= -100k and division k=-.0022 then plug into an equation 5=Ce^(300*-.0022) division 5/e^(.66)= C C= approx 9.77 so the model is p=9.77e^(-.0022t) confidence assessment: 3
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23:48:30 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 --> got it self critique assessment: 3
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