#$&* course Mth 272 September 17 around 10:55 004. `query 4
.............................................
Given Solution: `a Substituting the coordinates of the first and second points into the form y = C e^(k t) we obtain the equations .5 = C e^(3*k)and 5 = Ce^(4k) . Dividing the second equation by the first we get 5 / .5 = C e^(4k) / [ C e^(3k) ] or 10 = e^k so k = 2.3, approx. (i.e., k = ln(10) ) Thus .5 = C e^(2.3 * 3) .5 = C e^(6.9) C = .5 / e^(6.9) = .0005, approx. The model is thus close to y =.0005 e^(2.3 t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There was no actual question in the problem so I had to look at the solution to see what I needed to do.
.............................................
Given Solution: `a The details of the process: dy/dt = 5.2y. Divide both sides by y to get dy/y = 5.2 dt. This is the same as (1/y)dy = 5.2dt. Integrate the left side with respect to y and the right with respect to t: ln | y | = 5.2t +C. Therefore e^(ln y) = e^(5.2 t + c) so y = e^(5.2 t + c). This is the general function which satisfies dy/dt = 5.2 y. Now e^(a+b) = e^a * e^b so y = e^c e^(5.2 t). e^c can be any positive number so we say e^c = A, A > 0. y = A e^(5.2 t). This is the general function which satisfies dy/dt = 5.2 y. When t=0, y = 18 so 18 = A e^0. e^0 is 1 so A = 18. The function is therefore y = 18 e^(5.2 t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q4.6.5 Init investment $1000, rate 12%. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1000e^(0.12t) 1000e^(0.105t)=2*1000 e^(0.12t)=2 0.12t=ln(2) t=ln(2)/0.12 t=5.8 1000e^(0.12(10)) = 3,320 1000 e^(0.12(25)) = 20,087 confidence rating #$&* 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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)) = $20 087 Self-critique (if necessary): Once again the question is not there to know what to do for the problem. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q 4.6.8 demand fn p = C e^(kx) if when p=$5, x = 300 and when p=$4, x = 400 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5=Ce^(300k) 4-Ce^(400k) 5/4=e^(300k)/e^(400k) 5/4=e^(-100k) k=ln(5/4)/(-100) k=-0.0022 5=Ce^(300k) C=5/e^(300k) C=5/(e^(300ln(5/4)/-100)) C=5/[e^(-3ln(5/4))] C=9.8 p=9.8e^(-0.0022t) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): There was no stated question so I looked and saw to do it the way a previous one was done and continued that way ------------------------------------------------ Self-critique rating #$&* 3 "