course mth 272 I just wanted to remind you that I have submitted my proctor info to you and just want to make sure everything is ok with that. I plan on taking the test #1 on wed or thurs of this week at JTCC.
......!!!!!!!!...................................
08:40:29 4.6.06 (was 4.5.06) y = C e^(kt) thru (3,.5) and (4,5)
......!!!!!!!!...................................
RESPONSE --> dy/dt=ky y=e^kt 5=e^k(3) ln5 = 3k (1/3)ln5 = k k= .5364793041 y = 3e^.536479t confidence assessment: 1
.................................................
......!!!!!!!!...................................
08:48:13 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). **
......!!!!!!!!...................................
RESPONSE --> where did .5 come from?
.................................................
......!!!!!!!!...................................
08:57:18 4.6.10 (was 4.5.10) solve dy/dt = 5.2 y if y=18 when t=0
......!!!!!!!!...................................
RESPONSE --> 5.2=Ce^0 5.2=C find k y= 5.2e^kt 18=5.2e^k0 3.46=e^0k 1.24=k y= 5.2e^1.24t confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:03:32 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). **
......!!!!!!!!...................................
RESPONSE --> ok, I see that y = A e^(5.2 t) is the general function which satisfies dy/dt = 5.2 y because 5.2=k plug in 18 = A e^0. e^0 is 1 so A = 18 y =18e^(5.2t) self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:10:25 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%.
......!!!!!!!!...................................
RESPONSE --> use A=Pe^rt for 10 yr amt A=750e^.105(10) A = 2143.24 for 25 yr amt A = 750e^.105(25) A=10353.43 confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:12:47 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 *
......!!!!!!!!...................................
RESPONSE --> 3 self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:19:33 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
......!!!!!!!!...................................
RESPONSE --> i think you set up 5=Ce^300k and solve for C, but you dont know k, so im confused
.................................................
......!!!!!!!!...................................
09:28:29 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). **
......!!!!!!!!...................................
RESPONSE --> 5 = C e^(300 k) 4 = C e^(400 k) divide the first into the 2nd 5/4 = e^(300 k) / e^(400 k) 5/4 = e^(-100 k) k = ln(5/4) / (-100) = -.0022 then you substitue this into the first equation C = 5 / e^(300 k) = 5 / [ e^(300 ln(5/4) / -100 ) ] = 5 / [ e^(-3 ln(5/4) ] C=9.8 plug in p= 9.8e^(-.0022 t) self critique assessment: 2
.................................................