course Mth 272 Two or three of the questions...the responses(answers) did not relate to the question ask. «¤© {ÂÎÃeåX~å”UÞ~—ñÑÆÆÕîÓ…¤’̤Îassignment #004
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18:03:50 4.6.06 (was 4.5.06) y = C e^(kt) thru (3,.5) and (4,5)
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RESPONSE --> Using y = Ce^(kt) and the two ordered pairs. y = Ce^(kt) Solve for k. 5 = .5 e^k(1)
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18:26:27 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 --> I used mixed coordinates...i understand what I did wrong. self critique assessment: 2
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18:27:59 4.6.10 (was 4.5.10) solve dy/dt = 5.2 y if y=18 when t=0
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RESPONSE --> When k>0.....Exponential Growth y = 18e^(5.2t) confidence assessment: 2
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18:28:34 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: 2
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18:33:55 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 --> 10yrs: y = 1,000 e^((0.12)(10)) = $3,320.12 25 yrs: y = 1,000 e^((0.12)(25)) = $20,085.54 Doubled: 2,000 = 1,000 e^(0.12t) 2 = e^(0.12t) ln 2 = 0.12t (ln 2) / 0.12 = t 5.78 yrs = t confidence assessment: 2
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18:36:55 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 --> THis is not what I got for # 25...I am not sure where yu're getting (0.105). I answered #25. self critique assessment: 1
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18:39:30 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 --> Find k. p = Ce^(kt) 40 = 45e^(1000k) 8/9 = e(1000k) [ln (8/9)] / 1000 = k -0.000118 = k Then...p = 45e^[(1000)(-0.000118)] p = 39.99 confidence assessment: 1
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18:41:34 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 --> Sorry I'm lost. I'm looking at #44...there's not 300 and/or 400 in the question. I don't think this is the answer to Question #44. self critique assessment: 1
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