#$&* course Mth 272 06/06 7 pm uestion: `q4.6.1 (previously 4.6.06 (was 4.5.06)) y = C e^(kt) thru (3,.5) and (4,5)
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Given Solution: `a Substituting the coordinates of the first and second points into the form y = C e^(k t) we obtain the equations .5 = C e^(3*k)and 5 = Ce^(4k) . Dividing the second equation by the first we get 5 / .5 = C e^(4k) / [ C e^(3k) ] or 10 = e^k so k = 2.3, approx. (i.e., k = ln(10) ) Thus .5 = C e^(2.3 * 3) .5 = C e^(6.9) C = .5 / e^(6.9) = .0005, approx. The model is thus close to y =.0005 e^(2.3 t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: Ok ********************************************* Question: `q 4.6.2 (previously 4.6.10 (was 4.5.10)) solve dy/dt = 5.2 y if y=18 when t=0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dy/y= 5.2dt lny= 5.2t + C y= e^(5.2t + C) y= e^5.2t * e^C e^c= 18 y= e^(5.2t) * 18 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: Ok ********************************************* Question: `q4.6.5 (previously 4.6.25 (was 4.5.25)) Init investment $1000, rate 12%. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1000 e^(.12t) 1000 e^(.105 t) = 2 * 1000. e^(.12t) = 2 t= 5.78 years confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 gete^(.12 t) = 2. Taking the natural log of both sides.12t = ln(2) so thatt = 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): Self-critique Rating: Ok ********************************************* Question: `q 4.6.8 (previously 4.6.44 (was 4.5.42)) demand fn p = C e^(kx) if when p=$5, x = 300 and when p=$4, x = 400 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5= C e^(k*300)) = 1.25= C e^(-100 k) 4= C e^(k*400)) 1.25 = e^(-100 k ) ln 1.25= -100k k = -.002 5= C e^(-.002*300) C= 9.11 p= 9.11e^(-.002t)) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Self-critique Rating: Ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q4.6.5 (previously 4.6.25 (was 4.5.25)) Init investment $1000, rate 12%. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1000 e^(.12t) 1000 e^(.105 t) = 2 * 1000. e^(.12t) = 2 t= 5.78 years confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 gete^(.12 t) = 2. Taking the natural log of both sides.12t = ln(2) so thatt = 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): Self-critique Rating: Ok ********************************************* Question: `q 4.6.8 (previously 4.6.44 (was 4.5.42)) demand fn p = C e^(kx) if when p=$5, x = 300 and when p=$4, x = 400 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5= C e^(k*300)) = 1.25= C e^(-100 k) 4= C e^(k*400)) 1.25 = e^(-100 k ) ln 1.25= -100k k = -.002 5= C e^(-.002*300) C= 9.11 p= 9.11e^(-.002t)) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Self-critique Rating: Ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!