#$&* course Mth 174 8/2 4 Calculus IIAsst # 19
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Given Solution: dM/dt = 0.05M ** The equation is dM/dt = r * M. The question is not posed for the specific value r = .05. **
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11:39:48
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**** What is the solution to the equation?
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11:40:08 M = 1000e^(0.05*t) t is in years The equation is dM/dt = r * M. We separate variables to obtain dM / M = r * dt so that ln | M | = r * t + c and M = e^(r * t + c) = e^c * e^(r t), where c is an arbitrary real number. For any real c we have e^c > 0, and for any real number > 0 we can find c such that e^c is equal to that real number (c is just the natural log of the desired positive number). So we can replace e^c with A, where it is understood that A > 0. We obtain general solution M = A e^(r t) with A > 0. Specifically we have M ( 0 ) = 1000 so that 1000 = A e^(r * 0), which tells us that 1000 = A. So our function is M(t) = 1000 e^(r t).
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11:40:09
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**** Describe your sketches of the solution for interest rates of 5% and 10%.
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11:40:34 They're just exponential graphs The graph of 1000 e^(.05 t) is an exponential graph passing through (0, 1000) and (1, 1000 * e^.05), or approximately (1, 1051.27). The graph of 1000 e^(.10 t) is an exponential graph passing through (0, 1000) and (1, 1000 * e^.10), or approximately (1, 1105.17).
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11:40:34
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**** Does the doubled interest rate imply twice the increase in principle?
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11:40:42 No We see that at t = 1 the doubled interest rate r = .10 results in an increase of $105.17 in principle, which is more than twice the $51.27 increase we get for r = .05. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Query problem 11.5.25 was 11.5.22 (3d edition 11.5.20) At 1 pm power goes out with house at 68 F. At 10 pm outside temperature is 10 F and inside it's 57 F. Give the differential equation you would solve to obtain temperature as a function of time. Solve the equation to find the temperature at 7 am. **** What is your prediction of the temperature at 7 am and how did you get it? Would you revise your estimate up or down in considering the assumption you made in writing the differential equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dT / dt = -k (T - 10) T(t) = 10 + A e^(k t). T(0) = 68 T(9) = 57 68 = 10 + A e^(k * 0) 57 = 10 + A e^(k * 9) The second equation is: 57 = 10 + 58 e^(9 k) e^(9 k) = 47 / 58 9 k = ln(47 / 58) k = 1/9 * ln(47/58) = -.0234approx. T(t) = 10 + 58 * e^(-.0234 t) At 7 am t =18 so, T(18) = 10 + 58 * e^(-.0234 * 18) = 48, approx.. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Assuming that dT / dt = k * (T - 10) we find that T(t) = 10 + A e^(k t). Counting clock time t from 1 pm we have T(0) = 68 and T(9) = 57 giving us equations 68 = 10 + A e^(k * 0) and 57 = 10 + A e^(k * 9). The first equation tells us that A = 58. The second equation becomes 57 = 10 + 58 e^(9 k) so that e^(9 k) = 47 / 58 and 9 k = ln(47 / 58) so that k = 1/9 * ln(47/58) = -.0234, approx.. Our equation is therefore T(t) = 10 + 58 * e^(-.0234 t). At 7 am the clock time will be t = 18 so our temperature will be T(18) = 10 + 58 * e^(-.0234 * 18) = 48, approx.. At 7 a.m. the temperature will be about 48 deg.
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11:41:36
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**** What is your differential equation?
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11:41:47 dT/dt = -k(T - 10)
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11:41:48
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**** How did you solve your differential equation?
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11:42:03 I just worked it out by a general solution given in the book &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: **** 11.6.15 was 11.6.11 (3d edition 11.6.6). 20 cal/day maintains weight; rate of wt change is prop to difference with prop const 1/3500 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: W = weight and I = intake: dW/dt = w/3500 dW/dt = k ( I - 20 * W) so, dW/dt = 1/3500 ( I - 20 W) dW / (I - 20 W) = 1/3500 * dt Intergrate both sides to get: -1/20 ln | I - 20 W | = t / 3500 + c. ln | I - 20 W | = -20 t /3500 + c (-20 * c) | I - 20 W | = e^(-1/175 * t + c) | I - 20 W | = A e^(-1/175 * t), with A > 0. For I > 20 W I - 20 W = A e^(-1/175 * t) W = I / 20 - A e^(-1/175 * t) For I < 20 W -(I - 20 W) = -A e^(-1/175 * t) W = I / 20 + A e^(-1/175 * t) If W = 160 and I = 3000, I < 20 W W = I / 20 + A e^(-1/175 * t) t = 0 160 = 3000 / 20 + A e^(-1/175 * 0) 160 = 150 + A A = 10 Weight function = W(t) = 150 + 10 e^(-1/175 * t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: **** What is the differential equation you would solve to get W(t) for intake I cal/day?
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11:42:27 dW/dt = w/3500 ** The rate of change is is proportional to the difference between the number of calories consumed and the number required to maintain weight. The number of daily calories required to maintain weight is weight * 20, or using the notation W(t) for weight, the number of calories required to maintain weight is W(t) * 20. If the daily intake is I then the difference between the number of calories consumed and the number required to maintain weight is I - W(t) * 20. Thus to say that rate of change is is proportional to the difference between the number of calories consumed and the number required to maintain weight is to say that dW/dt = k ( I - 20 * W). We are told that k = 1/3500 so that dW/dt = 1/3500 ( I - 20 W). The equation is solved by separation of variables: dW / (I - 20 W) = 1/3500 * dt. Integrating both sides we get -1/20 ln | I - 20 W | = t / 3500 + c. We solve for W. First multiplying both sides by -20 we get ln | I - 20 W | = -20 t /3500 + c (-20 * c is still just an arbitrary constant so we still call the result c). | I - 20 W | = e^(-1/175 * t + c) or | I - 20 W | = A e^(-1/175 * t), with A > 0. If I > 20 W then we have I - 20 W = A e^(-1/175 * t), with A > 0 so that W = I / 20 - A e^(-1/175 * t), with A > 0. Thus if the person consumes more than 20 W calories per day weight will approach the limit I / 20 from below. If I < 20 W then we have -(I - 20 W) = -A e^(-1/175 * t), with A > 0 so that W = I / 20 + A e^(-1/175 * t), with A > 0. Thus if the person consumes fewer than 20 W calories per day weight will approach the limit I / 20 from above. If W = 160 when I = 3000 then I < 20 W and we have W = I / 20 + A e^(-1/175 * t), with A > 0. At t = 0 we have 160 = 3000 / 20 + A e^(-1/175 * 0), or so that 160 = 150 + A so we have A = 10. Now the weight function is W(t) = 150 + 10 e^(-1/175 * t). This graph starts at W = 150 when t = 0. The 'half-life' for e^(-1/175 t) occurs when -1/175 t = ln(1/2), at t = -175 * ln(1/2) = 121.3 approx.. At this point 10 e^(-1/175 * t) = 10 e^(-1/175 * 121.3) = 5 and W(t) = 150 + 5 = 155. The graph will therefore exponentially approach W = 160, passing thru points (0,150) and (121.3,155). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: **** (no longer present as of 4th edition but should be worked and self-critiqued) Query problem 11.6.16 (was 10.6.10) A chemical C is formed when a single molecule of substance A combines with a single molecule of substance B to form a single molecule of substance C. The rate at which molecules of C are formed is proportional to product of the amount of chemical A present, and the amount of chemical B present. The initial quantities are respectively a and b, and x is the amount of C present at time t. What is the differential equation for x = quantity of C at time t? If a and b are equal, then in terms of the unspecified proportionality constant k, what is the solution of this equation for x(0) = 0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dx/dt = k (a - x)*(b - x) a = b dx / dt = k ( a - x) ( b - x) = k ( a - x) ( a - x) dx / dt = k ( a-x)^2 dx / (a - x)^2 = k dt Integrate, 1/(a-x) = k t + c a - x = 1 / (k t + c) x = a - 1 / (kt + c) x(0) = 0 0 = a - 1 / (k * 0 + c) c = 1 / a. So, x = a - 1 / (k t + 1/a) x = a - a / (a k t + 1) x = a ( 1 - 1/(akt + 1) ). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: The problem states that C is formed by the combination of a molecule of A and a molecule of B, and that the rate of combination is proportional to the product of the numbers of molecules of A and B present. If x molecules of C have been formed, then x molecules of A and of B will have been used up. If the initial numbers of A and B are respectively a and b, then if x molecules of C have been formed there will be a - x molecules of A and b - x molecules of B. This product of the numbers of molecules of A and of B present will be (a - x) * (b - x), so the equation will be dx/dt = k (a - x)*(b - x). If a and b are the same then a = b and we can write dx / dt = k ( a - x) ( b - x) = k ( a - x) ( a - x). The equation is therefore dx / dt = k ( a-x)^2. Separating variables we have dx / (a - x)^2 = k dt. Integrating we have 1/(a-x) = k t + c so that a - x = 1 / (k t + c) and x = a - 1 / (kt + c). If x(0) = 0 then we have 0 = a - 1 / (k * 0 + c) so that c = 1 / a. Thus x = a - 1 / (k t + 1/a) = a - a / (a k t + 1) = a ( 1 - 1/(akt + 1) ). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Query problem 11.6.29 was 11.6.25 (3d edition 11.6.24) F = m g R^2 / (R + h)^2. Find the differential equation for dv/dt and show that the Chain Rule dv/dt = dv/dh * dh/dt gives you v dv/dh = -gR^2/(R+h)^2. Solve the differential equation, and use your solution to find escape velocity. Give your solution. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F = m*a a = dv/dt so, F = m*(dv/dt) m*(dv/dt) = -mgR^2/(R+h)^2 F = -mgR^2/(R+h)^2 dv/dt = -gR^2/(R+h)^2 dv/dt = dv/dh * dh/dt, as dh/dt is velocity dv/dt = dv/dh * v Substituting dv/dt, v*dv/dh = -gR^2/(R+h)^2 v dv = -gR^2/(R+h)^2 dh Integrate both sides, (v^2)/2 = -gR^2*[-1/(R+h)] + C (v^2)/2 = gR^2/(R+h) + C. v = v(0) at H = 0 (v(0)^2)/2 = gR^2/(R+0) + C (v(0)^2)/2 = gR + C C = (v(0)^2)/2 - gR. (v^2)/2 = gR^2/(R+h) + (v(0)^2)/2 - gR v^2 = 2*gR^2/(R+h) + v(0)^2 - 2*gR As h -> infinity, 2*gR^2/(R+h) -> 0 v^2 = v(0)^2 - 2*gR Minimum escape velocity = sqrt (2gR) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution:
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RESPONSE --> Good student solution: F = m*a and a = dv/dt, so F = m*(dv/dt) F = mgR^2/(R+h)^2 Substituting for F: m*(dv/dt) = -mgR^2/(R+h)^2 ( - because gravity is a force of attraction, so that the acceleration will be in the direction opposite the direction of increasing h) Dividing both sides by m: dv/dt = -gR^2/(R+h)^2 Using dv/dt = dv/dh * dh/dt, where dh/dt is just the velocity v: dv/dt = dv/dh * v Substituting for dv/dt in differential equation from above: v*dv/dh = -gR^2/(R+h)^2 so v dv = -gR^2/(R+h)^2 dh. Integral of v dv = Integral of -gR^2/(R+h)^2 dh Integral of v dv = -gR^2 * Integral of dh/(R+h)^2 (v^2)/2 = -gR^2*[-1/(R+h)] + C (v^2)/2 = gR^2/(R+h) + C. Since v = vzero at H = 0 (vzero^2)/2 = gR^2/(R+0) + C (vzero^2)/2 = gR + C C = (vzero^2)/2 - gR. Thus, (v^2)/2 = gR^2/(R+h) + (vzero^2)/2 - gR v^2 = 2*gR^2/(R+h) + vzero^2 - 2*gR As h -> infinity, 2*gR^2/(R+h) -> 0 So v^2 = vzero^2 - 2*gR. v must be >= 0 for escape, therefore vzero^2 must be >= 2gR (since a negative vzero is not possible). Minimum escape velocity occurs when vzero^2 = 2gR, Thus minimum escape velocity = sqrt (2gR). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Query problem 11.6.20 THIS IS THE FORMER PROBLEM, VERY UNFORTUNATELY OMITTED IN THE NEW EDITION. rate of expansion of universe: (R')^2 = 2 G M0 / R + C; case C = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dR/dx = `sqrt(2 G M0 / R ) dR/dx = 'sqrt( 2 G M0) / 'sqrt R dR / 'sqrt R = 'sqrt( 2 G M0) dx 2 'sqrt R = 'sqrt( 2 G M0) x 4 R = ( 2 G M0) x^x R = ( 2 G M0 x^2 ) / 4 R ' = `sqrt( 2 G M0 / R ) dR/dt = 'sqrt( 2 G M0) / 'sqrt R dR * sqrt(R) = sqrt(2 G M0) dt 2/3 R^(3/2) = sqrt(2 G M0) t + c R(0) = 0, c = 0, R = ( 3/2 sqrt( 2 G M0) t)^(2/3) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: what is your solution to the differential equation R' = `sqrt( 2 G M0 / R ), R(0) = 0?
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RESPONSE --> I am assuming R' = dR/dx so I have a variable of integration. dR/dx = `sqrt( 2 G M0 / R ) dR/dx = 'sqrt( 2 G M0) / 'sqrt R dR / 'sqrt R = 'sqrt( 2 G M0) dx 2 'sqrt R = 'sqrt( 2 G M0) x 4 R = ( 2 G M0) x^x R = ( 2 G M0 x^2 ) / 4 This then satisfies for x=0, R=0. R ' = `sqrt( 2 G M0 / R ) gives you dR/dt = 'sqrt( 2 G M0) / 'sqrt R so that dR * sqrt(R) = sqrt(2 G M0) dt and 2/3 R^(3/2) = sqrt(2 G M0) t + c. Since R(0) = 0, c = 0 and we have R = ( 3/2 sqrt( 2 G M0) t)^(2/3). This can be simplified, but the key is that R is proportional to t^(2/3), which increases without bound as t -> infinity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!