#$&* course Mth 174 12/6 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): So to find M with respect to t, t does not actually have to be in the diff equation?? example like with dm/dt = r*M, I see how everything is with respect to yrs or the time interval but it through me off having var r instead of t.
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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): There were several examples in the book just like the one asked. ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: !!!!!This problem is very similar to Population of an organism being prop to living capacity of the environment we went over in the notes!!! dW/dt = k(I – 20W), rearranged we get dW/(I – 20W) = k dt, now we solve =Int(dW/(I – 20W)) = (-1/20)ln|I – 20W| =Int(k dt) = kt Now we have, (-1/20) ln|I – 20W| = kt, so looking at question again if the prop constant = 1/3500 then we should have sub that value in before integrating but it shouldn’t matter if we plug it in now. (-1/20) ln|I – 20W| = (1/3500)t ln|I – 20W| = (1/3500)t/(-1/20) = -20t/3500 = approx. -.0057t |I – 20W| = e^(-.0057t) 20W = I – e^(-.0057t) W = (I – e^(-.0057t))/20 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 isto 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): I messed this one up bad. I forgot to add my constant C as part of the antiderivative of k dt. I believe I was close on everything else but let me know if I missed something else.
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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 I think I understand, so x was the total amount of var c in the chem process at clock time t. I believe you explained it exactly like that but for some reason I couldn’t quite grasp it. Then because a and b are equal at init clock time, that’s why rate of change was prop to (a – x)^2 ------------------------------------------------ Self-critique Rating:2 ********************************************* 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: I am having a problem figuring out where to start on this problem. I have a hard time setting up the problem. I really like working with the diff equations once you have them set up its really easy to understand and most of the time solve, and most of the time I understand what the equation is telling. Unlike a lot of the stuff we have covered the last two semesters, where I can work and solve the equation but have no idea what’s going on or what the solution is telling me. Anyway looking at solution to see the obvious solution (I’m sure) that I keep overlooking. confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): So using formula F = ma and knowing that a = `dv/`dt(change in vel. wrt change in time) we find that F = m*(`dv/`dt), and using F = mgR^2/(R + h)^2 we get.. F = mgR^2/(R + h)^2 = m*(`dv/`dt) = mgR^2/(R + h)^2, then dividing thru by m = `dv/`dt = gR^2/(R + h)^2, which is what we are to solve for. It makes me so mad when I look at solution and the logic is so “duh, how could anyone not see that”. It’s not hard working thru symbolic solutions but it”s very hard to me to find that starting point. ???Quick question, gR^2/(R + h)^2 = gR^2/R^2 + 2Rh + h^2 = g/2Rh + h^2, does this hold true or by doing this does it affect your solution you are trying to obtain. Or is it simply uncalled for and a waste of your time??????
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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): Didn’t understand we were to solve for R. ???Do you just assume certain diff equations are to be solved with respect to time if not stated otherwise, or is there a way that if not stated that we can find what our rate of change in this is with respect to change of that. I had a problem figuring out how to set the equation up in regards to the rate of change of the universe was wrt rate of change in what????? Self-critique Rating:2 ********************************************* Question: Problem (omitted from 4th and 5th editions but should be worked and self-critiqued) 11.6.13 from 3d edition Light intensity is defined to be light energy per unit of area. You don't need this definition to solve the following: If light intensity is absorbed by water in proportion to the amount of intensity present, with the rate measured in in units of intensity per unit of distance, and if 50% of the intensity is absorbed in the first 10 feet, how much is absorbed in the first 20 ft, and how much in the first 25 feet? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ???Again I am kinda assuming that because time is not mentioned and dist. is that rate of absorption is wrt dist. Guess it would have to be!!!!!???? dL/dx = kL(x – 10) ?????I am not sure how to set this one up so I’m going to look at solution and then work out. I’m not trying to be lazy and I hope I don’t appear to be taking the easy way out. I am trying to complete all assignments with not much time left. So instead of not even attempting some assignments I am working thru all of them at an accelerated speed. Still spending around an hr and a half per assignment though????? I am constantly over complicating things, diff equation was a lot simpler then I thought dI/dx = k*I We have worked this equation several times and we get Int(dI/I) = Int(k dx) =ln|I| = kx + c I = e^(kx + c) = ce^kx Now for x = 10, I = .5 .5 = ce^10k ??Again I’m stuck we can’t solve with two unknowns so I missed something somewhere???????? Ok I see where I = init absorption*e expression .5I0 = I0e^(k10) =2 = e^(k10) =ln|2| = k10 =ln|2|/10 = k k = approx. .0693 so I = I0e^(.0693x) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rate of absorption, in units of intensity per unit of distance, is proportional to the intensity of the light. If I is the intensity then the equation is dI/dx = k I. This is the same form as the equation governing exponential population growth and the solution is easily found by separating variables. We get I = C e^(-k x). If the initial intensity is I0 then the equation becomes I = I0 e^(-k x). If 50% is absorbed in the first 10 ft then the intensity at that position will be .50 I0 and we have .5 I0 = I0 e^(-k x) so that e^(-k * 10 ft) = .5 and k * 10 ft = ln(2) so that k = ln(2) / (10 ft). Then at x = 20 ft we have I = I0 * e^(- ln(2) / 10 ft * 20 ft) = I0 * e^-(2 ln(2) ) = .25 I0, i.e., 25% will be left At x = 25 ft we have I = I0 * e^(- ln(2) / 10 ft * 25 ft) = I0 * e^-(2.5 ln(2) ) = .18 I0 (approx), i.e., about 18% will be left. Note that the expressions e^(-2 ln(2)) and e^(-2.5 ln(2)) reduce to .5^2 and .5^(2.5), since e^(-ln(2)) = 1/2.
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16:50:31 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Apparently the last problem given is missing some information. Very interesting stuff. It's amazing what can be done with differential equations (and I'm sure this stuff is just the tip of the iceberg). I'm afraid that the tip of that iceberg is absorbing way too much energy and is melting way too fast.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ???Only question I have is how we ended up with –k ???? Rest of the solution is understood. ------------------------------------------------ Self-critique Rating:2.5 ********************************************* Question: "