#$&* course mth 174 6/10 6:45pm Calculus IIAsst # 19
.............................................
Given Solution: ** The equation is dM/dt = r * M. The question is not posed for the specific value r = .05. **
.........................................
11:39:48
......!!!!!!!!...................................
**** What is the solution to the equation?
......!!!!!!!!...................................
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).
.........................................
11:40:09
......!!!!!!!!...................................
**** Describe your sketches of the solution for interest rates of 5% and 10%.
......!!!!!!!!...................................
11:40:34 &&&***** to compare the graphs we take the value of m(t) at rate .05 and rate .10 At r=.05 So we get at t(0) 1000e^.05*0=1000 so this point is (0,1000) At t(1) we get 1000e^.05=1051.27 so this point is (1, 1051.77) At r=.10 We get at t(0) 1000e^0=1000 so this point is (0, 1000) At t(1) we get 1000e^.10=1105.17 so this point os (1, 1105.17) We can see that the graph when rate=.10 is increasing faster than the graph when rate is .05*********** 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).
.........................................
11:40:34
......!!!!!!!!...................................
**** Does the doubled interest rate imply twice the increase in principle?
......!!!!!!!!...................................
11:40:42 &&&****the doubled interest rate does not imply twice the increase in principle because the difference between t=1 at .05 and t=1 at .10 is a difference of around $50.00 which isn’t doubling the principle.******* 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:3 ********************************************* 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: We are told that at 1pm the power goes out and at this time t(0) the temperature in the house is 68 degrees. We are also told that at 10pm the temperature outside is 10 degrees and the temperature inside is 57 degrees, 10pm would represent t(8). We are first asked to make a prediction and tell how we made it for the temperature inside the house at 7am. 7am would be t(18) because it is 18 hours after clock time 0. So at 1pm which is clock time 0, the temperature is 68 degrees in the house. After 9 hours, the temperature is 57 degrees which is a drop of 11 degrees over 9 hours or about .818 degrees per hour. So to find an estimate of the temperature at 7am we would multiply 18*.818=14.72 and subtract from initial temperature of 68 degrees and get an estimate of the temperature inside the house at 7am to be 53 degrees. To find the diferential equation, we use the formula for newton’s law of heating/cooling which says that rate of change of temperature is equation to constant K *(temperature difference) So we have dT/dt=k*(t-10) rearranging we get dT/(t-10)=Kdt integrating both sides we get ln absolute value of (T-10)= Kt+c T-10=e^kt*e^c let A equal arbitrary e^c with c greater than 0 T=10+ae^kt We know that at t(0) the temperature in the house is 68 degrees so T(0) =68 we also know 9 hours later at t(9) the temperature is 57 degrees so t(9)=57 So from this we can make 2 equations 68=10+Ae^k(0) = 58 and 57=10+Ae^k*9 Solving for A in the first equation we get 68=10+Ae^k(0) = 58 so A=58 now since we know A we can plug its value into the 2nd equation getting 57=10+58e^9k 58e^95=47 E^9k=47/58 9k=ln of absolute value of .8103 9k=-.2104 K=-.0234 So know we have t(t)=10+ae^kt T(t)=10+58e^-.0234t So to solve to find the temperature in the house at 7am, which is 18 hours after time 0 and thus t(18) we have T(18)=10+58e^-.0234*18 48.06 so 48 degrees would be the temperature in the house at 7am. After solving the differential equation and getting the temperature at 7am to be 48 degrees my estimate was 53 degrees so I overestimated. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.
.........................................
11:41:36
......!!!!!!!!...................................
**** What is your differential equation?
......!!!!!!!!...................................
11:41:47 ****&&&&dT/dt = k(T - 10)****
.........................................
11:41:48
......!!!!!!!!...................................
Using newtons law of cooling and heating which is a constant K * the temperature change which is this case is K(t-10) because the temperature given at 10pm is 10 degrees
......!!!!!!!!...................................
11:42:03 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ###Am I correct on understanding newton’s law of heating and cooling that rate of change of temperature is equal to a constant K*temperature change??????? ------------------------------------------------ Self-critique Rating:3
.............................................
Given Solution: **** What is the differential equation you would solve to get W(t) for intake I cal/day?
......!!!!!!!!...................................
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): ### when it comes to the + or - from the absolute value I must have used the incorrect one because I got A=-10 when I know A should =10. At this point would I need to go back and solve for the opposite sign of the equation or would I be able to assume that if one sign gives me A=-10 the other sign would give me A=10. I wasn’t clear on this during the notes, I was thinking it didn’t make a difference which one you used?????? ------------------------------------------------ Self-critique Rating:3
.............................................
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): ##### I don’t follow this step 0 = a - 1 / (k * 0 + c) so that c = 1 / a. I get 0=(a-1)/c 0=a-1 a=1 because of the 0 I multiply by c and get 0 and thus can’t solve for c. How do you get c=1/a????? ------------------------------------------------ Self-critique Rating:3
.............................................
Given Solution:
......!!!!!!!!...................................
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): ### I followed and understood up until the last step Minimum escape velocity occurs when vzero^2 = 2gR, Thus minimum escape velocity = sqrt (2gR). How did we determine the minimum escape velocity is it because the v subzero ^2 -2gR is not affecting the h values as h approaches infinity of 2gR^2/(R+h) if so why don’t we include the (R+h) denominator when finding the minimum escape velocity???? ------------------------------------------------ Self-critique Rating:3
.............................................
Given Solution: what is your solution to the differential equation R' = `sqrt( 2 G M0 / R ), R(0) = 0?
......!!!!!!!!...................................
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): ###Is my solution ok or do I need to take further steps???? ------------------------------------------------ Self-critique Rating:3
.............................................
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.
.................................................
......!!!!!!!!...................................
16:50:31 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> 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.
.................................................
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ###I was surprised to see the many practical uses of differential equations, very enlightening query!!*** ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!