From the section DESCRIBING HOW THINGS CHANGE

Note that the solutions are written in the format you would actually submit as a solution, namely as a Maple file. Maple commands appear red. Maple responses are black with cursive type. Text comments appear black with typewriter text.

• 2.18:
  Begin by taking the hint: namely use Maple to find the general solution of Newton's Law of Cooling equation which describes how the temperature of a hot body cools to the ambient temperature (i.e. the surrounding temperature).
  

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  Let's begin by explaining what the above solution from Maple means. First, A is the ambient temperature. Our physical intuition tells us that coffee cools to the temperature A and no further, so the solution from Maple must predict a temperature dependence on time which starts at the initial coffee temperature and ends at A. That's precisely what we observe if the value of a is negative. In that case, at t=0, the exponential is 1 and T=A+K (note that we've replaced Maple's constant _C1 with our own constant K), so we define A+K as the initial temperature of the coffee. This defines the value of K. For example, if the coffee starts at 70 degrees Celsius and the room temperature is 20 degrees, then K must be 50. As time increases we can see that, if a is negative, the temperature decreases since the exponential decreases. For any time t > 0, T(t) < A + K. As t -> infinity, T(t) -> A as we know it should.

  The effect of adding cream is to suddenly lower the temperature of the coffee. We can signify this by saying that T is immediately reduced by temperature value C. Hence, in mathematical language, whenever you add cream, T(t) -> T(t) - C. Now, we write down a statement or equation which says what the temperature is for the Prime Minister's coffee (let's call it TPM(t)) is after ten minutes, just before the cream is added. By our general solution, it must be TPM(t) = A + K*exp(a*t) and just after the cream is added, it becomes TPM(t) = A + K*exp(a*t) - C.

  Now the President's coffee gets cooled at the beginning by the addition of cream, so, at t=0, the President's coffee temperature (let's call it TPR) is TPR(t=0) = A + K - C. This is the initial temperature from which the President's coffee temperature will start to cool. Since we already defined the constant that goes into our general description for cooling as the difference between ambient temperature and the starting temperature, the equation which describes the cooling of the President's coffee must be TPR(t) = A + (K - C)*exp(a*t).

  To determine which temperature is greater, we can simply subtract the equations to get TPM(t) - TPR(t) = [A + K*exp(a*t) - C] - [A + (K - C)*exp(a*t)] = C*exp(a*t) - C. Since C is positive and a is negative, the difference is always negative for any value of t greater than 0. Thus the President's coffee is always at greater temperature than the Prime Minister's.

  To ask whether this result is reasonable, we need only note that if we plot the temperature as a function of time, we find a decreasing exponential curve. Remember that a decreasing exponential curve flattens as it approaches its asymptotic value. By lowering the coffee temperature with cream before cooling begins, the President's coffee has a lower rate of cooling than the Prime Minister's, i.e. the Prime Minister's coffee temperature drop over ten minutes is greater than the coffee temperature drop of the President's coffee. When the Prime Minister adds cream at the end of the ten minutes the coffee must inevitably drop below the President's coffee temperature. Try various values of a, K, and C to convince yourself that this is true.

• 2.22:
  
  1. (a.) Since the acceleration is the derivative of velocity with respect to time, we can write down the differential equation describing the motion of the skydiver and ask Maple to dsolve it for us. The equation we want is given in the problem, so we just let the solution be V, the velocity as a function of time.
     

     Now to solve for the constant we need one known value of velocity for a known time. The obvious choice is that we assume that the velocity of the skydiver at time t = 0 is zero, so we can let Maple solve for us
     

     The values of g and vt, the terminal velocity are constant, hence the value of _C1 we need can be determined from the equation
     

     Which, for t=0, yields _C1 = -60. Hence, our expression for V and the plot of V are
     
     

     The plot looks as we expect. The velocity increases rapidly at first, but tapers off to the terminal velocity as time goes on. To get within 5 m/s of terminal velocity, the speed of the skydiver must be 60 - 5 = 55 m/s, so
     

     So after 15.2 seconds, the skydiver is at terminal velocity.

     (b.) The distance fallen as a function of time can be derived once the velocity as a function of time is known since the velocity is the derivative of distance with respect to time. Hence,
     

     Again, we need to solve for the constant _C1 by noting that the distance fallen is zero at time zero, so
     

     Since x(t=0) = 0, _C1 = -367.3469 and the equation for the distance fallen as a function of time and the plot of same are
     
     

     Again, the plot is as expected. Distance fallen rises quadratically for small times, but then increases linearly after a while since the velocity is nearly constant (terminal velocity). We can find the altitude of the skydiver at any time by subtracting distance fallen from 10,000 feet, the starting altitude. Maple will tell us the time at which the skydiver reaches 500 feet altitude. Note that we need to be careful to convert from feet to meters! (1 meter equals 3.281 feet)
     

     Given the quadratic dependence of altitude on time there are two mathematical solutions. Only the positive time is allowed in our universe so the maximum time of fall is 54.4 seconds.

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• 2.23:
  
  1. (a.) Since the acceleration is the derivative of velocity with respect to time, we can write down the differential equation describing the motion of the skydiver and ask Maple to dsolve it for us just as in the previous problem. Remember to clear the values of g and vt first (i.e. unassign them by g := 'g': vt := 'vt':). Also, we will want to put in the initial condition that v(t=0) = 0 here to save the trouble of solving for it later.
     

     The plot of velocity as a function of time is
     

     The plot looks as we expect. The velocity increases rapidly at first, but tapers off to the terminal velocity as time goes on. In comparison to the previous problem, we see that the rise to terminal velocity occurs faster here since the resistance force rises as the square of the velocity rather than just as the velocity. To get within 5 m/s of terminal velocity, the speed of the skydiver must be 60 - 5 = 55 m/s, so
     

     So after 9.6 seconds, the skydiver is at terminal velocity. This is considerably shorter than the time for the previous problem due to the sharper increase in resistance force with velocity.

     (b.) The distance fallen as a function of time can be derived once the velocity as a function of time is known since the velocity is the derivative of distance with respect to time. Again, we make use of the initial condition that x(0) = 0, i.e. the distance fallen at time t = 0 is 0. Hence,
     

     The plot of distance fallen is
     

     Again, the plot is as expected. Distance fallen rises quadratically for small times, but then increases linearly after a while since the velocity is nearly constant (terminal velocity). The time for quadratic behavior is small since the resistance force rises so rapidly. We can find the altitude of the skydiver at any time by subtracting distance fallen from 10,000 feet, the starting altitude. Maple will tell us the time at which the skydiver reaches 500 feet altitude. Note that we need to be careful to convert from feet to meters! (1 meter equals 3.281 feet). An fsolve does the trick
     

     So the time of fall is 52.5 seconds. Note that this is close to the time for the previous problem. This is expected since the effect of the v^2 resistance force is only to get the skydiver up to terminal velocity more quickly. Since the skydiver is high enough that most of the journey down is at constant velocity, then the fall time for the v^2 resistance is only slightly less than that for the v resistance. Note that this is why the v resistance is considered a "good enough" approximation in the previous problem.

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