Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Re: Dave's calculus problem
Posted:
Aug 3, 1997 6:44 AM


I agree with both of Joshua's answers (to tenths, since I used a spreadsheet with dt=.1).
Thanks, Joshua. Looks like superb analysis.
BTW, the graph of amount_in_liver vs. time is logistic (in appearance).
I built my spreadsheet around the following formulas:
(B5)Amt in muscle: =IF(B4<=0,0,B45*B$1) (C5)Amt in body: =B$4B5 (E5)Amt in liver: =0.5*D4*B$1+E4 (D5)Amt in blood: =C5E5
where B$1=dt=.1 and B$4=10
Dave Slomer AP Calculus and Computer Science Teacher Winton Woods HS, Cincinnati, OH 45240  > From: Joshua Zucker <zucker@csli.Stanford.EDU> > To: Multiple recipients of list <apcalc@ets.org> > Subject: Dave's calculus problem > Date: Thursday, July 31, 1997 3:43 PM > > I think Dave Slomer's calculus problem is a good one. > > I certainly hope it's not beyond the reach of the typical BC student > at least! > > Just for a refresher, the question asks: > > >Here's a good problem from Dick and Patton (page 547, sec 8.4): > > > >"An injection of a particular medication is absorbed into the blood at a > >rate of 5 mg/hr. The same medication is removed by the liver at a rate > >proportional to the amount in the blood (0.5 mg/hr per mg in the > >blood). If an injection of 10 mg is given, sketch the graph of the amount > >in the blood over the next 5 hours. When does the amount in the blood > >peak? How much is in the blood after 5 hours?" > > From t = 0 to 2, the differential equation (with A = amount in blood > in mg) is > > dA/dt = 5  .5A > and since A is less than 10, this is a strictly increasing function. > > From t = 2 to 5, the differential equation is > dA/dt = .5A > which is clearly strictly decreasing. > > So, we have two problems to solve: what happens from t = 0 to 2, and > what happens from 2 to 5. > > Both equations are variables separable (the second one is particularly > easy). > > Solving the first one gave an answer which surprised me: > > A = 10 (1  e^(t/2)) > > This sure doesn't look like something that would work in general! > OK, it does start at zero, but it seems like it would increase at > whatever rate it feels like. What happened to the 5? > > So to check my answer, I thought about changing the .5A term. What I > was worried about was what happens if it's  epsilon A instead. The > solution I got in terms of epsilon is > > A = 5/epsilon (1  e^(epsilon t)) > > That 5/epsilon sure looks worrisome, but then if you powerseries > expand e^epsilon t = 1  epsilon t, you get a nice perfect > firstorder result of > > A = 5 t > > So now I believe the solution (especially if you see that the second > order term is a justright 5 * epsilon * t^2). > > So back to business. We want to find A(2). Calculator time, unless > we happen to know the reciprocal of e offhand, > A(2) = 10*(1  e^(1)) = 6.32 approximately. > > Now the equation for t > 2 is just A = A_0 * (e^(deltat / 2)) > where deltat is the time since t = 2. So at t = 5 we have > > A = 6.32 * (e^(3/2)) = about 1.41 > > This makes some sense; half the stuff is being removed each hour, so > it should go down by a factor of 8, only the rate is constantly > slowing down, so the factor of 8 is an overestimate. > > So, a challenging but doable problem for a BC calc student I should > think. > > For a middle school student, of course you'd do it by Euler's method, > or slope fields :) I think a talented middle school algebra student > who's had a course heavy on graphing, slope, plotting points and so > forth would be able to understand an explanation of how to make a > numerical approximation to this kind of problem. Only supertop > students would be able to solve it on their own. > > Joshua Zucker



