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Re: Dave's calculus problem
Posted:
Aug 3, 1997 6:44 AM
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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,B4-5*B$1)
(C5)Amt in body: =B$4-B5
(E5)Amt in liver: =0.5*D4*B$1+E4
(D5)Amt in blood: =C5-E5
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 <ap-calc@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 power-series
> expand e^-epsilon t = 1 - epsilon t, you get a nice perfect
> first-order result of
>
> A = 5 t
>
> So now I believe the solution (especially if you see that the second
> order term is a just-right -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^(-delta-t / 2))
> where delta-t 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 super-top
> students would be able to solve it on their own.
>
> --Joshua Zucker
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