Fill in the Blank

Question

Your neighbor is growing a slightly different watermelon. It also has a rind whose thickness is one tenth of the radius of that watermelon. However, the rind of your neighbor's water melons grows at a constant rate of 20 cubic centimeters a week. The radius of your neighbor's watermelon after 5 weeks is _________ centimeters and at that time it is growing at ______ centimeters per week.

Answer 1

Hi Makena L.,

Actually, this is an interesting problem, unlike most of what's asked on Wyzant (i.e. students who didn't read the textbook, and want to be told what to answer!). The reason I find it interesting, is that it teaches you HOW to go about solving a problem. So -- First, you set up the mathematical situation, namely, the surface area of the melon times the rate of radius change of the rind (that's instantanteous rate of volume change of the rind!) is increasing at 20 cc/wk. So 4*.pi.*r^2 [note: can treat this as constant r, once the melon has a finite size!] * d(r/10)/dt = 20 . Where r = total radius of the melon. looks like r^3 is proportional to t; so

assume r = k*(t)^(1/3). Then r^2 = k^2* t^(2/3) . At t=5, have d(r/10)/dt = (k/30)t^(-2/3) =(k/30)*5^(-2/3), and r^2 = k^2*5^(2/3); Rate of rind volume change = 4/30*.pi.*k^2*5^(2/3)*(k/3)*5^(-2/3) = 4/30.pi.k^3 = 20, then k = (150*.pi.)^(1/3) . I think. I kept making silly arithmetic errors, and finally just solved for the constant k AFTER the rind-volume-rate-change calculation.

I suppose you can take it from there?

--Cheers, --Mr. d.

P.S. Something that initially threw me, was that it appeared that the rate of radius change would have had to

be infinite at t=0, from that constant rate of volume increase of rind at t=0. But, that's a consequence of the mathematical model, and real conditions don't have to obtain at boundaries in models!