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log growth prob
Posted:
Apr 28, 1998 4:32 PM


Ms. Ghitelman. (I concur with your results.) Concerning your separable differential equation dR/dt=0.001R(200R) with initial conditions R(0)=25. After separating, use partial fractions to integrate 1/(R(200R)) with respect to R and get (1/200)(lnRln(200R))or (1/200)ln(R/(200R)). So the solution to the differential equation is R/(200R)=e^(.2t+k). When t is 0, we have k=ln(1/7), so the implicit relationship between R and t is R/(200R)=(1/7)e^(0.2t). Now when R=50, t is about 4.24 . So it takes about 4.24 weeks to double from 25 rabbits to 50. So the (c) choice is the correct one (unless I errored somewhere along the way.)
Charlie ESA Cade, LA



