DISSOLVE BRIEF ABSTRACT Simple differential equation models for dissolution of spherical material in solution are compared. GENERAL INFORMATION FileName: DISSOLVE Full title: Comparing two models for a dissolving capsule Last Revision Date: 26 May 1996. Developer: Robert Feller Richmond High School, Richmond IN 47374 USA. Contact: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Aaron D. Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Aaron.Klebanoff@Rose-Hulman.Edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering. STATEMENT OF PROBLEM Part A - A spherical capsule has volume V and surface area S. Assume that this capsule dissolves in a certain solution such that the rate of change in volume, dV/dt, is directly proportional to the capsule's surface area. Show that dV/dt = - k (V^(2/3)) for some k > 0. Part B - Assuming k = .4836, and t is measured in minutes, if the original volume of the capsule is 300 ml., how long will it take the capsule to lose half its volume? Part C - Assuming the same size capsule as in part B, what if the change in volume of this second capsule were instead directly proportional to its volume, and after 2 minutes, only 75 ml of the capsule volume still remained. This might occur if the capsule is permeated by liquid and the small granuals contained in the capsule are each dissolving on their own. How long would it take this capsule to lose half its volume? Part D - If the two capsules in part B and C start to dissolve at the same time, when will the difference in their volumes be the greatest? Part E - For the two capsules in part B and C, when is the rate of change of the volumes of both capsules the same? KEYWORDS Differential equation model, exponential decay, optimization. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Students need to be able to solve or use CAS to solve first order differential equations and to optimiza a function of one variable. Time allotment - time management This should not take more than one hour if the students are proficient at solving the differential equations and comparing the answers in two cases. Expectations Students will be able to solve first order separable differential equation and manipulate the answers Future payoffs Several modeling hypotheses are made and no judgement on either is offered. Students should be encouraged to conjecture models and see the consequences of their conjuegure. Moreover, several conjectures might lead to reasonable physical solutions. Extensions Further complications of the models might be offered. References and Sources POSSIBLE SOLUTION(S) Part A Take the volume formula for a sphere and solve it for r. rsol=Solve[ V[t]==4/3 Pi r[t]^3, r[t] ] 3 1/3 1/3 {{r[t] -> (----) V[t] }, 4 Pi 1/3 3 1/3 1/3 {r[t] -> -((-1) (----) V[t] )}, 4 Pi 2/3 3 1/3 1/3 {r[t] -> (-1) (----) V[t] }} 4 Pi Then determine a formula for surface area in terms of V(t). We then substitute this into our differential equation model which says dV/dt is directly proportional to the capsule's surface area. SA[t_] = 4 Pi r[t]^2/.rsol[[1]] 1/3 2/3 (36 Pi) V[t] This yields a differential equation in V[t]. sol = DSolve[V'[t] == - k (36 Pi)^(1/3) V[t]^(2/3), V[t],t] 1/3 -(k (36 Pi) t) 3 {{V[t] -> (----------------- + C[1]) }, {V[t] -> 0}} 3 Which we solve and whose constant of integration we determine in terms of the initial condition. vsol[t_] = V[t]/.sol[[1]] 1/3 -(k (36 Pi) t) 3 (----------------- + C[1]) 3 csolve = Solve[vsol[0]==v0,C[1]] 1/3 1/3 1/3 {{C[1] -> v0 }, {C[1] -> -((-1) v0 )}, 2/3 1/3 {C[1] -> (-1) v0 }} And hence we have a model of the volume of the capsule as a function of time t. vol[t_] = vsol[t]/.csolve[[1]] 1/3 -(k (36 Pi) t) 1/3 3 (----------------- + v0 ) 3 Part B We enter given parameters into our volume formula. v[t_] = vol[t]/.{k->.4836,v0->300} 1/3 1/3 3 (300 - 0.1612 (36 Pi) t) solt = Solve[v[t]==0,t] {{t -> 8.58733}, {t -> 8.58733}, {t -> 8.58733}} endTime = t/.solt[[1]] 8.58733 Plot[v[t],{t,0,endTime},PlotRange->{0,300}] -Graphics- Thus we can determine how long it takes the capsule to drop from 300 ml to 150 ml, i.e half its volume. It appears to be 1.77 min. Solve[v[t]==150,t] {{t -> 1.77156}, {t -> 11.9952 - 5.90263 I}, {t -> 11.9952 + 5.90263 I}} Part C solC = DSolve[{vol2'[t]==-k2*vol2[t],vol2[0]==300},vol2[t],t] 300 {{vol2[t] -> -----}} k2 t E vol2[t_]=vol2[t]/.solC[[1]] 300 ----- k2 t E This is an exponential decay function. Now find k2. solk2 = NSolve[vol2[2]==75,k2] {{k2 -> 0.693147}} We seek the time it takes to go from 300 ml to 150 ml. That is apparently t = 1 min. v2[t_]=vol2[t]/.solk2[[1]] 300 ----------- 0.693147 t E FindRoot[v2[t]==150,{t,2}] {t -> 1.} Part D Let's look at the graph of the two functions and also the graph of their difference. Plot[{v[t],v2[t]},{t,0,endTime}] -Graphics- voldiff[t_]=v[t]-v2[t]; Plot[voldiff[t],{t,0,endTime}] -Graphics- To determine when the two volume difference is greatest we3 set the difference volume function's derivative equal to 0. soldiff = FindRoot[voldiff'[t]==0,{t,1.6}] {t -> 1.57179} This is the value of t that would represent the time the volumes differ the most. And they differ by 62.66 ml at time t = 1.57 min. voldiff[t]/.soldiff[[1]]//N 62.6627 Part E Since the first derivative gives us the rate of change of the volumes we seek when these two functions are equal. p3=Plot[{v'[t],v2'[t]},{t,0,endTime}] When the difference of the two functions is zero, the two functions have the same value. FindRoot[v'[t] == v2'[t],{t,2.5}] {t -> 1.57179} FindRoot[v'[t] == v2'[t],{t,9}] {t -> 9.10313} There are two values of t for which the first derivatives are equal but only one in the realistic domain before one of the capsule models predicts the capsule is all gone. And hence it is at this one time, t = 1.57 min, that the capsules are shrinking at the same rate. ISSUES IN SOLUTION