Expert: Paul Klarreich - 12/6/2007

Question
At time t, t is greater than or equal to zero, the volume of a sphere is increasing at a rate proportional to the reciprocal of its radius.  At t=0 the radius of sphere is 1 and at t=15 the radius is 2.  

A.) Find the radius of the sphere as a function of t.
B.) At time t will the volume of the sphere be 27 times its volume at t=0?

Answer
Questioner:   William
Category:  Calculus
Private:  No
 
Subject:  AP Calculus
Question:  At time t, t is greater than or equal to zero, the volume of a sphere is increasing at a rate proportional to the reciprocal of its radius.  At t=0 the radius of sphere is 1 and at t=15 the radius is 2.  

A.) Find the radius of the sphere as a function of t.
B.) At time t will the volume of the sphere be 27 times its volume at t=0?
...........................................................

Hi, William,

Your sentence, 'the volume of a sphere is increasing at a rate proportional to the reciprocal of its radius'
says:

dV    k
-- = ---
dt    r

That's a differential equation, which we usually solve by integrating. But V and r are not the same thing, so first we use the chain rule:

dV   dV dr
-- = -- --
dt   dr dt

and, since V of a sphere =
   4 pi r^3
V = --------
      3
dV
-- = 4pi r^2
dr

Now our DE looks like this:

       dr    k
4pi r^2 -- = ----
       dt    r

Separate the variables:

4pi r^3 dr = k dt

Integrate:

pi r^4 = kt + C, constant of integration.

Now we want values of k and C, and we need two facts, such as:
 
At t=0 the radius of sphere is 1

pi(1)^4 = k(0) + C

C = pi, so

pi r^4 = kt + pi

at t=15 the radius is 2.  

pi(2)^4 = k(15) + pi

16 pi = 15k + pi

15 pi = 15k

k = 1

OKAY, then:

pi r^4 = t + pi
     t + pi
r^4 = ------- or t/pi + 1
       pi

r = (t/pi + 1)^1/4   << your value of r as f(t)
===================================================
B.) At time t will the volume of the sphere be 27 times its volume at t=0?

HUH?  Did you mean to write:

At WHAT time t will the volume of the sphere be 27 times its volume at t=0?

At t = 0, r = 1 and  V = 4pi/3

27 times that is  V = 36pi

4pi r^3
-------- = 36pi
  3

r^3 = 27, or r = 3

Now solve  f(t) = 3:

(t/pi + 1)^1/4 = 3

t/pi + 1 = 3^4 = 81

t/pi = 80

t = 80 pi

AP Calculus, you said?  I wonder who writes this stuff.