Differential Eqns

• May 4th 2009, 06:44 PM
ny_chow
Differential Eqns
At time t, t $\ge$ 0, the volume of a sphere is increasing at a rate proportional to the reciprocal of its radius. At t = 0, the radius of the sphere is 1 and at t = 15, the radius is 2. (Volume of sphere is V = 4/3pir^3)

a) Find the radius of the sphere as a fuction of t

b) at what time t will the volume of the sphere be 27 times its volume @ t= 0?
• May 4th 2009, 07:33 PM
redsoxfan325
Quote:

Originally Posted by ny_chow
At time t, t $\ge$ 0, the volume of a sphere is increasing at a rate proportional to the reciprocal of its radius. At t = 0, the radius of the sphere is 1 and at t = 15, the radius is 2. (Volume of sphere is V = 4/3pir^3)

a) Find the radius of the sphere as a fuction of t

b) at what time t will the volume of the sphere be 27 times its volume @ t= 0?

a.) It's best to list know information first. We know that:
$\frac{dV}{dt}=\frac{k}{r}$ where $k$ is a constant.
$V=\frac{4}{3}\pi r^3$
$\frac{dV}{dr}=4\pi r^2$

$\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt} \implies \frac{k}{r} = 4\pi r^2\cdot\frac{dr}{dt} \implies \frac{dr}{dt}=\frac{k}{4\pi r^3}$

Separate the variables and integrate: $\int 4\pi r^3\,dr = \int k\,dt \implies \pi r^4 = kt+C$.

$t=0, r=1 \implies \pi=C$

$t=15, r=2 \implies 16\pi=15k+\pi \implies k=\pi$

So, $\pi r^4 = \pi t+\pi \implies \boxed{r = \sqrt[4]{t+1}}$

I have to go now. If the second part is still unanswered when I get back, I'll help you out then.
• May 4th 2009, 07:41 PM
ny_chow
i think i can do the second part on my own. it seems pretty straight forward. thanks!