# Thread: calculus ab word problem

1. ## calculus ab word problem

At time t, t is greater than or equal to 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 he radius is 2 (the volume V of a sphere with radius r is V=4/3 pi*r^3
a) Find the radius of this sphere as a function of t.
b) At what time t will the volume of the sphere be 27 times its volume at t=0?

I'm completely lost and don't even know where to start. i've been looking at this for about 15 and really need some help. Any pointers would be great

2. volume of a sphere is increasing at a rate proportional to the reciprocal of its radius
$\frac{dV}{dt} = \frac{k}{r}$

using the formula for a sphere, $V = \frac{4}{3} \pi r^3$ , ...

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

so ...

$4\pi r^2 \frac{dr}{dt} = \frac{k}{r}$

separate variables ...

$4\pi r^3 \, dr = k \, dt$

integrate and get r as a function of t

3. so when i integrate is it
pi*r^4 = kt+c
if i solve for c is it pi?

so r= the fourth root of [(1/pi) kt+1]

is that right or am i completely off track?

4. ## r = fourth root of [(1/pi) kt + 1]

You are on the right track. Just make sure that in the equation

r = fourth root of [(1/pi) kt + 1], the kt and 1 are separate, and only the kt is being multiplied to the (1/pi).

other then that, you are correct.

,

,

,

### at time t volume of sphere is increasing at rate proportional to reciprocal of its radius

Click on a term to search for related topics.