# Thread: Related Rates with Spheres

1. ## Related Rates with Spheres

Here's the question I have:

The radius of a sphere is increasing at a constant rate of 2cm/sec. At the instant when the volume of the sphere is increasing at 32pi cm^3/sec, the surface area of the sphere is

A) 8pi
B) 32pi/3
C) 16pi
D) 64pi
E) 256pi/3

So far I have dr/dt=2, and when dv/dt=32, SA=?

V=4/3pir^3
SA=4pir^2

I'm not sure how to relate these two equations. I assume I would solve for the radius of the sphere first, by finding the derivative of the Volume equation: dv/dt=4pir^2dr/dt, but I don't know what to do after that.

2. ## Re: Related Rates with Spheres

Given ...

$\dfrac{dr}{dt}=2 \, cm/sec$, $\dfrac{dV}{dt} = 32\pi \, cm^3/sec$

$\dfrac{dV}{dt} = \color{red}{4\pi r^2} \cdot \dfrac{dr}{dt}$

3. ## Re: Related Rates with Spheres

Originally Posted by Velmahr
Here's the question I have:

The radius of a sphere is increasing at a constant rate of 2cm/sec. At the instant when the volume of the sphere is increasing at 32pi cm^3/sec, the surface area of the sphere is

A) 8pi
B) 32pi/3
C) 16pi
D) 64pi
E) 256pi/3

So far I have dr/dt=2, and when dv/dt=32, SA=?

V=4/3pir^3
SA=4pir^2

I'm not sure how to relate these two equations. I assume I would solve for the radius of the sphere first, by finding the derivative of the Volume equation: dv/dt=4pir^2dr/dt, but I don't know what to do after that.
$\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}=32\pi$ solve for $r$

4. ## Re: Related Rates with Spheres

$\frac{dV}{dt}= \frac{d((4/3)\pi r^3)}{dt}= \frac{d((4/3)\pi r^3)}{dr}\frac{dr}{dt}= (4\pi r^2)(2)$.