# Homework Help: Related rates and a spherical weather balloon

1. Jul 20, 2011

### 1irishman

(b]1. The problem statement, all variables and given/known data[/b]

A spherical weather balloon has a radius of 1m when it is 1500m high.
You observe that the radius increases at a rate of 2cm/min as it continues to rise.
At what rate is the surface area increasing when the radius is 4m?

2. Relevant equations

I thought volume of a sphere might be useful to help solve this question.
V = 4/3pi(r^3)

3. The attempt at a solution
I was not sure if the first sentence of the question could help me solve this problem or not, so i decided to try and solve it without using the info from there.

It looks like from the given information that the following might be the case:

if i let r be the radius then r = 4
the rate at which the radius increases with respect to time could be shown as:
dr/dt = 2

Now, this is where i get stuck; i am not sure if i am correct in making my next step to differentiate the volume of a sphere function, but this is what i arrived at:
if V=4/3pi(r^3)
THEN
dV/dt=4pi(r^2)
I am stuck at this point, because i don't know how to take it further, help please?

Last edited: Jul 20, 2011
2. Jul 20, 2011

### gdbb

You need to know the equation for the surface area of a sphere:

$$SA = 4 \pi r^2$$

3. Jul 20, 2011

### 1irishman

thank you for the formula.

Are my calculations correct?

DS/dt=8(pi)(r)dr/dt
=8(pi)(4)(2)
=201.062cm^2/min

Last edited: Jul 20, 2011
4. Jul 22, 2011

### gdbb

Oh, almost! Try adding units into your calculation to see if you can figure out where you went wrong.

5. Jul 22, 2011

### 1irishman

2cm/min is 5.556X10^-6m/s^2

6. Jul 22, 2011

### 1irishman

ooops...4m = 400cm
so,
the result is 20106.2cm^2/min

7. Jul 22, 2011

### 1irishman

in the future, how do i know which unit to convert?