Related rates and a spherical weather balloon

[1irishman]

(b]1. Homework Statement [/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?

Homework Equations

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

 

[gdbb]

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

[tex]SA = 4 \pi r^2[/tex]

 

[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

 

[gdbb]

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

 

[1irishman]

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

 

[1irishman]

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

 

[1irishman]

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