Related rates and the volume of spheres

[howsockgothap]

Homework Statement

The volume of a spherical balloon is increasing at a rate of 4m^3/min. How fast is the surface area increasing when the radius is three meters?

Homework Equations

V=4/3piR^3
A=4piR^2

The Attempt at a Solution

V=s.a.*R/3
dv/dt=d(s.a.R/3)/dt
dv/dt=(d(s.a.R/3)/dR)*(dR/dt)

 

[Je m'appelle]

Try this

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

$$\frac{dA}{dt} = \frac{d}{dt}(4\pi r^2)$$

Knowing that

$$\frac{dV}{dt} = 4 \ m^3 \ min^{-1}$$

And

$$r = 3 \ m$$

It's a simple plug-in values problem, solve the first equation for dr/dt then plug-in the value found into the second equation in order to find dA/dt, there's no mistake.

Give it a try.