Related rates and the volume of spheres

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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)
 
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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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