A spherical snowball is being made so that its volume is increasing at the rate of 8ft^3/min/ Find the rate at which the radius is increasing when the snowball is 4t in diameter.

Question

A spherical snowball is being made so that its volume is increasing at the rate of 8ft^3/min/ Find the rate at which the radius is increasing when the snowball is 4t in diameter.

Show Complete Solution:

Defined variables-                                                                           

Numerical facts known about the variables and their

           derivatives with respect to t-                                                

The unknown (what you wish to find)-                                           

The equation to relate the variable that depend on t-                       

Final answer (includes the solution on differentiation)-                   

Interpretation-                                                                                  

Answer 1

Oh, related rate problems are fun.

The volume of a sphere is

V=4/3 pi r^3

If we differentiate both sides with respect t, we get

dV/dt=4/3 pi 3 r^2 dr/dt=4 pi r^2 dr/dt

We're looking to find the rate of change for the radius, dr/dt, so let's go ahead and solve for that now

dr/dt=1/(4 pi r^2) dV/dt

The problem asks us for the rate of change of the radius of the snowball when its diameter is 4t (what is t here? Do you mean 4 ft? That's what I'll assume for my answer), meaning the radius is 2ft. Substituting what we have then

dr/dt=1/(4*pi*(2)^2) * 8 = 8/(16pi)=1/(2pi) ft/min