Rates of Change Calculus Question

Calculus
Tutor: None Selected Time limit: 1 Day

The radius of a sphere is r cm at time t seconds. Find the radius when the rate of increase of the surface area is equal to the rate of increase of the radius. 

Nov 29th, 2014

Love it when calculus questions pop up here.

Ok, the surface area of a sphere (SA) is 4*pi*r^2.

SA = 4*pi*r^2

We want to take the derivative of this equation with respect to time:

dSA/dt = 4*pi*2*r*(dr/dt) = 8*pi*r*(dr/dt)

We know from the problem that the rate of increase of the surface area is equal to the rate of increase of the radius.  That is, dSA/dt = dr/dt.  So, I'm going to replace dSA/dt with dr/dt:

(dSA/dt) = 8*pi*r*(dr/dt)

(dr/dt) = 8*pi*r*(dr/dt)

Divide through by dr/dt:

1 = 8*pi*r

Solve for r:

r = 1/(8*pi)



Nov 29th, 2014

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Nov 29th, 2014
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Nov 29th, 2014
Dec 10th, 2016
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