SOLUTION: CALCULUS(MAXIMA AND MINIMA):THE VOLUME OF A SPHERE IS INCREASING AT THE RATE OF 6CM^2/HR.AT WHAT RATE IS ITS SURFACE AREA INCREASING WHEN THE RADIUS IS 40 CM?

Algebra ->  Test -> SOLUTION: CALCULUS(MAXIMA AND MINIMA):THE VOLUME OF A SPHERE IS INCREASING AT THE RATE OF 6CM^2/HR.AT WHAT RATE IS ITS SURFACE AREA INCREASING WHEN THE RADIUS IS 40 CM?      Log On


   

Question 1045812: CALCULUS(MAXIMA AND MINIMA):THE VOLUME OF A SPHERE IS INCREASING AT THE RATE OF 6CM^2/HR.AT WHAT RATE IS ITS SURFACE AREA INCREASING WHEN THE RADIUS IS 40 CM?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

.

Related rates:

V = (4/3)pir^3

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

6 =(4/3)pi* 2r^2 dr/dt

3/(4/3)pir^2 = dr/dt

9%2F%284pi%2Ar%5E2+%29= dr/dt

S = 4pir^2

dS/dt = 4pi 2r dr/dt  substitute for dr/dt to solve

dS/dt =4pi%2A2r+%289%2F%284pi%2Ar%5E2%29%29

dS/dt = 18/r

substitute: r = 40cm