at what rate is the volume of the sphere changing at the instant when the surface area is increasing at the rate 4 square centimeters per minute

Question

at what rate is the volume of the sphere changing at the instant when the surface area is increasing at the rate 4 square centimeters per minute and the radius is increasing at the rate of 0.1 centimeters per minute?

Answer 1

The volume (V) and surface Area (A) of a sphere are:

 

     V = (4/3)*pi*r³

     A = 4*pi*r²

 

Their rates of change wrt time (t) are (use the chain rule):

 

     dV/dt = (dV/dr)*(dr/dt) = 4*pi*r²*dr/dt

 

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

 

Now dA/dt is given as 4 cm²/sec and dr/dt is given as 0.1 cm/sec.  Substitute those values into the dA/dt equation and solve for r:

 

     dA/dt = 8*pi*r*dr/dt

     4 = 8*pi*r*(0.1)

     4/(8*pi*0.1) = r

     5/pi = r

 

Now substitute r=5/pi and dr/dt=0.1 into dV/dt:

 

      dV/dt = 4*pi*r²*dr/dt

      dV/dt = 4*pi*(5/pi)²*(0.1)

 

     dV/dt = 10/pi  cm³/sec

Answer 2

This is a related rates problem.

 

The area of a sphere, A, is   A = 4 pi r²    where r is the radius.

 

   Thus the rate of change of areas is    dA/dt =  8 pi r   (dr/dt).   So

 

        4 = 8 pi r (.1)     which implies that    r = 5/pi

 

The volume of a sphere is   (4 pi /3) r³  

     Thus  dV/dt = 4 pi r² (dr/dt) =  4 pi (5/pi)² (.1) =  10/ pi ~ 3.18