The radius of a cylinder is decreasing at the rate of 4 ft/min, while the height is increasing at the rate of 2 ft/min. What is the rate of change of the volume when the radius is 2 feet and the height is 6 feet?

I got -88pi ft^3/min. Is that correct?

1 Answer
Jan 26, 2018

The volume is decreasing at the rate of #=276.5ft^3mn^-1#

Explanation:

The radius of the cylinder is #=r#

The height of the cylinder is #=h#

The volume of the cylinder is

#V=pir^2h#

Differentiating with respect to #t#

#(dV)/dt=pih2r(dr)/dt+pir^2(dh)/dt#

#(dr)/dt=-4ftmn^-1#

#(dh)/dt=2ftmn^-1#

The radius is #r=2ft#

The height is #h=6ft#

#(dV)/dt=pi(2*2*6*(-4)+2^2*2)=-276.5ft^3mn^-1#