How do you find the rate at which the volume of a cone changes with the radius is 40 inches and the height is 40 inches, where the radius of a right circular cone is increasing at a rate of 3 inches per second and its height is decreasing at a rate of 2inches per second?

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Answer 1 · 2015-03-22T19:10:58

#V=1/3 pi r^2 h# Differentiate implicitly with respect to #t#

#d/dt (V)=d/dt(1/3 pi r^2h)=1/3 pi d/dt(r^2h)#

We'll need the product rule on the right.

#(dV)/(dt)=1/3 pi[2r(dr)/(dt)h+r^2(dh)/(dt)]#

(I use the product rule in the form #(FS)'=F'S+FS'#)

Substitute the given information and do the arithmetic.

#(dV)/(dt)=1/3 pi[2(40)(3)(40)+(40)^2(-2)]#

#=1/3 pi[6(40)^2-2(40)^2]=1/3 pi [4 (40)^2]=(6400 pi)/3# cu in / s