Calculus help: cone (v=1/3pir^2h) is increasing at the rate of 4pi in^3 / sec

[Juliaaa]

The volume v of a cone (v=1/3pir^2h) is increasing at the rate of 4pi cubic inches per second. At the instant when the radius of the cone is 2 inches, it's volume is 8pi cubic inches and the radius is increasing at 1/3 inches per second.
A. At the instant when the radius of the cone is 2 inches, what is the rate of change of the area of the base?
B. At the instant when the radius of the cone is 2 inches, what is the rate of change of its height h?
C. At the instant when the radius of the cone is 2 inches, what is the instantaneous rate of change of the area of its base with respect to its height h?

 

[Deleted member 4993]

The volume v of a cone (v=1/3pir^2h) is increasing at the rate of 4pi cubic inches per second. At the instant when the radius of the cone is 2 inches, it's volume is 8pi cubic inches and the radius is increasing at 1/3 inches per second.
A. At the instant when the radius of the cone is 2 inches, what is the rate of change of the area of the base?
B. At the instant when the radius of the cone is 2 inches, what is the rate of change of its height h?
C. At the instant when the radius of the cone is 2 inches, what is the instantaneous rate of change of the area of its base with respect to its height h?

V = π/3 * r² * h

dV/dt = π/3 * [2 * r * h * dr/dt + r² * dh/dt]

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