The radius r(t) of the base of a cone is increasing at a rate of 10 meters per second. The height h(t) of the cone is fixed at 6 meters. At a certain instant , the radius is 1 meter. What is the t 0 rate of change of the volume V(t) of the cone at that instant?

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Answer:

dV/dt = 40π

Step-by-step explanation:

We are told that The radius r(t) of the base of a cone is increasing at a rate of 10 meters per second. Thus;

dr/dt = 10 m

Height: h = 6 m

Volume of cone is given by the formula;

V = ⅓πr²h

dV/dr = ⅔πrh

We want to find the rate at which the volume is changing at radius of 1m.

Thus;

dV/dt = (dV/dr) × (dr/dt)

dV/dt = ⅔π(1 × 6) × (10)

dV/dt = 40π