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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
Question
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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Mathematics
3 years
2021-07-30T08:28:20+00:00
2021-07-30T08:28:20+00:00 1 Answers
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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