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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Amity 3 years 2021-07-30T08:28:20+00:00 1 Answers 55 views 0

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  1. RobertKer
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    2021-07-30T08:30:04+00:00
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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

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