Question

The volume of a cone of radius r and height h is given by V=πr²h³. If the radius and the height both increase at a constant rate of 12 cms/sec, at what rate, in cubic centimeters per second, is the volume increasing when the height is 9 centimeters and the radius is 6 centimeters?

Answer 1

Answer:

The volume of cone is increasing at a rate 1808.64 cubic cm per second.

Step-by-step explanation:

We are given the following in the question:

$\dfrac{dr}{dt} = 12\text{ cm per sec}\\\\\dfrac{dh}{dt} = 12\text{ cm per sec}$

Volume of cone =

$V = \dfrac{1}{3}\pi r^2 h$

where r is the radius and h is the height of the cone.

Instant height = 9 cm

Instant radius = 6 cm

Rate of change of volume =

$\dfrac{dV}{dt} = \dfrac{d}{dt}(\dfrac{1}{3}\pi r^2 h)\\\\\dfrac{dV}{dt} = \dfrac{\pi}{3}(2r\dfrac{dr}{dt}h + r^2\dfrac{dh}{dt})$

Putting values, we get,

$\dfrac{dV}{dt} = \dfrac{\pi}{3}(2(6)(12)(9) + (6)^2(12))\\\\\dfrac{dV}{dt} =1808.64\text{ cubic cm per second}$

Thus, the volume of cone is increasing at a rate 1808.64 cubic cm per second.