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

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

Explanation:

We are given the following in the question:

`(dr)/(dt) = 12\text{ cm per sec}\\\\(dh)/(dt) = 12\text{ cm per sec}`

Volume of cone =

`V = (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 =

`(dV)/(dt) = (d)/(dt)((1)/(3)\pi r^2 h)\\\\(dV)/(dt) = (\pi)/(3)(2r(dr)/(dt)h + r^2(dh)/(dt))`

Putting values, we get,

`(dV)/(dt) = (\pi)/(3)(2(6)(12)(9) + (6)^2(12))\\\\(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.