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:
[tex]\dfrac{dr}{dt} = 12\text{ cm per sec}\\\\\dfrac{dh}{dt} = 12\text{ cm per sec}[/tex]
Volume of cone =
[tex]V = \dfrac{1}{3}\pi r^2 h[/tex]
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 =
[tex]\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})[/tex]
Putting values, we get,
[tex]\dfrac{dV}{dt} = \dfrac{\pi}{3}(2(6)(12)(9) + (6)^2(12))\\\\\dfrac{dV}{dt} =1808.64\text{ cubic cm per second}[/tex]
Thus, the volume of cone is increasing at a rate 1808.64 cubic cm per second.
