Question

To make the cone, melted chocolate is pumped onto a huge cold plate at a rate of 2 ft3/sec. Due to the low temperature, the chocolate forms the shape of a cone as it solidifies quickly. If the height is always equal to the diameter as the cone is formed, how fast is the height of the cone changing when it is 5 ft high?

Answer 1

Answer:

The height of cone is increasing at a rate 0.102 feet per second.

Step-by-step explanation:

We are given the following in the question:

$\dfrac{dV}{dt} = 2\text{ cubic feet per second}$

Instant height = 5 feet

The height of the cone is always equal to the diameter.

Volume of cone =

$V = \dfrac{1}{3}\pi \dfrac{d^2}{4}h\\\\\text{where d is the diameter and h is the height of cone}\\\\V = \dfrac{1}{12}\pi h^3$

Rate of change of volume =

$\dfrac{dV}{dt} = \dfrac{d}{dt}(\dfrac{1}{12}\pi h^3)\\\\\dfrac{dV}{dt} =\dfrac{\pi}{4}h^2\dfrac{dh}{dt}$

Putting all the values, we get,

$2=\dfrac{\pi}{4}(5)^2\dfrac{dh}{dt}\\\\\Rightarrow \dfrac{dh}{dt} = \dfrac{8}{25\pi} =0.102$

Thus, the height of cone is increasing at a rate 0.102 feet per second.