Question

The radius of a right circular cone is increasing at a rate of 1.4 in/s while its height is decreasing at a rate of 2.4 in/s. At what rate is the volume of the cone changing when the radius is 140 in. and the height is 186 in.

Answer 1

Answer:

The volume is increasing at a rate of 27093 cubic inches per second.

Step-by-step explanation:

Volume of a cone:

THe volume of a cone, with radius r and height h, is given by:

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

In this question:

We have to differentiate implictly is function of t, so the three variables, V, r and h, are differenciated. So

$\frac{dV}{dt} = \frac{\pi r^2}{3}\frac{dh}{dt} + \frac{2\pi rh}{3}\frac{dr}{dt}$

The radius of a right circular cone is increasing at a rate of 1.4 in/s while its height is decreasing at a rate of 2.4 in/s.

This means that $\frac{dr}{dt} = 1.4, \frac{dh}{dt} = -2.4$

Radius is 140 in. and the height is 186 in.

This means that $r = 140, h = 186$

At what rate is the volume of the cone changing?

$\frac{dV}{dt} = \frac{\pi r^2}{3}\frac{dh}{dt} + \frac{2\pi rh}{3}\frac{dr}{dt}$

$\frac{dV}{dt} = \frac{\pi (140)^2}{3}(-2.4) + \frac{2\pi 140*186}{3}1.4$

$\frac{dV}{dt} = -0.8\pi(140)^2 + 62*2\pi*1.4*140$

$\frac{dV}{dt} = 27093$

Positive, so increasing.

The volume is increasing at a rate of 27093 cubic inches per second.