Question

The radius of a right circular cone is increasing at a rate of 1.1 in/s while its height is decreasing at a rate of 2.6 in/s. At what rate is the volume of the cone changing when the radius is 107 in. and the height is 151 in.

Answer 1

Answer:

The volume of the cone is increasing at a rate of 1926 cubic inches per second.

Step-by-step explanation:

Volume of a right circular cone:

The volume of a right circular cone, with radius r and height h, is given by the following formula:

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

Implicit derivation:

To solve this question, we have to apply implicit derivation, derivating the variables V, r and h with regard to t. So

$\frac{dV}{dt} = \frac{1}{3}\left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right)$

Radius is 107 in. and the height is 151 in.

This means that $r = 107, h = 151$

The radius of a right circular cone is increasing at a rate of 1.1 in/s while its height is decreasing at a rate of 2.6 in/s.

This means that $\frac{dr}{dt} = 1.1, \frac{dh}{dt} = -2.6$

At what rate is the volume of the cone changing when the radius is 107 in. and the height is 151 in.

This is $\frac{dV}{dt}$. So

$\frac{dV}{dt} = \frac{1}{3}\left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right)$

$\frac{dV}{dt} = \frac{1}{3}(2(107)(151)(1.1) + (107)^2(-2.6))$

$\frac{dV}{dt} = \frac{2(107)(151)(1.1) - (107)^2(2.6)}{3}$

$\frac{dV}{dt} = 1926$

Positive, so increasing.

The volume of the cone is increasing at a rate of 1926 cubic inches per second.