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 2.3 in/s. At what rate is the volume of the cone changing when the radius is 110 in. and the height is 151 in.

Answer 1

Answer:

The volume is changing at a rate given by:

$\frac{dV}{dt} =19559.56\,\,\frac{in^3}{s}$

Step-by-step explanation:

Let's recall the formula for the volume of acone, since it is the rate of the cone changing what we need to answer:

Volume of cone = $\frac{1}{3} B\,*\,H$

where B is the area of the base (a circle of radius R) which equals = $\pi\,R^2$

and where H stands for the cone's height.

We apply the derivative over time operator ($\frac{d}{dt}$) on both sides of the volume equation, making sure that we apply the rule for the derivative of a product:

$V=\frac{1}{3} B\,*\,H\\\\V= \frac{1}{3} \pi\,R^2\,H\\\frac{dV}{dt} =\frac{\pi}{3}\,( \frac{d(R^2)}{dt} H+R^2\,\frac{dH}{dt} )\\\frac{dV}{dt} =\frac{\pi}{3}\,( 2\,R\,\frac{dR}{dt}\, H+R^2\,\frac{dH}{dt} )\\\frac{dV}{dt} =\frac{\pi}{3}\,( 2\,(110\,in)(1.4\,\frac{in}{s} )\,(151\,in)+(110\.in)^2\,(-2.3)\frac{in}{s} )\\\frac{dV}{dt} =19559.56\,\,\frac{in^3}{s}$