Question

The radius of a right circular cone is increasing at a rate of 3 inches per second and its height is decreasing at a rate of 3 inches per second. At what rate is the volume of the cone changing when the radius is 40 inches and the height is 30 inches

Answer 1

Answer:

The volume of cone is increasing at a rate 2512 cubic inches per second.                                  

Step-by-step explanation:

We are given the following in the question:

$\dfrac{dr}{dt} = 3\text{ inches per second}\\\\\dfrac{dh}{dt} = -3\text{ inches per second}$

Radius = 40 inches    

Height = 30 inches

The volume of cone is given by:

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

Rate of change of volume is given by:

$\dfrac{dV}{dt} = \dfrac{1}{3}\pi (2r\dfrac{dr}{dt}h+\dfrac{dh}{dt}r^2)$

Putting the values, we get,

$\dfrac{dV}{dt} = \dfrac{1}{3}(3.14) \big(2(40)(3)(30)+(-3)(40)^2\big)\\\\\dfrac{dV}{dt} = \dfrac{1}{3}(3.14)(2400)\\\\\dfrac{dV}{dt} = 2512$

Thus, the volume of cone is increasing at a rate 2512 cubic inches per second.