Mathematics: 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

An equation for the absolute value that represents the midpoint of 10 and 15 would be (10 + 15)/2 = 12.5...

The radius of a right circular cone is increasing

madison6592

21.01.2022

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.

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