Question

The radius r(t)r(t)r, (, t, )of the base of a cylinder is increasing at a rate of 111 meter per hour and the height h(t)h(t)h, (, t, )of the cylinder is decreasing at a rate of 444 meters per hour. At a certain instant t_0t 0 ​ t, start subscript, 0, end subscript, the base radius is 555 meters and the height is 888 meters. What is the rate of change of the volume V(t)V(t)V, (, t, )of the cylinder at that instant?

Answer 1

Answer:

The volume is decreasing at a rate 20π cubic meter per hour.

Step-by-step explanation:

We are given the following in the question:

The radius is increasing at a rate 1 meter per hour.

$\dfrac{dr}{dt} = 1\text{ meter per hour}$                

The height is decreasing at a rate 4 meter per hour  

$\dfrac{dh}{dt} = -4\text{ meter per hour}$

At an instant time t,

r = 5 meter

h = 8 meters

Volume of cylinder =

$\pi r^2 h$

where r is the radius and h is the height of the cylinder.

Rate of change of volume is given by:

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

Putting all the values we get,

$\dfrac{dV}{dt} = 2\pi (5)(8)(1) + \pi (5)^2(-4)\\\\\dfrac{dV}{dt} = 80\pi - 100\pi = -20\pi \approx -62.8$

Thus, the volume is decreasing at a rate 20π cubic meter per hour or 62.8 cubic meter per hour.