Question

The radius of the base of a cylinder is increasing at a rate of 111 meter per hour and the height of the cylinder is decreasing at a rate of 444 meters per hour.At a certain instant, the base radius is 555 meters and the height is 888 meters.What is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)?

Answer 1

The volume is decreasing by the rate of 20 π cubic metre per hour.

Step-by-step explanation:

Since, the volume of a cylinder,

$V=\pi r^2h$

Where,

r = radius

h = height

Differentiating w. r. t. t (time)

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

We have,

$\frac{dr}{dt}=1\text{ m per hour}, \frac{dh}{dt}=-4\text{ m per hour}, r=5\text{ m},h=8\text{ m}$

By substituting the values,

$\frac{dV}{dt}=\pi(25(-4)+2(5)(8)(1))$

$\frac{dV}{dt}=\pi(-100+80)$

$\frac{dV}{dt}=-20\pi$

Hence, the volume is decreasing by the rate of 20 π cubic metre per hour.

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