Question

A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate is the height of the water changing when the height is 3m?

Answer 1

Answer:

Therefore the rate change of height is  $\frac{1}{\pi}$ m/s.

Step-by-step explanation:

Given that a vertical cylinder is leaking water at rate of 4 m³/s.

It means the rate change of volume is 4 m³/s.

$\frac{dV}{dt}=4 \ m^3/s$

The radius of the cylinder remains constant with respect to time, but the height of the water label changes with respect to time.

The height of the cylinder be h(say).

The volume of a cylinder is $V=\pi r^2 h$

                                                 $=( \pi \times 2^2\times h)\ m^3$

$\therefore V= 4\pi h$

Differentiating with respect to t.

$\frac{dV}{dt}=4\pi \frac{dh}{dt}$

Putting the value $\frac{dV}{dt}$

$\Rightarrow 4\pi \frac{dh}{dt}=4$

$\Rightarrow \frac{dh}{dt}=\frac{4}{4\pi}$

$\Rightarrow \frac{dh}{dt}=\frac{1}{\pi}$

The rate change of height does not depend on the height.

Therefore the rate change of height is  $\frac{1}{\pi}$ m/s.