Question

A cylindrical tank with radius 5 m is being filled with water at a rate of 3 m^3/min. How fast is the height of the water increasing

Answer 1

Answer:

$\frac{3}{25\pi}$ meters per minute

Step-by-step explanation:

Given:

$V=\pi r^2h$

$r=5$

$\frac{dV}{dt}=3$

$\frac{dh}{dt}=?$

Substitute r=5:

$V=\pi r^2h$

$V=\pi(5)^2h$

$V=25\pi h$

Differentiate both sides:

$\frac{d}{dt}V=\frac{d}{dt}(25\pi h)$

$\frac{dV}{dt}=25\pi\frac{dh}{dt}$

Solve for dh/dt:

$3=25\pi\frac{dh}{dt}$

$\frac{3}{25\pi}=\frac{dh}{dt}$

Therefore, the height of the water is increasing at a rate of $\frac{3}{25\pi}$ meters per minute.