Water is flowing into a conical cistern at the rate of 5 m^3/min. The height of the cone is 10 m and the radius of the upper base 5 m.

$\dfrac{y}{x} = \dfrac{10}{5}$

$y = 2x$
 

 

$V = \frac{1}{3}\pi x^2 y$

$V = \frac{2}{3}\pi x^3$

$\dfrac{dV}{dt} = 2\pi x^2 \dfrac{dx}{dt}$
 

When y = 6 m, x = 3 m
$5 = 2\pi (3^2) \dfrac{dx}{dt}$

$\dfrac{dx}{dt} = 0.0884 ~ \text{m/min}$

$\dfrac{y}{x} = \dfrac{10}{5}$

$y = 2x$

$\dfrac{dy}{dt} = 2 \dfrac{dx}{dt}$
 

 

When y = 6 m, x = 3 m
$Q = vA$

$Q = \dfrac{dy}{dt} \pi x^2$

$5 = \left( 2\dfrac{dx}{dt} \right) \pi (3^2)$

$\dfrac{dx}{dt} = 0.0884 ~ \text{m/min}$