Question

A conical tank (as in Example 2) has height 3 m and radius 2 m at the base. Water flows in at a rate of 2 m3/min. How fast is the water level rising when the level is 1 m?

Answer 1

Answer:

$\dfrac{dy}{dt}=1.432\ m/min$

Explanation:

given,

height of cone = 3 m

radius of cone = 2 m

flow rate = 2 m³/min

water level rate when height at 1 m = ?

volume of cone

$V = \dfrac{1}{3}\pi r^2h$

now,

$V = \dfrac{1}{3}\pi r'^2y$

where radius of cone at y depth

$r' = \dfrac{r}{h}y$

so,

$V = \dfrac{1}{3}\pi \dfrac{r^2}{h^2}y^3$

differentiating both side

$\dfrac{dV}{dt}= \pi \dfrac{r^2}{h^2}y^2\dfrac{dy}{dt}$

now, h = 1 m

$2 = \pi \dfrac{2^2}{3^2}\times 1^2\dfrac{dy}{dt}$

$\dfrac{dy}{dt}=1.432\ m/min$