# How do you find the rate at which water is pumped into an inverted conical tank that has a height of 6m and a diameter of 4m if water is leaking out at the rate of 10,000(cm)^3/min and the water level is rising 20 (cm)/min?

Mar 2, 2015

This question has already been answered although you seem to be missing the height of the water in the cone at the time the water level is rising $20 \frac{c m}{\min}$.
Assuming this question came from the same source, the specified height of water was $2 m$ or $200 c m$.

The cone has a radius of 2 m (half the diameter) and a height of 6 m
for a ratio of $\frac{r a \mathrm{di} u s}{h e i g h t} = \frac{1}{3}$

This ratio is constant for volumes of water contained in the cone,

Therefore the volume of the cone (or water in the cone), normally written as
$V \left(r , h\right) = \frac{\pi {r}^{2} h}{3}$

can be re-written as
$V \left(h\right) = \frac{\pi {\left(\frac{h}{3}\right)}^{2} h}{3}$
$= \frac{\pi {h}^{3}}{27}$

and therefore
$\frac{d V \left(h\right)}{\mathrm{dh}} = \frac{\pi}{9} {h}^{2}$ $\frac{c {m}^{3}}{c m}$

We are told
$\frac{d h}{\mathrm{dt}} = 20 \frac{c m}{\min}$

The increase in volume contained in the cone is given by
$\frac{d V}{\mathrm{dh}} \times \frac{d h}{\mathrm{dt}}$ at water level height of $200 c m$

$= \frac{\pi}{9} {\left(200 c m\right)}^{2} \times 20 \frac{c m}{\min}$
$= 2 , 792 , 527 \frac{c {m}^{3}}{\min}$ (approx. assuming I haven't slipped up somewhere)

The inflow of water must be the total of
the outflow (leakage)
plus
the amount needed to raise the water level:
$10 , 000 \frac{{\left(c m\right)}^{3}}{\min}$
$+ 2 , 792 , 527 \frac{{\left(c m\right)}^{3}}{\min}$

$= 2 , 802 , 527 \left({\left(c m\right)}^{3} / \min\right)$