# I have tried methods but still can't get correct answer, Can give me some helps??

## Water is leaking out of an inverted conical tank at a rate of 14800.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 10.0 meters and the diameter at the top is 3.0 meters. If the water level is rising at a rate of 16.0 centimeters per minute when the height of the water is 3.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute

Nov 2, 2017

#### Explanation:

The volume of water in the tank is

$V = \frac{9 \pi}{400} {h}^{3}$.

The volume of water is changing at a rate of

$\frac{\mathrm{dV}}{\mathrm{dt}} = \frac{27 \pi}{400} {h}^{2} \frac{\mathrm{dh}}{\mathrm{dt}}$

When $h = 3.5 m = 350 \left(c m\right)$ and $\frac{\mathrm{dh}}{\mathrm{dt}} = 16 \frac{c m}{\min}$, we get

$\frac{\mathrm{dV}}{\mathrm{dt}} = \frac{27 \pi}{400} \left({350}^{2}\right) \left(16\right) \frac{c {m}^{3}}{\min}$

The rate of change of volume is the difference between the rate at which water is being pumped in minus the rate of leakage.

So the rate at which it is being pumped in is:

$\frac{27 \pi}{400} \left({350}^{2}\right) \left(16\right) + 14800$ $\frac{c {m}^{3}}{\min}$

(do the arithmetic I get $430432.7$ $\frac{c {m}^{3}}{\min}$ )

Nov 2, 2017

I guess is 1.53xx10^3 cm^3//min

#### Explanation:

Yo have a large cone of $h = 10$ and $d = 3.0 m$.

The smaller cone (filled with water ) will have the same proportions as the larger one. So

$10 \to 3$
$3.5 \to d$, gives $d = \frac{3.5 \cdot 3}{10} = 1.05 m$

At this time (when the radius is 52.5 cm, the
$\Delta h = 16 c m / \min$. The added volume at this time is

$\pi \cdot \Delta h \cdot {52.5}^{2} = 138544 c {m}^{3} / \min$

Since the volume in and the volume out are constant we can say that the rate as the water is being pumped into the cone vessel is

$138544 + 14800 = 1.53 \times {10}^{3} c {m}^{3} / \min$