Will the Inverted Cone Water Tank Overflow?

[amcavoy]

A water tank is in the shape of an inverted cone with depth 10 meters and top radius 8 meters. Water is flowing into the tank at 0.1 cubic meters/min but leaking out at a rate of 0.001h^2 cubic meters/min, where h is the depth of the water in meters. Will the tank ever overflow?

Thoughts:

V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (\frac{4}{5}h)^{2}h

\frac{dV}{dt}=\frac{16}{25}\pi h^{2}\frac{dh}{dt}

Now I replace \frac{dV}{dt} with 0.1-0.001h^2. This is where I am stuck. Any suggestions?

Thanks.

 

[kleinwolf]

This just gives a separable diff equ...

It can be solved but why solve it ? :

If the water reaches a highest point this means that h(t) has a maximum...

dh/dt=0->h=10m...which curiously is exactly the height of the cone

Is it a maximum ? It turns out the second derivative is zero, the third derivative is zero...aso..

So does it overflow ?

 

[M98Ranger]

To determine if the tank will ever overflow, we need to find the depth of the water when the inflow rate equals the outflow rate. This is when 0.1-0.001h^2 = 0. This can be solved by setting the equation equal to 0 and using the quadratic formula to find the roots. The roots are h=10 and h=-10. Since the depth can't be negative, the only possible solution is h=10 meters. This means that the depth of the water will reach 10 meters, but not exceed it. Therefore, the tank will never overflow.