# Thread: Depth of water when the water level is falling the least rapidly

1. ## Depth of water when the water level is falling the least rapidly

Can someone help me with this problem? Thank you.

A spherical tank of radius 10ft is being filled with water. When it is completely full, a plug at its bottom is removed. According to Torricelli's law, the water drains in such a way that dV/dt = -k*sqrt(y), where V is the volume of water in the tank and k is a positive empirical constant.

a) Find dy/dt as a function of the depth y
b) Find the depth of water when the water level is falling at least rapidly (you will need to compute the derivative of dy/dt with respect to y)

Regarding a), I have done this:
V=1/3*pi*y^2(3a-y) where a is the radius
dV/dt=10*pi*2y(dy/dt) - (1/3)*pi*3y^2(dy/dt)
-k*sqrt(y) = (dy/dt)(20*pi*y - pi*y^2)
dy/dt=-[(k*sqrt(y))/pi*y(20-y)]

And this is what I came up for dy/dt as a function of the depth y

But I am not sure how to proceed with point b) Any help will be apprecited.

2. Maybe I am misunderstanding, but it appears you're using the equation for the volume of a cone........the tank is spherical.

dV equals the area of the liquid multiplied by dy.

Try to equate the area of the liquid in terms of h.

By the diagram, since the tank is a sphere, the surface area of the liquid is a circle with radius

$\displaystyle \sqrt{25-(y-5)^{2}}$

The area of a circle is $\displaystyle {\pi}r^{2}$, so we have the area of the surface of the liquid at time t as

$\displaystyle {\pi}(\sqrt{25-(y-5)^{2}})^{2}=10{\pi}y-{\pi}y^{2}$

So, we have $\displaystyle (10{\pi}y-{\pi}y^{2})\frac{dy}{dt}=-k\sqrt{y}$

$\displaystyle \frac{10{\pi}y-{\pi}y^{2}}{\sqrt{y}}dy=-kdt$

Now, we have a separated DE.

3. V=1/3*pi*y^2(3a-y) is the volume of a spherical tank that is partly filled - this way I have the depth of the sphere in the equation.

4. Oh, I see. At first glance it looked like the cone formula. My bad.

5. I just saw the rest of your reply. I see that I am on the right way.

I now wonder how to find the depth of water when the water level is falling at least rapidly. They suggest to compute the derivative of dy/dt with respect to y. Does this mean that I have to take the derivative of the derivative and why it will give me the depth of water when the water level is falling at least rapidly.

I just did the b). For those interested I took the derivative of dy/dt and found one critical value. Thank you for the help!