# Spherical Tank Filling / Related Rates Problem

• Jul 28th 2009, 11:25 AM
nickerus
Spherical Tank Filling / Related Rates Problem
This is straight out of a Calc 1 text, and I am having trouble figuring it out. I'm assuming I shouldn't have to integrate, since it's from calc 1.

"Water is being pumped into a spherical tank of radius 60 feet at the constant rate of 10 cubic feet per second. Find the rate at which the radius of the top level of water in the tank changes when the tank is half full."

It should be 0, since that's where the radius will begin to shrink again, but I can't prove it mathematically.
• Jul 28th 2009, 11:45 AM
ANDS!
You don't need to integrate anything.

First we want the volume of a sphere, which is:

$\displaystyle V = \frac{4}{3}\pi(r)^3$

We are tasked with finding a rate of change, which means we are going to be dealing with derivatives.

We are told that $\displaystyle \frac{d(V)}{dt}$ is equal to "10 cubic feet per second". We need to find $\displaystyle \frac{dr}{dt}$.

Taking the derivative of the volume of a sphere, and differentiating it with respect to time, will yield us an equation, that we can solve for $\displaystyle \frac{dr}{dt}$.

If you need further help lemme know.
• Jul 28th 2009, 01:14 PM
nickerus
This is exactly what I thought to do, but when solving for dr/dt, I got some random answer. I suppose I could be deriving incorrectly; I'll just show you my work.

$\displaystyle V=\frac{4}{3} \pi r^3$
$\displaystyle \frac{dV}{dt}=4 \pi r^2 \frac{dr}{dt}$
$\displaystyle 10=4 \pi r^2 \frac{dr}{dt}$
$\displaystyle \frac{dr}{dt}=\frac{10}{4 \pi r^2}$

and at r=60, where the tank is half full, this comes out to about .00022105, which is close to 0, but not 0. There lies my problem.
• Jul 28th 2009, 02:01 PM
ANDS!
Hopefully my logic isn't wrong here:

Well if you plug in

$\displaystyle 10=4\pi r^2 \frac{dr}{dt}$

with $\displaystyle \frac{dr}{dt}$ equal to zero, you will immediately see that we've got ourselves a computational problem.

I understand the "logic" problem you are having with this, in that at some point the radius has to STOP expanding, and start contracting. I mean there has to be a maximum to the radius right?

If we look at an equation that we define as the radius (using available information), we can see that:

$\displaystyle r=\sqrt[3]{\frac{3(V)}{2\pi}}$ which is the radius defined by the volume of semi-circle, then we can see from the graph that ther is no absolute maximum or minimum for the radius. If there were, then we could say that the rate of change of the radius never hits zero, but comes quite close.
• Jul 28th 2009, 02:26 PM
nickerus
The answer in the back of the book is zero, and it follows logically that it would be zero. For instance, if we graphed $\displaystyle r(t)$, we would see that $\displaystyle r'(t)$ has to be zero at some point (the half-way point). It curves up, slowly levels off (where $\displaystyle r'(t)=0$), and then descends.

I think the problem is in the logic of taking the derivative of the Volume formula. If we were asking, for instance, the rate of expansion of the sphere being filled with gas, then this could apply; but we are asking based on the pooling of water at the bottom of a spherical tank, and the radius of the surface area of said pool.

I googled around and found something about a spherical cap, a term I'd never heard before: Spherical Cap -- from Wolfram MathWorld. I think this may apply here in some way, mostly just because it's impossible for the rate of change of our radius to never be zero.
• Jul 28th 2009, 02:54 PM
ANDS!
What equation are you using with the radius defined in terms of time?
• Jul 28th 2009, 03:15 PM
Calculus26
The fallacy in the previous posts is that V =4/3 pi *r^3

As the tank fills the volume of fluid is not a sphere

See attachment for solution