Divide By Zero

Oct 5th, 2009 by Nick

I try not to do this in class – at least not intentionally. However, today in my math competition class we spent a little time working with formulas for Volume and Surface Area. I wanted to ask a set of tough questions. The final question on the sheet is below

The volume of sphere $_1$ is 100 $\text{u}^3$ larger than sphere $_2$ . How much larger is the radius of sphere $_1$ than sphere $_2$ ?

My thinking is that the solutions could be represented on the graph with x as the initial radius and y as the increment necessary to create a difference of 100. I am getting different outputs from the graphing software on my computer and the output from Wolfram Alpha and I’d like to be able to infer coordinates from the graph.

How would you do this?

Posted in everything else

7 Responses to “Asking questions to which you do not know the answer”

on 05 Oct 2009 at 9:02 pm1Jason Dyer

Hmm….

Let r be the radius of sphere 1 and z be the radius of sphere 2.
(4/3 pi r^3) – 100 = (4/3 pi z^3)
r^3 – z^3 = 75/pi
(r-z)(r^2+rz+z^2) = 75/pi
r – z = (75/pi) / (r^2+rz+z^2)

I don’t know if it’s possible to do better.

If you want a graph, try this one.
on 05 Oct 2009 at 10:13 pm2Justin Lanier

Perhaps I’m not understanding your intent, but as stated there isn’t a fixed numerical answer to your problem. The difference between the radii could be as large as the cube root of 75/pi (if the smaller sphere is a point) and anything on down toward zero (because as the inner sphere gets bigger, the shell of volume 100 you’ll have to add on to make the larger sphere will get thinner and thinner–larger surface area implies smaller thickness). If that’s the right interpretation of your question and you buy that argument, maybe graphing isn’t necessary? If an algebraic relationship describing how the difference in the radii depends on the inner radius is what you want, I doubt that things get any better than what Jason has put down.
Is that useful?
on 05 Oct 2009 at 10:55 pm3Mr. H

Looks like you just about covered it.

Graphing the y (increment) as function of x(radius) pretty much gives you the graph.

The only thing I see as an opportunity is to help students build intuition.

If sphere2 has radius zero, what should radius of sphere1 be to get 100?

The solution is about finding the thickness y of a shell with inner radius x so that the shell has a volume of 100.

Explore the effect of increasing inner radius size on the thickness of such a shell. Look at the graph and see if it matches up to intuition.

Explore same concept with different geometric shapes.

Given a hollow cone filled with water and 100 cc of oil so that the oil floats on top. How thick is this film of oil if you’re given the height/distance from the surface of water to the tip of the cone? If additional water is poured into this cone does the film become thinner or thicker?
on 06 Oct 2009 at 12:05 am4Nick

Thanks Jason. Would it be possible to state z as a function of r? So, simply to work from
r^3 – z^3 = 75/pi
r^3 = 75/pi + z^3
r = (75/pi + z^3)^(1/3)
Which looks similar to the solution I linked to originally. This would be useful with the Ti-83/84′s the kids have.
on 06 Oct 2009 at 12:14 am5Nick

Mr. H – maybe I’m missing something, but I don’t see why your equation and mine should work out differently. I really thought I could arrive at $y = \sqrt[3]{\frac{75}{\pi}+x^3}-x$ from the equation I began $\frac{4}{3}\pi{(x+y)}^3 - \frac{4}{3}\pi{x}^3=100$ . Why aren’t these equivalent? Here’s why I think they are.
on 06 Oct 2009 at 8:15 am6Jason Dyer

Nick, the graph you mention is the same one I link to at Wolfram Alpha. That’s what I originally typed in to think about the problem.
on 06 Oct 2009 at 10:59 am7Mr. H

I thought you had a different equation when I clicked on the link to wolfram. It might have something to do with URL parsing.

You’re right. That’s actually the same equation that I started with.

I wrote y as function of x so that you can graph that on a graphing calculator. You can use the trace feature or the table feature to figure out specific points on the graph. Which is what I would do if I wanted to infer coordinates from the graph.