Polynomial Problem: Solving for Radius of Gas Tank Volume

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Homework Help Overview

The discussion revolves around a problem involving the volume of a gas tank, which consists of a right cylinder and two hemispherical ends. The original poster is attempting to find the radius of the tank given its total volume of 50 cubic meters and its length of 10 meters.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formulation of a cubic equation derived from the volume of the tank. There are attempts to isolate the variable r and questions about the methods available for solving cubic equations, including the cubic formula and numerical methods like Newton's method.

Discussion Status

Several participants have contributed different perspectives on solving the cubic equation. Some have pointed out potential errors in the original setup, while others suggest numerical methods as viable alternatives. The discussion reflects a mix of algebraic and numerical approaches without reaching a consensus on a single method.

Contextual Notes

Participants note that the problem is from a pre-calculus context, and there is an emphasis on understanding the underlying mathematics rather than memorizing procedures. There are indications of uncertainty regarding the application of the cubic formula and the effectiveness of numerical methods.

mouseman
I'm stumped! I'm on this question in my math book that reads something like this:

"A gas tank that is 10 meters in length (end to end) consists of a right-cylinder and is capped at either end by a hemisphere. What is the radius of the tank if the volume is 50 cubic meters?"

Ok i got as far as
4/3[pi]r^2(r + 15/2) = 50

but I can't seem to figure out how to isolate r. I know I'm overlooking something mundane, but can someone give me a hint?

Tanks! Ha ha ha! (get it? Tanks?)

Sorry.
 
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You have a cubic equation:
r3+ (15/2)r2- 75/(2[pi])=0.

I don't think there is a simple way to solve that. There is, of course, the "cubic formula" but that's not going to be easy here. You could also use a numerical method like "Newton's method".
 
Originally posted by HallsofIvy
You have a cubic equation:
r3+ (15/2)r2- 75/(2[pi])=0.

Ok yeah I got that too.

Originally posted by HallsofIvy
I don't think there is a simple way to solve that. There is, of course, the "cubic formula" but that's not going to be easy here. You could also use a numerical method like "Newton's method".

By "cubic formula" you mean x3 + y3 = (x + y)(x2 - 2xy + y2) or whatever it is?
 
Last edited by a moderator:
This is from a pre-calculus book in a chapter before one titled "Finding factors and zeros of polynomials." (i.e. polynomial division)
 
As far as I can see, you've made an error at the beginning.
Sphere volume: 4/3*Pi*r^3
Cylinder vol.: Pi*r^2*h
The equation should be: 4/3*Pi*r^3+Pi*r^2*(10-2*r)=50
... (errors possible)
r^3-15*r^2+75/Pi=0

After this... the "easy way" is with the cubic formula. The one you'll find useful.(googled)
 
hi buddy
the answer using calculator is r= 1.1731 meters to four places of decimals.you can do it without using calculator by pen-and-paper iteration(i used calc. for exactly the same thing).
1)start with r=1.
2)use r^2 = 2/15*(75/2pi - r^3) to get new r
3)repeat 2) until convergence occurs(in some cases it may not converge but here it does converges.see any text-book for more info on convergence)

numerical methods are much more useful nowadays than analytic ones due to their (almost) infinite range of application .
 
Yeah I used a calculator to find it out too but I was hoping I could do it algebraically...
See I'm just trying to understand the math, I don't want to end up memorizing it. I just figured with the information in all the previous chapters I could do it without any "advanced" math.
But maybe my search is in vain.
 

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