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The curve is the intersection of: $$4x=(y+z)^2$$ $$4x^2+3y^2=3z^2$$

And the interval of curve length is from $O(0,0,0)$ to $M(x,y,z)$

The answer is $\sqrt2 z$

My substitution is $u=y+z$ and $v=z-y$, then I put them into these two equations generating the relation of $u$ and $v$. However when I go down the track, trying to write the $dL$ to get the length, I found that the integral is complex to resolve. Did I go wrong or something else?

Zhen Zhang
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1 Answers1

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By inspection, one branch of the solution is $x=0$ and $y=-z$. This is simply a line with with slope $-1$ in the $yz$-plane and, of course, the segment from the origin to a point $(0,-z,z)$ along this line has length $\sqrt{2}\,z$.

There is, however, another branch along which the problem could be harder. The two branches are shown in the figure below. My solution is for the diagonal line.

enter image description here

Mark McClure
  • 30,510
  • Thank you! And this is really a tricky skill. I search the wolfram for the $4x=(y+z)^2$ and it show solution is $x=0, y=-z$. I feel pretty strange what does the solution means here? I suppose x be any positive number. – Zhen Zhang Jul 01 '14 at 03:05
  • @Sacheo You're welcome, though I wouldn't be so fast to accept this answer - I think it's missing something and someone else may spot it.. You can always upvote without accepting. – Mark McClure Jul 01 '14 at 03:12