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I am doing calculus integration, and need to show my work for Horizontal slicing (even though Vertical slicing is far easier).

The equation is $$y= x/\sqrt{2x^2+1}$$

I need to rewrite the equation so that it is $x=\;...$ in order to horizontally slice it (in other words, it should be rewritten so that it is dependent on $y$). This isn't exactly a calculus question, although it is being used for calculus. I'm probably missing something that is pretty obvious. Any help would be greatly appreciated!

2 Answers 2

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Heres a tip: Take reciprocals. Notice $$\left( \frac{1}{y}\right)^2=\left( \frac{\sqrt{2x^2+1}}{x}\right)^2=\frac{2x^2+1}{x^2}=2+\frac{1}{x^2}$$

Then we get $$\frac{1}{x^2}=\frac{1}{y^2}-2$$

Take reciprocals again and we find $$x^2=\frac{1}{\frac{1}{y^2}-2}$$

Take square roots, and we are finished.

Hope that helps

Edit: Just to make things complete I decided to add the final line: $$x=\pm \sqrt{\frac{y^2}{1-2y^2}}$$

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    Thanks for the help! I can't believe that I missed this solution with the reciprocal.2011-02-24
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    @Isaac: You are correct! Should be fixed.2011-02-24
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    Yes, you got it.2011-02-24
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If you are working on an interval where $x\ge 0$, $$\begin{align} \frac{x}{\sqrt{2x^2+1}}&=\sqrt{\frac{x^2}{2x^2+1}} \\ &=\sqrt{\frac{2x^2+1-1}{2(2x^2+1)}} \\ &=\sqrt{\frac{1}{2}\left(1-\frac{1}{2x^2+1}\right)} \end{align}$$ Now that there's only one instance of $x$ on the right side, you should be able to more easily solve for $x$.