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I have the following literal equation that needs to be solved for h:

$$S= \pi r \sqrt{r^2+h^2} $$

I isolated the square root and got this:

$$\frac{S}{ \pi r} = \sqrt{r^2+h^2}$$

Then I squared both sides to eliminate the square root on the right:

$$\frac{S^2}{ \pi^2 r^2} =r^2+h^2$$

Then I isolated the h term:

$$\frac{S^2}{ \pi^2 r^2} - r^2 =h^2$$

Now I am not quite sure what to do. I assume that it is more complex than just square rooting everything to get:

$$\frac{S}{ \pi r} -r=h$$

An algebraic calculator says the solution is:

$$h= \frac{ \sqrt{- \pi^2r^4+S^2} }{ \pi r} and \; h= -\frac{ \sqrt{- \pi^2r^4+S^2} }{ \pi r}$$

But I don't understand how it computed that solution. Could anyone give me a step-by-step explanation of your solution?

ProfessorStealth
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  • It should be $h=\pm \sqrt{\frac{S^2}{\pi^2 r^2}-r^2}$ – Alex Jan 26 '14 at 00:12
  • @Alex It's the same thing after putting everything with the same denominator. – Git Gud Jan 26 '14 at 00:12
  • @Git Gud Yes, but that's where he made the mistake. – Alex Jan 26 '14 at 00:13
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    @ProfessorStealth You're making two mistakes. One is assuming that $x^2=a\implies x=\sqrt a$. Correct would be $x^2=a\implies x=\pm \sqrt a$. The other is that $\sqrt{x+y}\neq \sqrt{x}+\sqrt y$. – Git Gud Jan 26 '14 at 00:15

1 Answers1

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$$S= \pi r \sqrt{r^2+h^2} $$ $$\frac{S}{\pi r}=\sqrt{r^2+h^2} $$ $$\frac{S^2}{(\pi r)^2}=r^2+h^2 $$ $$\frac{S^2}{(\pi r)^2}-r^2=h^2 $$ $$h=\pm\sqrt{\frac{S^2}{(\pi r)^2}-r^2}$$ $$h=\pm\sqrt{\frac{S^2-(\pi r)^2r^2}{(\pi r)^2}}$$ $$h=\pm\frac{\sqrt{S^2-\pi^2 r^4}}{\pi r}$$

Adi Dani
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  • Thank you for the solution. But how would I go about giving $r^2$ a common denominator with the other terms? Also, would this be necessary or just additional work? – ProfessorStealth Jan 26 '14 at 00:24