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In a book there is a derivation for y' that comes from

$$\frac{y'}{[1+(y')^2]^{1/2}} = c,$$

where $c$ is a constant. The result they had was

$$y' = \sqrt{\frac{c^2}{1-c^2}}.$$

How did they get this? I tried expanding the square, and other tricks, I cant seem to get their result.

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This formula holds if $c\geq 0$. More precise result is $y'=\frac{c}{\sqrt{1-c^2}}$ –  Norbert Jan 25 '12 at 21:43

1 Answer 1

up vote 4 down vote accepted

Square both sides, solve for ${y^{\prime}}^2$ and then take the square root.

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Id done that, I just forgot to reverse the division of y'^2's, so they would cancel out and give 1+1/y'^2 = 1/c^2. Now it works. –  BB_ML Feb 8 '11 at 11:31

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