Find the radius of curvature of the parabola y^2 = 4px at (0,0).

Please teach me how to solve this question. Thank you very much.

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- Feb 8th 2007, 07:05 AMJenny20radius of curvature
Find the radius of curvature of the parabola y^2 = 4px at (0,0).

Please teach me how to solve this question. Thank you very much. - Feb 8th 2007, 11:07 AMJenny20
Please check my work. The answer of this question is 2 lpl.

I got the answer is 2p. Could you please help me find out the reason? Thank you very much. - Feb 8th 2007, 11:07 AMCaptainBlack
In this case take the curvature as:

$\displaystyle

\kappa=x''/(1+x')^{3/2}\,

$

where we consider $\displaystyle x\,$ as a function of $\displaystyle y\,$ and differentiation is with respect to $\displaystyle y\,$

This has $\displaystyle x\,$ and $\displaystyle y\,$ interchanged compared to the usual expression but the absolute value of the curvature is the same either way, and this way is easier to handle.

Here:

$\displaystyle x=y^2/(4p)\,$

so:

$\displaystyle \frac{dx}{dy}=x'=y/(2p)\,$,

and:

$\displaystyle

\frac{d^2x}{dy^2}=x''=1/(2p)\,

$

so at $\displaystyle (0, 0)\,$, $\displaystyle x'=0\,$, so:

$\displaystyle

\kappa=\frac{[1/(2p)]}{1}=1/(2p)\,

$

so the radius of curvature $\displaystyle r=|1/\kappa|=2|p|\,$.

RonL