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.
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