Radius of Curvature

**britmath** · September 5th 2012, 01:39 PM

I'm having some difficulties with this seemingly basic question:

Find the radius of curvature at the point $\left(\sqrt{2},2\right)$ on the curve with equation $y=x^2$ .

The steps I'm comfortable with are:

$\frac{\mathrm{d}y}{\mathrm{d}x}=2x$

and obviously the differential relations

$\frac{\mathrm{d}y}{\mathrm{d}x}=\tan\psi \qquad\frac{\mathrm{d}x}{\mathrm{d}s}=\cos\psi \qquad\frac{\mathrm{d}y}{\mathrm{d}s}=\sin\psi$

therefore

$\begin{align*} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2&=4x ^2\\ \left( \frac{\mathrm{d}s}{\mathrm{d}x}\right)^2&=1+ \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2 \end{align*}$

Given that $\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2$ is $4y$ , I've considered proceeding using each of:

$s&=\int \left[ 1+ \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^2\right]^{\frac{1}{2}} \mathrm{d}x$

and

$\begin{align*} s&=\int \left[ 1+ \left( \frac{\mathrm{d}x}{\mathrm{d}y} \right)^2\right]^{\frac{1}{2}} \mathrm{d}y \end{align*}$

but neither method seems to help me, as I'm not sure how to deal with the integral of the square root in either case (typically in these types of questions the square root can be eliminated, or it is in a simple form $\sqrt{ax+b}$ ).

Any help on what I'm missing (I'm guessing its either an identity or substitution, but I can't see it) will be greatly appreciated (especially as a similar question is causing me the same basic problem!). Thanks!

**MaxJasper** · September 5th 2012, 02:40 PM

Use this for radius of curvature:

$\rho =|\frac{\left(y'^2+1\right)^{3/2}}{y''}| = \frac{27}{2}$

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