# Thread: Radius of curvature problem

1. ## Radius of curvature problem

I'm having a little trouble finding the radius of curvature for problems of the following form

$y^n=f(x)$

For examples. I'm asked to find the radius of curvature at the point (0,0) for the curve $y^2=4ax$

So

$2y\, y' = 4a \implies y' = 2ay^{-1}$

and

$y'' = -2ay^{-2}\, y'$

So

$\rho = \frac{[1+(y')^2]^{\frac{3}{2}}}{y''} = \frac{[1+4a^2y^{-2}]^{\frac{3}{2}}}{-2ay^{-2}\, y'}$

it's at this point where I'm sure I've messed up. If anyone could tell me how to solve these types of questions I'd be very grateful

Regards

Stonehambey

2. This is a parabola.

Let y=t, then $x=\frac{t^{2}}{4ax}$

and ${\kappa}(t)=\frac{\frac{1}{|2a|}}{[\frac{t^{2}}{4a^{2}}+1]^{\frac{3}{2}}}$

$t=0 \;\ when \;\ (x,y)=(0,0), \;\ so \;\ {\kappa}(0)=\frac{1}{|2a|}, \;\ {\rho}=2|a|$

3. Originally Posted by galactus
This is a parabola.

Let y=t, then $x=\frac{t^{2}}{4ax}$

and ${\kappa}(t)=\frac{\frac{1}{|2a|}}{[\frac{t^{2}}{4a^{2}}+1]^{\frac{3}{2}}}$

$t=0 \;\ when \;\ (x,y)=(0,0), \;\ so \;\ {\kappa}(0)=\frac{1}{|2a|}, \;\ {\rho}=2|a|$
Hi, thanks for the reply, but I'm afraid I cannot follow your answer (and I would very much like to, since it's correct). Would you mind explaining the steps at all?

Regards,

Stonehambey

4. Ah, so I did a little reading this morning, and I can now work through it parametrically. However this question was in the exercise right after the cartesian method had been explained, the parametric method hadn't even been covered yet. The logical conclusion is that you can solve this using cartesian method. But I'm not sure how since I keep getting situations where I'm dividing by zero!

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