• Mar 13th 2009, 11:54 AM
Stonehambey
I'm having a little trouble finding the radius of curvature for problems of the following form

$\displaystyle y^n=f(x)$

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

So

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

and

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

So

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

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
• Mar 13th 2009, 12:08 PM
galactus
This is a parabola.

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

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

$\displaystyle t=0 \;\ when \;\ (x,y)=(0,0), \;\ so \;\ {\kappa}(0)=\frac{1}{|2a|}, \;\ {\rho}=2|a|$
• Mar 13th 2009, 12:31 PM
Stonehambey
Quote:

Originally Posted by galactus
This is a parabola.

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

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

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

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
• Mar 14th 2009, 02:48 AM
Stonehambey
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! (Headbang)