1. ## plane curvature

Let (x(s), y(s)) be a plane curve parametrized by arclength. Show that the curvature is given by the formula:

C = |x'y'' - x''y'|

The standard formula for curvature is:

||dT/ds||,

where T is the unit tangent vector and equal to:

x'(t) / ||x'(t)||

I don't know where to start here, or even if these are the formulas needed to get to the first equation. Any help would be appreciated!

2. Originally Posted by gralla55
Let (x(s), y(s)) be a plane curve parametrized by arclength. Show that the curvature is given by the formula:

C = |x'y'' - x''y'|

The standard formula for curvature is:

||dT/ds||,

where T is the unit tangent vector and equal to:

x'(t) / ||x'(t)||

I don't know where to start here, or even if these are the formulas needed to get to the first equation. Any help would be appreciated!

We can express the curvature of a plane curve as $\displaystyle \kappa (t)=\frac{\|r'(t)\times r''(t)\|}{\|r'(t)\|^3}$ , so evaluate this taking into consideration that $\displaystyle \|r'(t)\|=1$ if we use the arclength parametrization.

Tonio

3. Could you elaborate? Thanks

4. Originally Posted by gralla55
Let (x(s), y(s)) be a plane curve parametrized by arclength. Show that the curvature is given by the formula:

C = |x'y'' - x''y'|

The standard formula for curvature is:

||dT/ds||,

where T is the unit tangent vector and equal to:

x'(t) / ||x'(t)||

I don't know where to start here, or even if these are the formulas needed to get to the first equation. Any help would be appreciated!
$\displaystyle r=xi + yj$

$\displaystyle r=xi + yj$

$\displaystyle r=xi + yj$

$\displaystyle Curvature = C = ||\frac{d (T(t))}{ds} || = ||r''(s)||$

$\displaystyle C(t) = \frac{ ||r'(t) X r''(t)||}{ ||r'(t)||3 }$

Where,

$\displaystyle ||r'(t)||^3 =1^3$

$\displaystyle C= ||r'(t)$ X $\displaystyle r''(t)||$

Which is just the cross product from what we wrote above, which is

$\displaystyle C=|xy -xy`|$