# plane curvature

• Apr 18th 2010, 08:55 PM
gralla55
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!
• Apr 19th 2010, 02:33 AM
tonio
Quote:

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 $\kappa (t)=\frac{\|r'(t)\times r''(t)\|}{\|r'(t)\|^3}$ , so evaluate this taking into consideration that $\|r'(t)\|=1$ if we use the arclength parametrization.

Tonio
• Apr 19th 2010, 06:59 PM
gralla55
Could you elaborate? Thanks
• Apr 19th 2010, 08:00 PM
AllanCuz
Quote:

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!

$r=xi + yj$

$r=xi + yj$

$r=xi + yj$

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

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

Where,

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

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

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

$C=|xy -xy`|$