# A straightforward question (but help needed!)

• Jan 11th 2012, 04:27 PM
maxgunn555
A straightforward question (but help needed!)
Recall how to calculate the arc-length and the curvature of a parametric curve $\delta:\ I\rightarrow \mathbb{R^2}$

Any help as to the steps involved greatly appreciated (running low on time before the exam day after tomorrow).
• Jan 11th 2012, 07:31 PM
chisigma
Re: A straightforward question (but help needed!)
If the curvature is expressed in the form $y=f(x)\ ,\ a, then the arc lenght is ...

$L= \int_{a}^{b} \sqrt{1+ (y^{'})^{2}}\ dx$ (1)

Kind regards

$\chi$ $\sigma$
• Jan 12th 2012, 03:26 PM
maxgunn555
Re: A straightforward question (but help needed!)
Thanks.
I don't quite follow the 'y = f(x)'. i thought parametric curves looked like eg x = 2t +3, y = t^2 +2 e.t.c...
• Jan 12th 2012, 08:35 PM
chisigma
Re: A straightforward question (but help needed!)
If the curve is defined in parametric form $x=x(t)\ ,\ y=y(t)\ ,\ a , then ...

$L=\int_{a}^{b} \sqrt {[x^{'}(t)]^{2} + [y^{'}(t)]^{2}}\ dt$ (1)

Kind regards

$\chi$ $\sigma$
• Jan 13th 2012, 05:21 AM
maxgunn555
Re: A straightforward question (but help needed!)
I'm sure now that's the formula i was looking for. Thanks A lot.