Curvature

O

odyssey

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Give a formula, in terms of \(\displaystyle x(t)\) and \(\displaystyle y(t)\) and and their derivatives, for the curvature for a curve in \(\displaystyle \mathbb{R}^2\) given by \(\displaystyle \alpha(t) = (x(t),y(t)).\)

How can I do this problem? Do I just apply the curvature formula and substitute \(\displaystyle \alpha(t)\) with \(\displaystyle (x(t),y(t)).\)
 
There are several ways to describe curvature depending on the type of framing you use. There's also signed curvature and unsigned curvature.

If you use TNB framing, arclength paramaterized, then the curvature is the "coefficient" of the vector N, i.e. \(\displaystyle T'\cdot N\)
 
Assume you are arc-length parameterized (re-paramaterize so that your curve is \(\displaystyle \alpha:[0,L]\to \mathbb{R}^2\)), then

\(\displaystyle T(s) = \dfrac{\alpha'(s)}{||\alpha'(s)||}\)

\(\displaystyle N(s) = \dfrac{T'(s)}{||T'(s)||} \iff T'(s) = ||T'(s)||\cdot N(s)\)

and

\(\displaystyle T'(s) = \kappa N(s)\)

so

\(\displaystyle \kappa = T'(s)\cdot N(s) = ||T'(s)||\).

If this is not satisfactory to you, then I would need what your professor defines as curvature to do anything else.
 
Daon2,

I got the unit tangent to be \(\displaystyle T(t) = \frac{f'(t)}{||f'(t)|| }= \frac{\langle x'(t), y'(t)\rangle}{\sqrt{(x'(t))^2 +(y'(t))^2}}\) but I am having trouble finding \(\displaystyle T'(t).\)
How can I take the derivative of that?
 
The square root you have is the definition of \(\displaystyle ds/dt\). So if you are arc-length paramaterized, \(\displaystyle ds/dt=1\). So

\(\displaystyle T'(s) = \dfrac{dT}{ds} = (x''(s), y''(s))\). If you are not arclength parameterized, then

\(\displaystyle T'(t) = \dfrac{dT}{dt} = \left(\dfrac{\alpha'(t)}{ds/dt}\right)' = \dfrac{\frac{ds}{dt}\alpha''(t) - \alpha'(t)\frac{d^2s}{dt^2}} {\left(\frac{ds}{dt}\right)^2}\)

So

\(\displaystyle \dfrac{dT}{ds} = \dfrac{\frac{dT}{dt}}{\frac{ds}{dt}} = \dfrac{\frac{ds}{dt}\alpha''(t) - \alpha'(t)\frac{d^2s}{dt^2}} {\left(\frac{ds}{dt}\right)^3}\)
 
Daon2,

I am a bit confused. How did you get

\(\displaystyle \dfrac{dT}{ds} = \dfrac{\frac{dT}{dt}}{\frac{ds}{dt}} = \dfrac{\frac{ds}{dt}\alpha''(t) - \alpha'(t)\frac{d^2s}{dt^2}} {\left(\frac{ds}{dt}\right)^3}?\)

It seems all that has changed is that there is an exponent of 3 in the denominator instead of 2?

Also, how can I take the derivative of the form you provided

\(\displaystyle \frac{ds}{dt}\alpha''(t) - \alpha'(t)\frac{d^2s}{dt^2} = [\sqrt{(x'(t))^2 +(y'(t))^2}][\langle x''(t), y''(t) \rangle] - [\langle x'(t), y'(t) \rangle][ \sqrt{(x'(t))^2 +(y'(t))^2}]''\)

that is what I am having the most difficulty in.
 
correct, i divided by another ds/dt to obtain dT/ds. And there shouldn't be a double-prime on that last square root.

\(\displaystyle \frac{d^2s}{dt^2} = \frac{d}{dt}\sqrt{[x'(t)]^2+[y'(t)]^2}\)

So you se the chain rule:

\(\displaystyle \dfrac{d}{dt}\sqrt{f(t)} = \dfrac{f'(t)}{2\sqrt{f(t)}}\)
 
Last edited:
Ah, I never thought to use the chain rule. Dumb mistake by me. Thank you very much!
 
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