Curvature math problem


by roam
Tags: curvature, math
roam
roam is offline
#1
Apr25-12, 03:38 AM
P: 884
1. The problem statement, all variables and given/known data

I am trying to calculate the curvature of a curve given by the position function:

[itex]\vec{r}(t)= sin(t) \vec{i} + 2 \ cos (t) \ vec{j}[/itex]

The correct answer must be:

[itex]\kappa (t) = \frac{2}{(cos^2(t) + 4 \ sin^2 (t))^{3/2}}[/itex]

I tried several times but I can't arrive at this answer.


2. Relevant equations

Curvature is given by:

[itex]\kappa(t) = \frac{||T'(t)||}{||r'(t)||}[/itex]

Where

[itex]T (t) = \frac{r'(t)}{||r'(t)||}[/itex]

3. The attempt at a solution

[itex]r'(t) = \left\langle cos(t), \ -2 sin(t) \right\rangle[/itex]

[itex]||r'(t)|| = \sqrt{cos^2(t) + (2 \ sin(t))^2} = \sqrt{3 \ sin^2t+1}[/itex]

Therefore

[itex]T = \frac{cos(t)}{\sqrt{3 \ sin^2 (t) +1}} \ \vec{i}, \ \frac{-2 \ sin(t)}{\sqrt{3 \ sin^2 (t) +1}} \vec{j}[/itex]

[itex]\frac{dT}{dt} = \frac{-4 \ sin(t)}{(3 \ sin^2(t) +1)^{(3/2)}} \ \vec{i} + \frac{-2 \cos (t)}{(3 \ sin^2(t) +1)^{(3/2)}} \ \vec{j}[/itex]

[itex]||T'(t)|| = \sqrt{\left( \frac{-4 \ sin(t)}{3 \ sin^2(t) + 1)^{3/2}} \right)^2 + \left( \frac{-2 \cos (t)}{(3 \ sin^2(t) + 1)^{3/2}} \right)^2}[/itex]

[itex]= \sqrt{ \frac{16 \ sin^2(t)}{(3 \ sin^2 t)^3} + \frac{4 \ cos^2(t)}{(3 \ sin^2 t)^3}}[/itex]

Putting this in the equation given

[itex]\kappa = \sqrt{ \frac{16 \ sin^2(t)}{(3 \ sin^2 t)^3} + \frac{4 \ cos^2(t)}{(3 \ sin^2 t)^3}}. \frac{1}{\sqrt{3 \ sin^2t+1}}[/itex]

But I can't see how this can be simplified to get to the correct answer. I appreciate any guidance.
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hamsterman
hamsterman is offline
#2
Apr25-12, 07:05 AM
P: 74
You dropped +1 from the bottom, while calculating ||T'||. The top equals [itex]12sin^2(t)+4 = 4(3sin^2(t)+1)[/itex] and it all works out nicely.
roam
roam is offline
#3
Apr25-12, 07:44 AM
P: 884
Quote Quote by hamsterman View Post
You dropped +1 from the bottom, while calculating ||T'||. The top equals [itex]12sin^2(t)+4 = 4(3sin^2(t)+1)[/itex] and it all works out nicely.
Thank you very much! I got it! :)


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