Register to reply

Curvature math problem

by roam
Tags: curvature, math
Share this thread:
roam
#1
Apr25-12, 03:38 AM
P: 895
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.
Phys.Org News Partner Science news on Phys.org
NASA team lays plans to observe new worlds
IHEP in China has ambitions for Higgs factory
Spinach could lead to alternative energy more powerful than Popeye
hamsterman
#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
#3
Apr25-12, 07:44 AM
P: 895
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! :)


Register to reply

Related Discussions
Mean curvature Differential Geometry 10
Spacetime curvature and the force pulling an object down the curvature Special & General Relativity 8
Gaussian Curvature, Normal Curvature, and the Shape Operator Calculus & Beyond Homework 0
GR and Curvature Special & General Relativity 3
Geodesic Curvature (Curvature of a curve) Differential Geometry 8