# Curvature math problem

by roam
Tags: curvature, math
 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: $\vec{r}(t)= sin(t) \vec{i} + 2 \ cos (t) \ vec{j}$ The correct answer must be: $\kappa (t) = \frac{2}{(cos^2(t) + 4 \ sin^2 (t))^{3/2}}$ I tried several times but I can't arrive at this answer. 2. Relevant equations Curvature is given by: $\kappa(t) = \frac{||T'(t)||}{||r'(t)||}$ Where $T (t) = \frac{r'(t)}{||r'(t)||}$ 3. The attempt at a solution $r'(t) = \left\langle cos(t), \ -2 sin(t) \right\rangle$ $||r'(t)|| = \sqrt{cos^2(t) + (2 \ sin(t))^2} = \sqrt{3 \ sin^2t+1}$ Therefore $T = \frac{cos(t)}{\sqrt{3 \ sin^2 (t) +1}} \ \vec{i}, \ \frac{-2 \ sin(t)}{\sqrt{3 \ sin^2 (t) +1}} \vec{j}$ $\frac{dT}{dt} = \frac{-4 \ sin(t)}{(3 \ sin^2(t) +1)^{(3/2)}} \ \vec{i} + \frac{-2 \cos (t)}{(3 \ sin^2(t) +1)^{(3/2)}} \ \vec{j}$ $||T'(t)|| = \sqrt{\left( \frac{-4 \ sin(t)}{3 \ sin^2(t) + 1)^{3/2}} \right)^2 + \left( \frac{-2 \cos (t)}{(3 \ sin^2(t) + 1)^{3/2}} \right)^2}$ $= \sqrt{ \frac{16 \ sin^2(t)}{(3 \ sin^2 t)^3} + \frac{4 \ cos^2(t)}{(3 \ sin^2 t)^3}}$ Putting this in the equation given $\kappa = \sqrt{ \frac{16 \ sin^2(t)}{(3 \ sin^2 t)^3} + \frac{4 \ cos^2(t)}{(3 \ sin^2 t)^3}}. \frac{1}{\sqrt{3 \ sin^2t+1}}$ But I can't see how this can be simplified to get to the correct answer. I appreciate any guidance.
 P: 74 You dropped +1 from the bottom, while calculating ||T'||. The top equals $12sin^2(t)+4 = 4(3sin^2(t)+1)$ and it all works out nicely.
P: 895
 Quote by hamsterman You dropped +1 from the bottom, while calculating ||T'||. The top equals $12sin^2(t)+4 = 4(3sin^2(t)+1)$ and it all works out nicely.
Thank you very much! I got it! :)

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