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.
Re: Curvature 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.