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Curvature math problem 
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#1
Apr2512, 03:38 AM

P: 916

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. 


#2
Apr2512, 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.



#3
Apr2512, 07:44 AM

P: 916




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