Curvature of the curve r(t) = sin(t)i + 2cos(t)j

[roam]

Homework Statement

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.

Homework Equations

Curvature is given by:

$\kappa(t) = \frac{||T'(t)||}{||r'(t)||}$

Where

$T (t) = \frac{r'(t)}{||r'(t)||}$

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.

 

[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.

 

[roam]

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! :)