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### Section 1-10 : Curvature

1. Find the curvature of $$\vec r\left( t \right) = \left\langle {\cos \left( {2t} \right), - \sin \left( {2t} \right),4t} \right\rangle$$.

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We have two formulas we can use here to compute the curvature. One requires us to take the derivative of the unit tangent vector and the other requires a cross product.

Either will give the same result. The real question is which will be easier to use. Cross products can be a pain to compute but then some of the unit tangent vectors can be quite messy to take the derivative of. So, basically, the one we use will be the one that will probably be the easiest to use.

In this case it looks like the unit tangent vector won’t be that bad to work with so let’s go with that formula. Here’s the unit tangent vector work.

$\begin{array}{c}\vec r'\left( t \right) = \left\langle { - 2\sin \left( {2t} \right), - 2\cos \left( {2t} \right),4} \right\rangle \hspace{0.5in}\left\| {\vec r'\left( t \right)} \right\| = \sqrt {4{{\sin }^2}\left( {2t} \right) + 4{{\cos }^2}\left( {2t} \right) + 16} = \sqrt {20} = 2\sqrt 5 \\ \vec T\left( t \right) = \frac{1}{{2\sqrt 5 }}\left\langle { - 2\sin \left( {2t} \right), - 2\cos \left( {2t} \right),4} \right\rangle = \left\langle { - \frac{{\sin \left( {2t} \right)}}{{\sqrt 5 }}, - \frac{{\cos \left( {2t} \right)}}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right\rangle \end{array}$ Show Step 2

Now, what we really need is the magnitude of the derivative of the unit tangent vector so here is that work,

$\vec T'\left( t \right) = \left\langle { - \frac{2}{{\sqrt 5 }}\cos \left( {2t} \right),\frac{2}{{\sqrt 5 }}\sin \left( {2t} \right),0} \right\rangle \hspace{0.25in}\hspace{0.25in}\left\| {\vec T'\left( t \right)} \right\| = \sqrt {\frac{4}{5}{{\cos }^2}\left( {2t} \right) + \frac{4}{5}{{\sin }^2}\left( {2t} \right)} = \frac{2}{{\sqrt 5 }}$ Show Step 3

The curvature is then,

$\kappa = \frac{{\left\| {\vec T'\left( t \right)} \right\|}}{{\left\| {\vec r'\left( t \right)} \right\|}} = \frac{{{}^{2}/{}_{{\sqrt 5 }}}}{{2\sqrt 5 }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{5}}}$

So, in this case, the curvature will be independent of $$t$$. That won’t always be the case so don’t expect this to happen every time.