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Section 12.10 : Curvature

2. Find the curvature of \(\vec r\left( t \right) = \left\langle {4t, - {t^2},2{t^3}} \right\rangle \).

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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 will involve lots of quotients that would probably be unpleasant to take the derivative of. So, let’s go with the cross product formula this time.

We’ll need the first and second derivative of the vector function. Here are those.

\[\vec r'\left( t \right) = \left\langle {4, - 2t,6{t^2}} \right\rangle \hspace{0.5in}\vec r''\left( t \right) = \left\langle {0, - 2,12t} \right\rangle \] Show Step 2

Next, we need the cross product of these two derivatives. Here is that work.

\[\vec r'\left( t \right) \times \vec r''\left( t \right) = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\4&{ - 2t}&{6{t^2}}\\0&{ - 2}&{12t}\end{array}} \right| = - 24{t^2}\,\vec i - 8\vec k - 48t\,\vec j + 12{t^2}\,\vec i = - 12{t^2}\,\vec i - 48t\,\vec j - 8\vec k\] Show Step 3

We now need a couple of magnitudes.

\[\left\| {\vec r'\left( t \right) \times \vec r''\left( t \right)} \right\| = \sqrt {144{t^4} + 2304{t^2} + 64} \hspace{0.25in}\hspace{0.25in}\left\| {\vec r'\left( t \right)} \right\| = \sqrt {16 + 4{t^2} + 36{t^4}} \]

The curvature is then,

\[\kappa = \frac{{\left\| {\vec r'\left( t \right) \times \vec r''\left( t \right)} \right\|}}{{{{\left\| {\vec r'\left( t \right)} \right\|}^3}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\sqrt {144{t^4} + 2304{t^2} + 64} }}{{{{\left( {16 + 4{t^2} + 36{t^4}} \right)}^{\frac{3}{2}}}}}}}\]

A fairly messy formula here, but these will often be that way.