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Section 1-8 : Tangent, Normal and Binormal Vectors

2. Find the unit tangent vector for the vector function : \(\vec r\left( t \right) = t{{\bf{e}}^{2t}}\,\vec i + \left( {2 - {t^2}} \right)\vec j - {{\bf{e}}^{2t}}\vec k\)

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From the notes in this section we know that to get the unit tangent vector all we need is the derivative of the vector function and its magnitude. Here are those quantities.

\[\vec r'\left( t \right) = \left( {{{\bf{e}}^{2t}} + 2t{{\bf{e}}^{2t}}} \right)\,\vec i - 2t\,\vec j - 2{{\bf{e}}^{2t}}\vec k\] \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left( {{{\bf{e}}^{2t}} + 2t{{\bf{e}}^{2t}}} \right)}^2} + {{\left( { - 2t} \right)}^2} + {{\left( { - 2{{\bf{e}}^{2t}}} \right)}^2}} = \sqrt {{{\left( {{{\bf{e}}^{2t}} + 2t{{\bf{e}}^{2t}}} \right)}^2} + 4{t^2} + 4{{\bf{e}}^{4t}}} \] Show Step 2

The unit tangent vector for this vector function is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec T\left( t \right) = \frac{1}{{\sqrt {{{\left( {{{\bf{e}}^{2t}} + 2t{{\bf{e}}^{2t}}} \right)}^2} + 4{t^2} + 4{{\bf{e}}^{4t}}} }}\left( {\left( {{{\bf{e}}^{2t}} + 2t{{\bf{e}}^{2t}}} \right)\,\vec i - 2t\,\vec j - 2{{\bf{e}}^{2t}}\vec k} \right)}}\]

Note that because of the “mess” with this one we didn’t distribute the magnitude through to each term as we usually do with these. This problem is a good example of just how “messy” these can get.