Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 12.8 : Tangent, Normal and Binormal Vectors
1. Find the unit tangent vector for the vector function : \(\vec r\left( t \right) = \left\langle {{t^2} + 1,3 - t,{t^3}} \right\rangle \)
Show All Steps Hide All Steps
Start SolutionFrom 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.
\[\vec r'\left( t \right) = \left\langle {2t, - 1,3{t^2}} \right\rangle \] \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left( {2t} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( {3{t^2}} \right)}^2}} = \sqrt {1 + 4{t^2} + 9{t^4}} \] Show Step 2The unit tangent vector for this vector function is then,
\[\vec T\left( t \right) = \frac{1}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}\left\langle {2t, - 1,3{t^2}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {\frac{{2t}}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}, - \frac{1}{{\sqrt {1 + 4{t^2} + 9{t^4}} }},\frac{{3{t^2}}}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}} \right\rangle }}\]