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Section 1-8 : Tangent, Normal and Binormal Vectors

4. Find the tangent line to \(\displaystyle \vec r\left( t \right) = \left\langle {7{{\bf{e}}^{2 - t}},\frac{{16}}{{{t^3}}},5 - t} \right\rangle \) at \(t = 2\).

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Start Solution

First, we’ll need to get the tangent vector to the vector function. The tangent vector is,

\[\vec r'\left( t \right) = \left\langle { - 7{{\bf{e}}^{2 - t}}, - \frac{{48}}{{{t^4}}}, - 1} \right\rangle \]

Note that we could use the unit tangent vector here if we wanted to but given how messy those tend to be we’ll just go with this.

Show Step 2

Now we actually need the tangent vector at the value of \(t\) given in the problem statement and not the “full” tangent vector. We’ll also need the point on the vector function at the value of \(t\) from the problem statement. These are,

\[\begin{align*}\vec r\left( 2 \right) & = \left\langle {7,2,3} \right\rangle \\ \vec r'\left( 2 \right) & = \left\langle { - 7, - 3, - 1} \right\rangle \end{align*}\] Show Step 3

To write down the equation of the tangent line we need a point on the line and a vector parallel to the line. Of course, these are just the quantities we found in the previous step.

The tangent line is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {7,2,3} \right\rangle + t\left\langle { - 7, - 3, - 1} \right\rangle = \left\langle {7 - 7t,2 - 3t,3 - t} \right\rangle }}\]