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### Section 12.8 : 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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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,

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