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### Section 1-8 : Tangent, Normal and Binormal Vectors

3. Find the tangent line to $$\vec r\left( t \right) = \cos \left( {4t} \right)\vec i + 3\sin \left( {4t} \right)\vec j + {t^3}\vec k$$ at $$t = \pi$$.

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Start Solution

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

$\vec r'\left( t \right) = - 4\sin \left( {4t} \right)\vec i + 12\cos \left( {4t} \right)\vec j + 3{t^2}\vec k$

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( \pi \right) & = \cos \left( {4\pi } \right)\vec i + 3\sin \left( {4\pi } \right)\vec j + {\pi ^3}\vec k = \vec i + {\pi ^3}\vec k\\ \vec r'\left( \pi \right) & = - 4\sin \left( {4\pi } \right)\vec i + 12\cos \left( {4\pi } \right)\vec j + 3{\pi ^2}\vec k = 12\vec j + 3{\pi ^2}\vec k\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,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \vec i + {\pi ^3}\vec k + t\left( {12\vec j + 3{\pi ^2}\vec k} \right) = \vec i + 12t\,\vec j + \left( {{\pi ^3} + 3{\pi ^2}t} \right)\vec k}}$