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Section 9.7 : Tangents with Polar Coordinates

2. Find the tangent line to $$r = \theta - \cos \left( \theta \right)$$ at $$\displaystyle \theta = \frac{{3\pi }}{4}$$.

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First, we’ll need to following derivative,

$\frac{{dr}}{{d\theta }} = 1 + \sin \left( \theta \right)$ Show Step 2

Next using the formula from the notes on this section we have,

\begin{align*}\frac{{dy}}{{dx}} & = \frac{{\displaystyle \frac{{dr}}{{d\theta }}\sin \theta + r\cos \theta }}{{\displaystyle \frac{{dr}}{{d\theta }}\cos\theta - r\sin \theta }}\\ & \\ & = \frac{{\left( {1 + \sin \left( \theta \right)} \right)\sin \theta + \left( {\theta - \cos \left( \theta \right)} \right)\cos \theta }}{{\left( {1 + \sin \left( \theta \right)} \right)cos\theta - \left( {\theta - \cos \left( \theta \right)} \right)\sin \theta }}\end{align*}

This is a somewhat messy derivative (these often are) and, at least in this case, there isn’t a lot of simplification that we can do…

Show Step 3

Next, we’ll need to evaluate both the derivative from the previous step as well as $$r$$ at $$\theta = \frac{{3\pi }}{4}$$.

${\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \,\frac{{3\pi }}{4}}} = 0.2843 \hspace{0.75in} {\left. r \right|_{\theta = \frac{{3\pi }}{4}}} = 3.0633$

You can see why we need both of these right?

Show Step 4

Last, we need the $$x$$ and $$y$$ coordinate that we’ll be at when $$\theta = \frac{{3\pi }}{4}$$. These values are easy enough to find given that we know what $$r$$ is at this point and we also know the polar to Cartesian coordinate conversion formulas. So,

$x = r\cos \left( \theta \right) = 3.0633\cos \left( {\frac{{3\pi }}{4}} \right) = - 2.1661\hspace{0.25in}y = r\sin \left( \theta \right) = 3.0633\sin \left( {\frac{{3\pi }}{4}} \right) = 2.1661$

Of course, we also have the slope of the tangent line since it is just the value of the derivative we computed in the previous step.

Show Step 5

The tangent line is then,

$\require{bbox} \bbox[2pt,border:1px solid black]{{y = 2.1661 + 0.2843\left( {x + 2.1661} \right) = 0.2843x + 2.7819}}$