Section 9.7 : Tangents with Polar Coordinates
1. Find the tangent line to \(r = \sin \left( {4\theta } \right)\cos \left( \theta \right)\) at \(\displaystyle \theta = \frac{\pi }{6}\).
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Start SolutionFirst, we’ll need to following derivative,
\[\frac{{dr}}{{d\theta }} = 4\cos \left( {4\theta } \right)\cos \left( \theta \right) - \sin \left( {4\theta } \right)\sin \left( \theta \right)\] Show Step 2Next using the formula from the notes on this section we have,
\[\begin{align*}\frac{{dy}}{{dx}} & = \frac{{\displaystyle \frac{{dr}}{{d\theta }}\sin \theta + r\cos \theta }}{{\displaystyle \frac{{dr}}{{d\theta }}\cos\theta - r\sin \theta }}\\ & \\ & = \frac{{\left( {4\cos \left( {4\theta } \right)\cos \left( \theta \right) - \sin \left( {4\theta } \right)\sin \left( \theta \right)} \right)\sin \theta + \left( {\sin \left( {4\theta } \right)\cos \left( \theta \right)} \right)\cos \theta }}{{\left( {4\cos \left( {4\theta } \right)\cos \left( \theta \right) - \sin \left( {4\theta } \right)\sin \left( \theta \right)} \right)cos\theta - \left( {\sin \left( {4\theta } \right)\cos \left( \theta \right)} \right)\sin \theta }}\end{align*}\]This is a very messy derivative (these often are) and, at least in this case, there isn’t a lot of simplification that we can do…
Show Step 3Next, we’ll need to evaluate both the derivative from the previous step as well as \(r\) at \(\theta = \frac{\pi }{6}\).
\[{\left. {\frac{{dy}}{{dx}}} \right|_{\theta = \,\frac{\pi }{6}}} = \frac{1}{{3\sqrt 3 }} \hspace{0.75in} {\left. r \right|_{\theta = \frac{\pi }{6}}} = \frac{3}{4}\]You can see why we need both of these right?
Show Step 4Last, we need the \(x\) and \(y\) coordinate that we’ll be at when \(\theta = \frac{\pi }{6}\). 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,
\[x = r\cos \left( \theta \right) = \frac{3}{4}\cos \left( {\frac{\pi }{6}} \right) = \frac{{3\sqrt 3 }}{8} \hspace{0.75in} y = r\sin \left( \theta \right) = \frac{3}{4}\sin \left( {\frac{\pi }{6}} \right) = \frac{3}{8}\]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 5The tangent line is then,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{y = \frac{3}{8} + \frac{1}{{3\sqrt 3 }}\left( {x - \frac{{3\sqrt 3 }}{8}} \right) = \frac{1}{{3\sqrt 3 }}x + \frac{1}{4}}}\]