Consider the polar curve \(r=\cos (\theta )\). The graph is show below.

We want to find all the horizontal and vertical tangent lines to this curve.

First we find the \(\theta \) values where \(r=\cos (\theta )\) has a horizontal tangent line and give the equation of the line.

Since our curve can be generated by letting \(\theta \) vary from over the interval \([0, \pi )\), we can assume the \(\theta \) values lie in \([0, \pi )\). List the \(\theta \) values in order from smallest to largest below:

When \(\theta =\answer {\frac {\pi }{4}}\) the equation of the tangent line is \(\answer {y=\frac {1}{2} }\)

and when \(\theta =\answer {\frac {3\pi }{4}}\) the equation of the tangent line is \(\answer {y=-\frac {1}{2}}\).

Consider the formulas for changing from polar coordinates to Cartesian coordinates:

\begin{align*} x&=r\cos (\theta ) \\ y&=r\sin (\theta ) \end{align*}

Using the equation of our curve \(r=\cos (\theta )\), we can substitute for \(r\) to obtain:

\begin{align*} x&=\answer {\cos (\theta )\cos (\theta )} \\ y&=\answer {\cos (\theta )\sin (\theta ) } \end{align*}

Note that we have expressed both the \(x\) and \(y\) coordinates of the points of the curve in terms of functions of a single parameter \(\theta \).

Since we have a parametric description of our curve in terms of \(\theta \), we can use the chain rule to express \(\dd [y]{x}\) in terms of \(\theta \) to get \(\dd [y]{x}=\frac { dy/d\theta }{dx/d\theta }\).

Calculating we get

\begin{align*} \dd [x]{\theta }&=\answer { -2\cos (\theta )\sin (\theta ) } \\ \dd [y]{\theta }&=\answer { \cos ^2(\theta )-\sin ^2(\theta ) } \end{align*}

Thus we find \(\dd [y]{x}=\answer { \frac { \cos ^2(\theta )-\sin ^2(\theta )}{-2\cos (\theta )\sin (\theta )}}\).

The tangent line is horizontal when the slope is \(0\). Therefore we need to find where \(\dd [y]{x}\) is equal to \(0\).

Setting the numerator of \(\dd [y]{x}\) equal to \(0\) gives us \(\cos ^2(\theta )-\sin ^2(\theta )=0\).

We can rewrite this equation in terms of tangent as \(\answer { \tan ^2(\theta )=1 }\).

Taking square roots of both sides this becomes two equations

\begin{align*} \tan (\theta )&=1 \\ \tan (\theta )&=-1 \end{align*}

Notice that our curve \(r=\cos (\theta )\) can be generated by letting \(\theta \) vary over the interval \([0, \pi )\).

Therefore we need to solve these equations for \(\theta \) values in \([0, \pi )\).

The only solution for \(\tan (\theta )=1\) in this interval is \(\theta =\answer {\frac {\pi }{4}}\) and the only solution for \(\tan (\theta )=-1\) in this interval is \(\theta =\answer {\frac {3\pi }{4}}\).

Now we want to find the \(\theta \) values where \(r=\cos (\theta )\) has a vertical tangent line and give the equation of the line.

Since our curve can be generated by letting \(\theta \) vary from over the interval \([0, \pi )\), we can assume the \(\theta \) values lie in \([0, \pi )\). List the \(\theta \) values in order below:

When \(\theta =\answer {0}\) the equation of the tangent line is \(\answer {x=1 }\)

and when \(\theta =\answer { \frac {\pi }{2}}\) the equation of the tangent line is \(\answer {x=0}\).

In the hint for exercise 1 we derived a parametric description of our curve in terms of \(\theta \):

\begin{align*} \dd [x]{\theta }&=\answer { -2\cos (\theta )\sin (\theta ) } \\ \dd [y]{\theta }&=\answer { \cos ^2(\theta )-\sin ^2(\theta ) } \end{align*}

Where we calculated \(\dd [y]{x}=\answer { \frac { \cos ^2(\theta )-\sin ^2(\theta )}{-2\cos (\theta )\sin (\theta )}}\).

The tangent line is vertical when the slope becomes infinite. We can check for this by looking for where the denominator of \(\dd [y]{x}\) is equal to \(0\).

Setting the denominator equal to \(0\) gives us \(-2\cos (\theta )\sin (\theta )=0\).

Setting \(\cos (\theta )=0\) gives us \(\theta =\answer { \frac {\pi }{2}}\) and setting \(\sin (\theta )=0\) gives us \(\theta =\answer { 0}\) (use \(\theta \) values that lie in \([0, \pi )\)).