# Tangent Line in Polar Coordinates

Let's see how to derive equation of tangent line when we are given equation of curve ${r}={f{{\left(\theta\right)}}}$ in polar coordinates.

We will proceed in the same fashion as with tangent lines to parametric curves, because polar coordinates in some sense similar to parametric curves.

We know that ${x}={r}{\cos{{\left(\theta\right)}}}$ and ${y}={r}{\sin{{\left(\theta\right)}}}$.

Since we are given that ${r}={f{{\left(\theta\right)}}}$ we should treat ${r}$ as a function of $\theta$.

To find slope of tangent line, we need to find $\frac{{{d}{y}}}{{{d}{x}}}$ but in terms of ${r}$ and $\theta$. This can be easily done with chain rule:

$\frac{{{d}{y}}}{{{d}{x}}}=\frac{{{d}{y}}}{{{d}\theta}}\cdot\frac{{{d}\theta}}{{{d}{x}}}=\frac{{{d}{y}}}{{{d}\theta}}\cdot\frac{{1}}{{\frac{{{d}{x}}}{{{d}\theta}}}}$.

We have that $\frac{{{d}{y}}}{{{d}\theta}}=\frac{{{d}{r}}}{{{d}\theta}}{\sin{{\left(\theta\right)}}}+{r}{\cos{{\left(\theta\right)}}}$ and $\frac{{{d}{x}}}{{{d}\theta}}=\frac{{{d}{r}}}{{{d}\theta}}{\cos{{\left(\theta\right)}}}-{r}{\sin{{\left(\theta\right)}}}$.

So, we obtained following fact.

Fact. Slope of tangent line in polar coordinates is $\frac{{{d}{y}}}{{{d}{x}}}=\frac{{\frac{{{d}{r}}}{{{d}\theta}}{\sin{{\left(\theta\right)}}}+{r}{\cos{{\left(\theta\right)}}}}}{{\frac{{{d}{r}}}{{{d}\theta}}{\cos{{\left(\theta\right)}}}-{r}{\sin{{\left(\theta\right)}}}}}$.

Example. Find equation of tangent line to ${r}={1}+{2}{\cos{{\left(\theta\right)}}}$ at $\theta=\frac{\pi}{{4}}$.

We have that $\frac{{{d}{r}}}{{{d}\theta}}=-{2}{\sin{{\left(\theta\right)}}}$, so

$\frac{{{d}{y}}}{{{d}{x}}}=\frac{{-{2}{\sin{{\left(\theta\right)}}}{\sin{{\left(\theta\right)}}}+{\left({1}+{2}{\cos{{\left(\theta\right)}}}\right)}{\cos{{\left(\theta\right)}}}}}{{-{2}{\sin{{\left(\theta\right)}}}{\cos{{\left(\theta\right)}}}-{\left({1}+{2}{\cos{{\left(\theta\right)}}}\right)}{\sin{{\left(\theta\right)}}}}}=$

$=\frac{{{\cos{{\left(\theta\right)}}}+{2}{\cos{{\left({2}\theta\right)}}}}}{{-{\sin{{\left(\theta\right)}}}-{2}{\sin{{\left({2}\theta\right)}}}}}$.

So, slope of tangent line at point $\theta=\frac{\pi}{{4}}$ is ${m}=\frac{{{d}{y}}}{{{d}{x}}}{\mid}_{{\theta=\frac{\pi}{{4}}}}=\frac{{{\cos{{\left(\frac{\pi}{{4}}\right)}}}+{2}{\cos{{\left(\frac{\pi}{{2}}\right)}}}}}{{-{\sin{{\left(\frac{\pi}{{4}}\right)}}}-{2}{\sin{{\left(\frac{\pi}{{2}}\right)}}}}}=\frac{{1}}{{7}}{\left({1}-{2}\sqrt{{{2}}}\right)}$.

Now we need corresponding ${x}$ and ${y}$ coordinates when $\theta=\frac{\pi}{{4}}$:

${x}={r}{\cos{{\left(\theta\right)}}}={\left({1}+{2}{\cos{{\left(\theta\right)}}}\right)}{\cos{{\left(\theta\right)}}}={\left({1}+{2}{\cos{{\left(\frac{\pi}{{4}}\right)}}}\right)}{\cos{{\left(\frac{\pi}{{4}}\right)}}}=\frac{{{1}+\sqrt{{{2}}}}}{\sqrt{{{2}}}}$.

${y}={r}{\sin{{\left(\theta\right)}}}={\left({1}+{2}{\cos{{\left(\theta\right)}}}\right)}{\sin{{\left(\theta\right)}}}={\left({1}+{2}{\cos{{\left(\frac{\pi}{{4}}\right)}}}\right)}{\sin{{\left(\frac{\pi}{{4}}\right)}}}=\frac{{{1}+\sqrt{{{2}}}}}{\sqrt{{{2}}}}$.

So, equation of tangent line that passes through point $\theta=\frac{\pi}{{4}}$ is ${y}=\frac{{1}}{{7}}{\left({1}-{2}\sqrt{{{2}}}\right)}{\left({x}-\frac{{{1}+\sqrt{{{2}}}}}{\sqrt{{{2}}}}\right)}+\frac{{{1}+\sqrt{{{2}}}}}{\sqrt{{{2}}}}$ or approximately ${y}=-{0.2612}{x}+{2.153}$.

As with parametric curves there are curves that have several tangent line at one point. For example, in above example, such point is (0,0). The corresponding value(s) of $\theta$ we can find by solving equation ${1}+{2}{\cos{{\left(\theta\right)}}}={0}$. This gives two solutions on interval ${\left[{0},{2}\pi\right]}$: $\frac{{{2}\pi}}{{3}}$ and $\frac{{{4}\pi}}{{3}}$, so there will be two tangent lines at (0,0).

Similarly to parametric curves, curve in polar coordinates has

• horizontal tangent when $\frac{{{d}{y}}}{{{d}\theta}}={0}$ (provided that $\frac{{{d}{x}}}{{{d}\theta}}\ne{0}$).
• vertical tangent when $\frac{{{d}{x}}}{{{d}\theta}}={0}$ (provided that $\frac{{{d}{y}}}{{{d}\theta}}\ne{0}$).