Finding the tangent line to the polar curve

Steps for finding the tangent line to a polar curve at a particular point

We’ll find the equation of the tangent line to a polar curve in much the same way that we find the tangent line to a cartesian curve.

We’ll follow these steps:

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1. Find the slope of the tangent line $m$, using the formula

$m = \frac{dy}{dx} = \frac {\frac{dr}{d\theta}\sin{\theta}+r\cos{\theta}} {\frac{dr}{d\theta}\cos{\theta}-r\sin{\theta}}$

remembering to plug the value of $\theta$ at the tangent point into $dy/dx$ to get a real-number value for the slope $m$.

2. Find $x_1$ and $x2$ by plugging the value of $\theta$ at the tangent point into the conversion formulas

$x=r\cos{\theta}$

$y=r\sin{\theta}$

3. Plug the slope $m$ and the point $(x_1,y_1)$ into the point-slope formula for the equation of a line

$y-y_1=m(x-x_1)$

How to find the equation of the tangent line

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Building the tangent line equation step-by-step

Example

Find the tangent line to the polar curve at the given point.

$r=1+2\cos{\theta}$

at $\theta=\frac{\pi}{4}$

We’ll start by calculating $dr/d\theta$, the derivative of the given polar equation, so that we can plug it into the formula for the slope of the tangent line.

$r=1+2\cos{\theta}$

$\frac{dr}{d\theta}=-2\sin{\theta}$

Plugging $dr/d\theta$ and the given polar equation $r=1+2\cos{\theta}$ into the formula for $dy/dx$, we get

$m=\frac{dy}{dx}=\frac{(-2\sin{\theta})\sin{\theta}+(1+2\cos{\theta})\cos{\theta}}{(-2\sin{\theta})\cos{\theta}-(1+2\cos{\theta})\sin{\theta}}$

$m=\frac{dy}{dx}=\frac{-2\sin^2{\theta}+\cos{\theta}+2\cos^2{\theta}}{-2\sin{\theta}\cos{\theta}-\sin{\theta}-2\sin{\theta}\cos{\theta}}$

$m=\frac{dy}{dx}=\frac{-2\sin^2{\theta}+\cos{\theta}+2\cos^2{\theta}}{-4\sin{\theta}\cos{\theta}-\sin{\theta}}$

Plugging the value of $\theta=\pi/4$ into the slope equation, we’ll get a real-number value for the slope $m$.

$m=\frac{dy}{dx}=\frac{-2\sin^2{\frac{\pi}{4}}+\cos{\frac{\pi}{4}}+2\cos^2{\frac{\pi}{4}}}{-4\sin{\frac{\pi}{4}}\cos{\frac{\pi}{4}}-\sin{\frac{\pi}{4}}}$

$m=\frac{dy}{dx}=\frac{-2\left(\frac{\sqrt{2}}{2}\right)^2+\frac{\sqrt{2}}{2}+2\left(\frac{\sqrt{2}}{2}\right)^2}{-4\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}}$

$m=\frac{dy}{dx}=\frac{-2\left(\frac{2}{4}\right)+\frac{\sqrt{2}}{2}+2\left(\frac{2}{4}\right)}{-4\cdot\frac{2}{4}-\frac{\sqrt{2}}{2}}$

$m=\frac{dy}{dx}=\frac{-1+\frac{\sqrt{2}}{2}+1}{-2-\frac{\sqrt{2}}{2}}$

$m=\frac{dy}{dx}=\frac{\frac{\sqrt{2}}{2}}{-\frac{4}{2}-\frac{\sqrt{2}}{2}}$

$m=\frac{dy}{dx}=\frac{\frac{\sqrt{2}}{2}}{\frac{-4-\sqrt{2}}{2}}$

$m=\frac{dy}{dx}=\frac{\sqrt{2}}{2}\left(\frac{2}{-4-\sqrt{2}}\right)$

$m=\frac{dy}{dx}=\frac{\sqrt{2}}{-4-\sqrt{2}}$

If we want to get rid of the square root in the denominator, we can multiply by the conjugate.

$m=\frac{dy}{dx}=\frac{\sqrt{2}}{-4-\sqrt{2}}\left(\frac{-4+\sqrt{2}}{-4+\sqrt{2}}\right)$

$m=\frac{dy}{dx}=\frac{-4\sqrt{2}+2}{16-2}$

$m=\frac{dy}{dx}=\frac{-4\sqrt{2}+2}{14}$

$m=\frac{dy}{dx}=\frac{-2\sqrt{2}+1}{7}$

$m=\frac{dy}{dx}=\frac{1-2\sqrt{2}}{7}$

We’ll find the equation of the tangent line to a polar curve in much the same way that we find the tangent line to a cartesian curve.

Now we want to find $x_1$ and $y_1$ by plugging the value of $\theta$ at the tangent point and the given polar equation $r=1+2\cos{\theta}$ into the conversion formulas

$x=r\cos{\theta}$

$x_1=\left(1+2\cos{\frac{\pi}{4}}\right)\cos{\frac{\pi}{4}}$

$x_1=\left[1+2\left(\frac{\sqrt{2}}{2}\right)\right]\frac{\sqrt{2}}{2}$

$x_1=\left(1+\sqrt{2}\right)\frac{\sqrt{2}}{2}$

$x_1=\frac{\sqrt{2}+2}{2}$

$x_1=\frac{2+\sqrt{2}}{2}$

and

$y=r\sin{\theta}$

$y_1=\left(1+2\cos{\frac{\pi}{4}}\right)\sin{\frac{\pi}{4}}$

$y_1=\left[1+2\left(\frac{\sqrt{2}}{2}\right)\right]\frac{\sqrt{2}}{2}$

$y_1=\left(1+\sqrt{2}\right)\frac{\sqrt{2}}{2}$

$y_1=\frac{\sqrt{2}+2}{2}$

$y_1=\frac{2+\sqrt{2}}{2}$

Plugging $m$ and $\left(x_1,y_1\right)$ into the point-slope formula for the equation of a line, we get

$y-y_1=m(x-x_1)$

$y-\frac{2+\sqrt{2}}{2}=\frac{1-2\sqrt{2}}{7}\left(x-\frac{2+\sqrt{2}}{2}\right)$

$y-\frac{2+\sqrt{2}}{2}=\frac{1-2\sqrt{2}}{7}x-\frac{2+\sqrt{2}-4\sqrt{2}-4}{14}$

$y-\frac{2+\sqrt{2}}{2}=\frac{1-2\sqrt{2}}{7}x-\frac{-3\sqrt{2}-2}{14}$

$y-\frac{2+\sqrt{2}}{2}=\frac{1-2\sqrt{2}}{7}x+\frac{3\sqrt{2}+2}{14}$

$2y-\left(2+\sqrt{2}\right)=\frac{2-4\sqrt{2}}{7}x+\frac{3\sqrt{2}+2}{7}$

Eliminate the fractions by multiplying through by $7$.

$14y-7\left(2+\sqrt{2}\right)=\left(2-4\sqrt{2}\right)x+3\sqrt{2}+2$

$14y=\left(2-4\sqrt{2}\right)x+3\sqrt{2}+2+7\left(2+\sqrt{2}\right)$

$14y=\left(2-4\sqrt{2}\right)x+3\sqrt{2}+2+14+7\sqrt{2}$

$14y=\left(2-4\sqrt{2}\right)x+16+10\sqrt{2}$

$14y-\left(2-4\sqrt{2}\right)x=16+10\sqrt{2}$

$7y-\left(1-2\sqrt{2}\right)x=8+5\sqrt{2}$

The equation of the tangent line is $7y-\left(1-2\sqrt{2}\right)x=8+5\sqrt{2}$.

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