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}???.
