Steps for finding the tangent line to a polar curve in ~ a certain point

We’ll find the equation that the tangent heat to a polar curve in lot the same way that we uncover the tangent heat to a cartesian curve.

We’ll follow these steps:


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1. Discover the slope of the tangent line ???m???, making use of the formula

???m = \fracdydx = \frac \fracdrd\theta\sin\theta+r\cos\theta \fracdrd\theta\cos\theta-r\sin\theta???

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

You are watching: Slope of tangent line to polar curve

2. Find???x_1??? and???x2??? by plugging the value of???\theta??? in ~ the tangent suggest into the counter formulas

???x=r\cos\theta???

???y=r\sin\theta???

3. Plug the slope???m??? and also the point???(x_1,y_1)??? into the point-slope formula because that the equation that a line

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


How to find the equation the the tangent line



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Building the tangent heat 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\pi4???

We’ll begin by calculating ???dr/d\theta???, the derivative of the provided polar equation, so that we deserve to plug it right into the formula for the slope of the tangent line.

???r=1+2\cos\theta???

???\fracdrd\theta=-2\sin\theta???

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

???m=\fracdydx=\frac(-2\sin\theta)\sin\theta+(1+2\cos\theta)\cos\theta(-2\sin\theta)\cos\theta-(1+2\cos\theta)\sin\theta???

???m=\fracdydx=\frac-2\sin^2\theta+\cos\theta+2\cos^2\theta-2\sin\theta\cos\theta-\sin\theta-2\sin\theta\cos\theta???

???m=\fracdydx=\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 steep equation, we’ll acquire a real-number worth for the slope ???m???.

???m=\fracdydx=\frac-2\sin^2\frac\pi4+\cos\frac\pi4+2\cos^2\frac\pi4-4\sin\frac\pi4\cos\frac\pi4-\sin\frac\pi4???

???m=\fracdydx=\frac-2\left(\frac\sqrt22\right)^2+\frac\sqrt22+2\left(\frac\sqrt22\right)^2-4\cdot\frac\sqrt22\cdot\frac\sqrt22-\frac\sqrt22???

???m=\fracdydx=\frac-2\left(\frac24\right)+\frac\sqrt22+2\left(\frac24\right)-4\cdot\frac24-\frac\sqrt22???

???m=\fracdydx=\frac-1+\frac\sqrt22+1-2-\frac\sqrt22???

???m=\fracdydx=\frac\frac\sqrt22-\frac42-\frac\sqrt22???

???m=\fracdydx=\frac\frac\sqrt22\frac-4-\sqrt22???

???m=\fracdydx=\frac\sqrt22\left(\frac2-4-\sqrt2\right)???

???m=\fracdydx=\frac\sqrt2-4-\sqrt2???

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

???m=\fracdydx=\frac\sqrt2-4-\sqrt2\left(\frac-4+\sqrt2-4+\sqrt2\right)???

???m=\fracdydx=\frac-4\sqrt2+216-2???

???m=\fracdydx=\frac-4\sqrt2+214???

???m=\fracdydx=\frac-2\sqrt2+17???

???m=\fracdydx=\frac1-2\sqrt27???


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Now we want to find ???x_1??? and ???y_1??? by plugging the value of ???\theta??? at the tangent allude and the offered polar equation ???r=1+2\cos\theta??? into the conversion formulas

???x=r\cos\theta???

???x_1=\left(1+2\cos\frac\pi4\right)\cos\frac\pi4???

???x_1=\left<1+2\left(\frac\sqrt22\right)\right>\frac\sqrt22???

???x_1=\left(1+\sqrt2\right)\frac\sqrt22???

???x_1=\frac\sqrt2+22???

???x_1=\frac2+\sqrt22???

and

???y=r\sin\theta???

???y_1=\left(1+2\cos\frac\pi4\right)\sin\frac\pi4???

???y_1=\left<1+2\left(\frac\sqrt22\right)\right>\frac\sqrt22???

???y_1=\left(1+\sqrt2\right)\frac\sqrt22???

???y_1=\frac\sqrt2+22???

???y_1=\frac2+\sqrt22???

Plugging ???m??? and also ???\left(x_1,y_1\right)??? into the point-slope formula because that the equation the a line, we get

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

???y-\frac2+\sqrt22=\frac1-2\sqrt27\left(x-\frac2+\sqrt22\right)???

???y-\frac2+\sqrt22=\frac1-2\sqrt27x-\frac2+\sqrt2-4\sqrt2-414???

???y-\frac2+\sqrt22=\frac1-2\sqrt27x-\frac-3\sqrt2-214???

???y-\frac2+\sqrt22=\frac1-2\sqrt27x+\frac3\sqrt2+214???

???2y-\left(2+\sqrt2\right)=\frac2-4\sqrt27x+\frac3\sqrt2+27???

Eliminate the fractions by multiplying through by ???7???.

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???14y-7\left(2+\sqrt2\right)=\left(2-4\sqrt2\right)x+3\sqrt2+2???

???14y=\left(2-4\sqrt2\right)x+3\sqrt2+2+7\left(2+\sqrt2\right)???

???14y=\left(2-4\sqrt2\right)x+3\sqrt2+2+14+7\sqrt2???

???14y=\left(2-4\sqrt2\right)x+16+10\sqrt2???

???14y-\left(2-4\sqrt2\right)x=16+10\sqrt2???

???7y-\left(1-2\sqrt2\right)x=8+5\sqrt2???

The equation the the tangent heat is ???7y-\left(1-2\sqrt2\right)x=8+5\sqrt2???.