15. Polar Coordinates

Polar Curve as a Parametric Curve

Notice that any polar curve r=f(θ)r=f(\theta) can be regarded as a parametric curve with parameter θ\theta and coordinate functions: x(θ)=rcosθ=f(θ)cosθy(θ)=rsinθ=f(θ)sinθ x(\theta)=r\cos\theta=f(\theta)\cos\theta \qquad \qquad y(\theta)=r\sin\theta=f(\theta)\sin\theta Normally, we remove the reference to ff, writing r=r(θ)r=r(\theta) and: x(θ)=r(θ)cosθy(θ)=r(θ)sinθ x(\theta)=r(\theta)\cos\theta \qquad \qquad y(\theta)=r(\theta)\sin\theta Then the position vector of the curve is r(θ)=(x(θ),y(θ))=(r(θ)cosθ,r(θ)sinθ) \vec{r}(\theta)=(x(\theta),y(\theta)) =(r(\theta)\cos\theta,r(\theta)\sin\theta)

r(θ)\vec r(\theta) is the position vector (Notice the arrow over the r\vec r.) telling us the position in the plane for each value of the parameter θ\theta, whereas r(θ)r(\theta) is simply the function defining the polar curve. Don't get confused.

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We will use this formulation in two of our applications of polar curves: slope and arclength.

c. Slope of a Polar Graph

Recall that the slope of a parametric curve r(θ)=(x(θ),y(θ))\vec{r}(\theta)=(x(\theta),y(\theta)) is: dydx=dydθdxdθ \dfrac{dy}{dx} =\dfrac {\;\dfrac{dy}{d\theta}\;} {\;\dfrac{dx}{d\theta}\;} In fact, applying the product rule to a polar curve gives: dxdθ=drdθcosθr(θ)sinθdydθ=drdθsinθ+r(θ)cosθ\begin{aligned} \dfrac{dx}{d\theta} &=\dfrac{dr}{d\theta}\cos\theta-r(\theta)\sin\theta\\ \dfrac{dy}{d\theta} &=\dfrac{dr}{d\theta}\sin\theta+r(\theta)\cos\theta \end{aligned} Consequently:

Slope of a Polar Curve
The slope of a polar curve r=r(θ)r=r(\theta) is dydx=drdθsinθ+r(θ)cosθdrdθcosθr(θ)sinθ \dfrac{dy}{dx} =\dfrac {\;\dfrac{dr}{d\theta}\sin\theta+r(\theta)\cos\theta\;} {\;\dfrac{dr}{d\theta}\cos\theta-r(\theta)\sin\theta\;} provided the denominator is non-zero.

Caution: Remember, dydθ\dfrac{dy}{d\theta} is in the numerator and dxdθ\dfrac{dx}{d\theta} is in the denominator. It is easy to get these backwards.

Find the slope of the limacon r=1+2cosθr=1+2\cos\theta at:

  1. θ=π3\theta=\dfrac{\pi}{3}
  2. θ=2π3\theta=\dfrac{2\pi}{3}

Notice that θ=2π3\theta=\dfrac{2\pi}{3} is a point where the curve passes through the origin since cos2π3=12\cos\dfrac{2\pi}{3}=\dfrac{-1}{2} and r=1+2(12)=0r=1+2\left(\dfrac{-1}{2}\right)=0.

slopelimacon

Since drdθ=2sinθ\dfrac{dr}{d\theta}=-2\sin\theta, the slope is dydx=2sinθsinθ+(1+2cosθ)cosθ2sinθcosθ(1+2cosθ)sinθ\begin{aligned} \dfrac{dy}{dx} &=\dfrac{-2\sin\theta\sin\theta+(1+2\cos\theta)\cos\theta} {-2\sin\theta\cos\theta-(1+2\cos\theta)\sin\theta} \\ \end{aligned}

  1. At θ=π3=60\theta=\dfrac{\pi}{3}=60^\circ, sinθ=32\sin\theta=\dfrac{\sqrt{3}}{2} and cosθ=12\cos\theta=\dfrac{1}{2}. So: dydxθ=π/3=2(32)(32)+(1+2(12))(12)2(32)(12)(1+2(12))(32)=32+1323=12332=133\begin{aligned} \left.\dfrac{dy}{dx}\right|_{\theta=\pi/3} &=\dfrac{\;-2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) +\left(1+2\left(\frac{1}{2}\right)\right)\left(\frac{1}{2}\right)\;} {\;-2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) -\left(1+2\left(\frac{1}{2}\right)\right)\left(\frac{\sqrt{3}}{2}\right)\;} \\ &=\dfrac{\;-\,\frac{3}{2}+1\;}{\;-\,\frac{\sqrt{3}}{2}-\sqrt{3}\;} =\dfrac{-\,\frac{1}{2}}{-\,\frac{3\sqrt{3}}{2}}=\dfrac{1}{3\sqrt{3}} \end{aligned}
  2. At θ=2π3=120\theta=\dfrac{2\pi}{3}=120^\circ, sinθ=32\sin\theta=\dfrac{\sqrt{3}}{2} and cosθ=12\cos\theta=\dfrac{-1}{2}. So: dydxθ=2π/3=2(32)(32)+(1+2(12))(12)2(32)(12)(1+2(12))(32)=32+0320=33=3\begin{aligned} \left.\dfrac{dy}{dx}\right|_{\theta=2\pi/3} &=\dfrac{\;-2\left(\frac{\sqrt{3}}{2}\;\right)\left(\frac{\sqrt{3}}{2}\right) +\left(1+2\left(-\,\frac{1}{2}\right)\right)\left(-\,\frac{1}{2}\right)\;} {\;-2\left(\frac{\sqrt{3}}{2}\right)\left(-\,\frac{1}{2}\right) -\left(1+2\left(-\,\frac{1}{2}\right)\right)\left(\frac{\sqrt{3}}{2}\right)\;} \\ &=\dfrac{\;-\frac{3}{2}+0\;}{\;\frac{\sqrt{3}}{2}-0\;} =\dfrac{-3}{\sqrt{3}}=-\sqrt{3} \end{aligned}

In this example, notice that tan2π3=3\tan\dfrac{2\pi}{3}=-\sqrt{3} which is the slope at θ=2π3\theta=\dfrac{2\pi}{3}. This is no coincidence. Look at the figure. Notice that the curve comes into the origin tangent to the line θ=2π3\theta=\dfrac{2\pi}{3} and the slope of that line is tan2π3\tan\dfrac{2\pi}{3}.

If θ=θo\theta=\theta_o is a point where the polar curve r=r(θ)r=r(\theta) passes through the origin, i.e. r(θo)=0r(\theta_o)=0, then the slope at θ=θo\theta=\theta_o is: dydxθ=θo=tanθo \left.\dfrac{dy}{dx}\right|_{\theta=\theta_o}=\tan\theta_o

Proof

At θ=θo\theta=\theta_o, r(θo)=0r(\theta_o)=0. So dydxθ=θo=drdθ(θo)sinθo+r(θo)cosθodrdθ(θo)cosθor(θo)sinθo=drdθ(θo)sinθodrdθ(θo)cosθo=tanθo\begin{aligned} \left.\dfrac{dy}{dx}\right|_{\theta=\theta_o} &=\dfrac {\;\dfrac{dr}{d\theta}(\theta_o)\sin\theta_o+r(\theta_o)\cos\theta_o\;} {\;\dfrac{dr}{d\theta}(\theta_o)\cos\theta_o-r(\theta_o)\sin\theta_o\;} \\ &=\dfrac {\;\dfrac{dr}{d\theta}(\theta_o)\sin\theta_o\;} {\;\dfrac{dr}{d\theta}(\theta_o)\cos\theta_o\;} =\tan\theta_o \end{aligned}

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Find the slope of the 3-leaf rose r=sin3θr=\sin3\theta at:

  1. θ=π6\theta=\dfrac{\pi}{6}

  2. θ=π3\theta=\dfrac{\pi}{3}

slope3rose

Answer

dydxθ=π/6=3\left.\dfrac{dy}{dx}\right|_{\theta=\pi/6}=-\sqrt{3}

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Solution

Since drdθ=3cos3θ\dfrac{dr}{d\theta}=3\cos3\theta, the slope is: dydx=3cos3θsinθ+sin3θcosθ3cos3θcosθsin3θsinθ \dfrac{dy}{dx} =\dfrac{3\cos3\theta\sin\theta+\sin3\theta\cos\theta} {3\cos3\theta\cos\theta-\sin3\theta\sin\theta} At θ=π6=30\theta=\dfrac{\pi}{6}=30^\circ: sinθ=12,cosθ=32,sin3θ=1,cos3θ=0 \sin\theta=\dfrac{1}{2}, \qquad \cos\theta=\dfrac{\sqrt{3}}{2}, \qquad \sin3\theta=1, \qquad \cos3\theta=0 So: dydxθ=π/6=cosθsinθ=3212=3 \left.\dfrac{dy}{dx}\right|_{\theta=\pi/6} =\dfrac{\cos\theta}{-\sin\theta} =\dfrac {\;\dfrac{\sqrt{3}}{2}\;}{\;\dfrac{-1}{2}\;} =-\sqrt{3}

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Hint

Notice that θ=π3\theta=\dfrac{\pi}{3} is a point where the curve passes through the origin.

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Answer

dydxθ=π/3=3\left.\dfrac{dy}{dx}\right|_{\theta=\pi/3}=\sqrt{3}

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Solution

Since θ=π3\theta=\dfrac{\pi}{3} is a point where the curve passes through the origin: dydxθ=π/3=tanπ3=3 \left.\dfrac{dy}{dx}\right|_{\theta=\pi/3} =\tan\dfrac{\pi}{3}=\sqrt{3}

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