# Polar coordinates

## Cartesian coordinates and polar coordinates

• Let's review how we set up the Cartesian coordinate system.
• First, we select the origin $$O$$.
• Then we select a half-line, which will be the positive $$x$$-axis. Usually this points to the right.
• The half line that has a right angle counterclockwise from the positive $$x$$-axis gives the positive $$y$$-axis. Usually this points upward.
• Now to find the Cartesian coordinates $$(x,y)$$ of a point $$P$$, we draw the right triangle $$OQP$$, where
• $$Q$$ is on the $$x$$-axis, and
• $$OQP$$ is the right angle.
• Note that this defines the triangle in a unique way. $$Q$$ is called the projection of $$P$$ onto the $$x$$-axis.
• We have $$|x|=|OQ|$$. $$x>0$$ if $$\vec{OQ}$$ points in the positive $$x$$-direction, otherwise $$x<0$$.
• We have $$|y|=|QP|$$. $$y>0$$ if $$\vec{QP}$$ points in the positive $$y$$-direction, otherwise $$y<0$$.
• The polar coordinate system also uses the origin $$O$$ and the half-line giving the positive $$x$$-direction.
• In the context of polar coordinates, the positive $$x$$-axis is also called the polar axis.
• Now to find the polar coordinates $$(r,\theta)$$ of a point $$P$$:
• $$r$$ is the distance $$|OP|$$.
• $$\theta$$ is the angle from the polar axis to $$\vec{OP}$$.
• If $$O=P$$, that is $$r=0$$, then $$\theta$$ is any angle. It's convention to make it $$\theta=0$$.
• We let $$(-r,\theta)$$ be same point as $$(r,\theta+\pi)$$.
• Note that unlike Cartesian coordinates, multiple polar coordinates can give the same point
• The alternative notation gives $$P(-r,\theta)=P(r,\theta+\pi)$$.
• We also have $$(r,\theta)=(r,\theta+2\pi)$$.
• Also, we have $$(0,\theta)=(0,\eta)$$ for any two angles $$\theta$$ and $$\eta$$.
• Note that by construction we have the following conversion formulae. \begin{align*} x=r\cos\theta\quad&y=r\sin\theta\\ r=\sqrt{x^2+y^2}\quad&\theta=\begin{cases} \tan^{-1}(y/x) & x>0 \\ \tan^{-1}(y/x)+\pi & x<0\\ \pi/2 & x=0,\,y>0\\ 3\pi/2 & x=0,\,y<0\\ 0 & x=0,\,y=0. \end{cases} \end{align*}
• In the formula for $$\theta$$, it's important to make $$-\pi/2\le\tan^{-1}(y/x)\le\pi/2$$.
• Exercises. 10.3: 3, 4, 5, 6.

## Polar equations.

• Just as with Cartesian equations, it's possible to gives curves with polar equations.

• $$r=2$$.
• This collects the points with polar coordinates $$(2,\theta)$$ for some $$\theta$$.
• That is, points with distance from the origin $$|OP|=2$$.
• This gives the circle with centre the origin and radius 2.
• We can also use the conversion formula: \begin{align*} \sqrt{x^2+y^2}=&2\\ x^2+y^2=4. \end{align*}
• $$\theta=\pi/3$$.
• This collects the points with polar coordinates $$(r,\pi/3)$$ for some $$r$$.
• That is, we need the slope of the line $$\overline{OP}$$ to be $$\pi/3$$.
• Since the origin can have polar coordinates $$(0,\pi/3)$$ it is included
• Therefore, this is the line through $$O$$ with slope $$\tan\pi/3=\sqrt3$$.
• A possible Cartesian equation is: $y=\sqrt3x.$
• $$r=2\cos\theta$$.
• If you can't see from the equation how this curve looks like (I can't), then you can transform it to Cartesian coordinates.
• We know that $$x=r\cos\theta$$, and we'd much rather use that than deal with $$\cos\tan^{-1}(y/x)$$.
• Since $$(0,\pi/2)$$ is a point, it's OK to multiply the equation with $$r$$: \begin{align*} r=&2\cos\theta\\ r^2=&2r\cos\theta\\ x^2+y^2=&2x\\ (x-1)^2+y^2=&1 \end{align*}
• We see that the curve is the circle with centre $$(1,0)$$ and radius $$1$$.
• Exercises. 10.3: 15, 17, 19, 21, 23, 25, 65.

## Tangents to polar curves.

• Consider a polar curve $$C:r=f(\theta)$$.
• Using the conversion formula, we can get a parametrization for it: $x=f(\theta)\cos\theta,\,y=f(\theta)\sin\theta$
• To find the slopes of tangent lines, we can use the formula for parametric curves: $\frac{\mathrm dy}{\mathrm dx}=\frac{y'(\theta)}{x'(\theta)}=\frac{f'(\theta)\sin\theta+f(\theta)\cos\theta}{f'(\theta)\cos\theta-f(\theta)\sin\theta}.$
• Consider the curve $$r=1+\sin\theta$$.
• Click here for the graph of $$r=f(\theta)$$.
• You can see that $$r=0$$ when $$\theta=-\pi/2$$, then it starts to increase, hitting $$r=1$$ at $$\theta=0$$, having a maximum of $$r=2$$ at $$\theta=\pi/2$$, and then starting to decrease, hittin $$r=1$$ at $$\theta=\pi$$.
• Click here for a plot of the curve. It's called a cardioid for its heart shape.
• To find the slopes of its tangent lines, we calculate the derivatives of the coordinate functions: \begin{align*} y'(\theta)=&\cos(\theta)\sin(\theta)+\cos\theta+\sin\theta\cos\theta=\cos\theta(1+2\sin\theta),\\ x'(\theta)=&\cos^2\theta-\sin\theta-\sin^2\theta=1-\sin\theta-2\sin^2\theta=(1+\sin\theta)(1-2\sin\theta) \end{align*}
• By substitution, we find that the slope of the tangent line at $$\theta=\pi/3$$ is: $\frac{y'(\theta=\pi/3)}{x'(\theta=\pi/3)}=\frac{(1+\sqrt3)/2}{(1+\sqrt3/2)(1-\sqrt3)}=\frac{1+\sqrt3}{(2+\sqrt3)(1-\sqrt3)}=\frac{1+\sqrt3}{-1-\sqrt3}=-1$
• It has horizontal tangents where $$y'(\theta)=\cos\theta(1+2\sin\theta)=0$$. That is, either $$\theta=\pi/2,3\pi/2$$, or $$\theta=-\pi/6,-5\pi/6$$.
• Exercises. 10.3: 55, 57, 61, 63, 66.