- 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.

- First, we select the
- 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*.

- In the context of polar coordinates, the positive \(x\)-axis is also called the
- 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.

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.

- 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.