Class 16 - Polar coordinates (10/10/2017, Tuesday)

2nd Gateway

You must pass the second Gateway by Oct 13, Friday. Otherwise, your grade will be damaged a lot!

Webwork

• 8.3 due 10/18 Wed

New team

Seat with your new team members. Introduce yourselves to each other.

Today we will roughly cover Section 8.3 of your book.

Polar coordinate: a new presentation of points on plane

You are probably used to write $(1, \sqrt{3})$ to mean a point on a plane. This means that if I give you an $x$-coordinate $1$ and an $y$-coordinate $\sqrt{3}$, then you can give me exactly one point on the plane that have the specified coordinates.

There is another way for me to give you one point on the plane. If I tell you the distance from the origin, say $2$, and the angle from the $x$-axis in the counterclockwise direction, say $\pi/3$, then you realize that that must give you the same point $(1, \sqrt{3})$.

Exercise. Really check what you get are the same point.

In general, if I give you the distance $r \geq 0$ and the angle $0 \leq \theta < 2\pi,$ then the point that have these quantities is exactly $(r \cos(\theta), r \sin(\theta)).$ That is, the relationship between the polar coordinate $(r, \theta)$ and the rectangular coordinate $(x, y)$ is given by:

$x = r \cos(\theta)$ and $y = r\sin(\theta).$

Exercise. Given above, write down $r$ and $\tan(\theta)$ in terms of $x$ and $y.$

Area using polar coordinate

Now, let's consider how to compute area using polar coordinates. Again, we slice the desired area into pieces, but this time we cut the angle into many pieces. Hence, we consider $\Delta \theta$ and each slice will have the area of $\pi r^{2} \Delta\theta/(2\pi) = (1/2)r^{2} \Delta\theta.$ If $r = f(\theta)$, a function in $\theta$, then the area we want to compute enclosed by the graph $r = f(\theta)$ is can be approximated by

\[\sum_{\theta} (1/2) f(\theta)^{2} \Delta \theta.\]

As always, when the function $f$ is smooth enough, then we can take $\Delta\theta \rightarrow 0$ and this would give us the following expression for the actual area:

\[\int_{a}^{b} (1/2)f(\theta)^{2} d\theta,\]

where $\theta = a$ is the starting angle and $\theta = b$ is the ending angle.

Parametrization tricks: Derivative and Arc length

When we have $r = f(\theta)$ in polar coordinate, then $x = f(\theta)\cos(\theta)$ and $y = f(\theta)\sin(\theta),$ so we can think of $(x, y) = (x(\theta), y(\theta)$ as a parameterization in $0 \leq \theta < 2\pi.$ Then we may take the derivative of $y$ in $x$ as follows:

\[\dfrac{dy}{dx} = \dfrac{y'(\theta)}{x'(\theta)}.\]

Exercise. Compute the right-hand side carefully.

Moreover, the arc length made by the graph is from $\theta = a$ to $\theta = b$ is given by

\[\int_{a}^{b}\sqrt{x'(\theta)^{2} + y'(\theta)^{2}}d\theta.\]

Exercise. Simplify the integral above.

Exercise*. Solve #1 of Exam 2 Fall 2016.