# Recitation 28

Polar coordinates: We extend the meaning of polar coordinates (r, \theta) to the case in which r is negative by agreeing that, the points (-r, \theta) and (r,\theta), lie on the same line through O and at the same distance r from O, but on opposite sides of O. If r>0, the point (r,\theta) lies in the same quadrant as \theta; if r<0, it lies in the quadrant on the opposite side of the pole. Notice that (-r,\theta) represents the same point as (r,\theta+\pi).

Symmetry: When we sketch polar curves it is sometimes helpful to take advantage of symmetry.

1. If a polar equation is unchanged when \theta is replaced by -\theta, the curve is symmetric about the polar axis.
2. If the equation is unchanged when r is replaced by -r, or when \theta is replaced by \theta+\pi, the curve is symmetric about the pole. (This means that the curve remains unchanged if we rotate it through 180^\circ about the origin.)
3. If the equation is unchanged when \theta is replaced by \pi-\theta, the curve is symmetric about the vertical line \theta=\pi/2.

Example 1: Determine the symmetry of the curves (a) r = 2; (b) \theta = 1; (c) r = 2\cos\theta; (d) r=1+\sin\theta; (e) r=\cos 2\theta.

Hint: The curve is symmetric about (a) the polar axis, the pole and the vertical line; (b) the pole; (c) the polar axis; (d) the vertical line; (e) the polar axis, the pole and the vertical line.

Example 2: Sketch the region in the plane consisting of points whose polar coordinates satisfy the given conditions 2 < r < 3, 5\pi / 3 \leq \theta \leq 7\pi/3.

Hint: 5\pi / 3 \leq \theta \leq 7\pi/3 is equivalent to -\pi / 3 \leq \theta \leq \pi/3.

Example 3: Find a formula for the distance between the points with polar coordinates (r_1, \theta_1) and (r_2, \theta_2).

Solution: The distance is equal to \begin{aligned}\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} & = \sqrt{(r_1\cos\theta_1 - r_2\cos\theta_2)^2+(r_1\sin\theta_1 - r_2\sin\theta_2)^2} \\ & = \sqrt{r_1^2 + r_2^2 - 2r_1r_2(\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2)} \\ & = \sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_1-\theta_2)}.\end{aligned}

Example 4: Identify the curve r^2\cos 2\theta = 1 by finding a Cartesian equation for the curve.

Solution: Note that \cos^2\theta = \cos^2\theta - \sin^2\theta and x = r\cos\theta, y=r\sin\theta. The Cartesian equation is x^2 - y^2 = 1, and so the curve is a hyperbla.

Example 5: Find a polar equation for the curve represented by the given Cartesian equation x^2 + y^2 = 2cx.

Solution: Note that x^2 + y^2 = r^2 and x=r\cos\theta. The polar equation is r^2 = 2cr\cos\theta, and so r=2c\cos\theta (note that the case r=0 is already included).

Example 6: Sketch the curve with the given polar equation r=2\cos 4\theta by first sketching the graph of r as a function of \theta in Cartesian coordinates.

Hint: Sketch first the graph of r=2\cos4\theta for 0 \leq \theta \leq 2\pi.

Petals: r=\cos(k\theta) is an equation of a rose. If k is even, the rose has 2k petals. If k is odd, the rose has k petals.