# Recitation 29

Example 1: Investigate the family of polar curves given by r=1+c\sin\theta. How does the shape change as c changes? (These curves are called limaçons, after a French word for snail, because of the shape of the curves for certain values of c.)

Hint: The curve is symmetric about the vertical line \theta=\pi/2. The curve given by c and the one given by -c are symmetric about the polar axis. If c=0, the curve is a circle. If c=1, the curve has a “cusp” at the origin. If c>1, the curve crosses itself at the origin.

Example 2: The Cartesian coordinates of a point are given (2,-2) (i) Find polar coordinates (r,\theta) of the point, where r>0 and 0\leq\theta < 2\pi. (ii) Find polar coordinates (r,\theta) of the point, where r<0 and 0\leq\theta<2\pi.

Solution: (i) (2\sqrt{2}, 7\pi/4); (ii) (-2\sqrt{2}, 3\pi/4).

Example 3: Find the points on the given curve where the tangent line is horizontal or vertical. r=3\cos\theta.

Solution: The curve (in Cartesian coordinates) can be parametrized by \theta: x=r\cos\theta = 3\cos^2\theta, y=r\sin\theta=3\sin\theta\cos\theta. The point where the tangent line is horizontal (respectively, vertical) is given by \theta such that dy/d\theta = 3\cos^2\theta - 3\sin^2\theta = 0 (respectively, dx/d\theta = -6\sin\theta\cos\theta = 0), i.e. \theta = n\pi/4, where n is odd (respectively, \theta = n\pi/4, where n is even). The corresponding points are, in polar coordinates, (3/\sqrt{2}, \pm\pi/4) (respectively, (0,0), (3,0)).

Remark: The curve is actually, in Cartesian coordinates, a circle centered at (3/2, 0) with radius 3/2.

Example 4: Find the area of the region enclosed by one loop of the curve r=1+2\sin\theta (inner loop).

Solution: The inner loop is given by 7\pi/6 \leq \theta \leq 11\pi/6 (that is exactly when r\leq 0). The are enclosed by the inner loop is \int_{7\pi/6}^{11\pi/6}\frac{1}{2}(1+2\sin\theta)^2d\theta = \pi - \frac{3}{2}\sqrt{3}.

Example 5: Find the area of the region that lies inside the first curve and outside the second curve. r=3\cos\theta, r=1+\cos\theta.

Solution: The intersections (r,\theta) of the curves satisfy r=3\cos\theta=1+\cos\theta, and so the intersections are (3/2, \pm\frac{\pi}{3}). The area that lies inside the first curve and outside the second curve is then given by A=\int_{-\pi/3}^{\pi/3}\frac{1}{2}[(3\cos\theta)^2-(1+\cos\theta)^2]d\theta = \pi

Example 6: Find all points of intersection of the given curves r=\sin\theta, r=\sin 2\theta.

Solution: The intersection (r,\theta) stisfies r=\sin\theta=\sin2\theta=2\sin\theta\cos\theta, and so either \sin\theta = 0 or \cos\theta=1/2. Therefore the intersections are (0,0), (\sqrt{3}/2, \pi/3), (\sqrt{3}/2, 2\pi/3).

Example 7: Find the exact length of the polar curve r=2\cos\theta, 0\leq\theta\leq\pi.

Solution: The length of the curve is \int_0^\pi \sqrt{(2\cos\theta)^2+(2\sin\theta)^2}d\theta =2\pi.