1 10 points by finding a cartesian equation for the curve...

1. (10 points) By finding a Cartesian equation for the curve, identify the curve given in polar coordinates by (a) $r = 4 \sin \theta$ (b) $r = \frac{1}{2 \cos \theta - 7 \sin \theta}$ 2. (20 points) Consider the polar curve $r = \sin(2\theta)$, for $0 \le \theta \le 2\pi$. (a) Sketch the polar curve (b) Find the area enclosed by the polar curve

Transcript

00:01 Okay, so what we have is r equals 1 minus 2 sine of theta.

00:05 And in part a, what you want to do is take that 1 minus 2 sine of theta and just set it equal to 0.

00:11 So what you can do is subtract 1 to the right side.

00:15 0 minus 1 is negative 1, then divide by negative 2.

00:19 And what you need to ask yourself then is where does sine of theta equal positive 1 half? so hopefully you have the unit circle memorized because that happens over here at root 3 over 2, 1 half, but again over here in quadrant 2 and those two angles hopefully you have memorized are pi over 6 and 5 pi over 6 because we don't really care that the x -coordinate is negative, we care that sine is positive 1 half.

00:57 So the symmetry of the curve might actually be helpful if you just graph.

01:04 So maybe i'll just graph it first.

01:07 So i like to just draw the polar axis.

01:10 And if we plugged in 0 in for theta, well, sine of 0 is 0.

01:19 So 1 minus 0 gives me 1.

01:21 So we are to the right 1, and that's your order pair 1, 0.

01:27 At pi over 6, we're already at 0.

01:29 0 and then if i had pi over 2 in there so pi over 2 is straight up 0 1 and we end up with 1 minus 2 because 2 times 1 is 1 is 2 1 minus 2 is negative 1 so what that means is we need to be going down what's this value 0 sorry pi over 2 negative 1 and then it loops back around for the the y value goes out to negative one.

02:07 Let's see, if we continue to loop all the way to here, zero, negative one, we end up with one minus two times negative one, so one plus two is three.

02:18 So this is going to continue to loop down here...

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