Diary outline for Math 152-pt3,

# Diary outline for Math 152-pt3, - Math 152 diary spring...

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Unformatted text preview: Math 152 diary, spring 2009: third section Earlier material Much earlier material In reverse order: the most recent material is first. Monday, May 4 (Lecture #26) Today we concentrate on calculus in polar coordinates. I'll first sketch some "classical" polar coordinate curves, and then I'll do some computations with arc length and area. In class I did examples with cosine. Here are similar examples with sine. A collection of examples r=3+sin(θ) Let's consider r=3+sin(θ). Since the values of sine are all between –1 and 1, r will be between 2 and 4. Any points on this curve will have distance to the origin between 2 and 4 (the green and red circles on the accompanying graph). When θ=0 (the positive x-axis) r is 3. As θ increases in a counterclockwise fashion, the value of r increases to 4 in the first quadrant. In the second quadrant, r decreases from 4 to 3. In the third quadrant, corresponding the sine's behavior (decrease from 0 to –1) r decreases from 3 to 2. In all of this {in|de}crease discussion, the geometric effect is that the distance to the origin changes. We're in a situation where the central orientation is what matters, not up or down or left or right . Finally, in the fourth quadrant r increases from 2 to 3, and since sine is periodic with period 2Π , the curve joins its earlier points. The picture to the right shows the curve in black. I'd describe the curve as a slightly flattened circle. The flattening is barely apparent to the eye, but if you examine the numbers, the up/down diameter of the curve is 6, and the left/right diameter is 6.4. Converting to rectangular coordinates A naive person might think, "Well, I could convert the equation r=3+sin(θ) to rectangular coordinates and maybe understand it better." Except under rare circumstances (I'll show you one below), the converted equation is very irritating and difficult to understand. For example, Let's start with r=3+sin(θ) and multiply by r. The result is r 2 =3r+r·sin(θ). I multiplied by r so that I would get some stuff I'd recognize from the polar/rectangular conversion equations. r 2 is x 2 +y 2 and r·sin(θ) is y. So I have x 2 +y 2 =3r+y, or x 2 +y 2 –y=3r. I would rather avoid square roots so I will square this, and get (x 2 +y 2 –y) 2 =9r 2 =9(x 2 +y 2 ). This is a polynomial equation in x and y of highest degree 4, defining this curve implicitly . I don't get much insight from this. r=2+sin(θ) Now consider r=2+sin(θ). Again, the values of sine are all between –1 and 1, so r will be between 1 and 3. Any points on this curve will have distance to the origin between 1 and 3. We can begin (?) the curve at θ=0 when r=2, and spin around counterclockwise. The distance to the origin increases to r=3 at θ=Π/2 (the positive y-axis). The distance to the origin decreases back to r=2 when θ=Π (the negative x-axis). The curve gets closest to the origin when θ=3Π/2 (the negative y-axis) when r=1....
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Diary outline for Math 152-pt3, - Math 152 diary spring...

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