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 r²=3r+r·sin(θ). I multiplied by r so that I would get some stuff I'd recognize from the polar/rectangular conversion equations. r² is x²+y² and r·sin(θ) is y. So I have x²+y²=3r+y, or x²+y²–y=3r. I would rather avoid square roots so I will square this, and get (x²+y²–y)²=9r²=9(x²+y²). 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. Finally, r increases (as θ increases in the counterclockwise fashion) to r=3 again when θ=2Π.
Here the "deviation" from circularity in the curve is certainly visible. The bottom seems especially dented.

r=1+sin(θ)
We decrease the constant a bit more, and look at r=1+sin(θ). The values of sine are all between –1 and 1, so r will be between 0 and 2. The (red) inner circle has shrunk to a point. This curve will be inside a circle of radius 2 centered at the origin. We begin our sweep of the curve at 0, when r is 1. Then r increases to 2, and the curve goes through the point (0,2). In the θ interval from Π/2 to Π, sin(θ) decreases from 1 to 0, and the curves moves closer to the origin as r decreases from 2 to 1. Something rather interesting now happens as θ travels from Π to 3Π/2 and then from 3Π/2 to 2Π. The rectangular graph of 1+sine, shown here, decreases down to 0 and then increases to +1. The polar graph dips to 0 and then goes back up to 1. The dip to 0 in polar form is geometrically a sharp point! I used "!" here because I don't believe this behavior is easily anticipated. The technical name for the behavior when r=3Π/2 is cusp.
This curve is called a cardioid from the Latin for "heart" because if it is turned upside down, and if you squint a bit, maybe it sort of looks like the symbolic representation of a heart. Maybe.

r=1/2+sin(θ)
Let's consider r=1/2+sin(θ). The values of sine are all between –1 and 1, so r will be between –1/2 and 3/2. The (red) inner circle actually had "radius" –1/2, and it consists, of course, of points whose distance to the pole, (0,0), is 1/2. When θ is 0, r is 1/2. In the first two quadrants, 1/2+sin(θ) increases from 1/2 to 3/2 and then backs down to 1/2. In the second two quadrants, when θ is between Π and 2Π, more interesting things happen.
The rectangular graph on the interval [0,2Π] of sine moved up by 1/2 shows that this function is 0 at two values, and is negative between two values. The values are where 1/2+sin(θ)=0 or sin(θ)=–1/2. The values of θ satisfying that equation in the interval of interest are Π+Π/6 and 2Π–Π/6. The curves goes down to 0 distance from the origin at Π+Π/6, and then r is negative until 2Π–Π/6. The natural continuation of the curve does allow negative r's, and the curve moves "behind" the pole, making a little loop inside the big loop. Finally, at 2Π–Π/6, the values of r become positive, and the curve links up to the start of the big loop.
This curve is called a limaçon. The blue lines are lines with θ=Π+Π/2 and θ=2Π–Π/6. These lines, for the θ values which cross the pole, are actually tangent to the curve at the crossing points.

r=0+sin(θ)
Let's try a last curve in this family, with the constant equal to 0. What does r=sin(θ) look like? A graph is shown to the right.
There are several interesting features of this graph. First, this is a polar curve which does have a nice rectangular (xy) description. If we multiply r=sin(θ) by r, we get r²=r·sin(θ), so that x²+y²=y. This is x²+y²–y=0 or, completing the square, x²+y²–2(1/2)y+(1/2)²–(1/2)²=0 so that (x–0)²+(y–1/2)²=(1/2)². This is a circle of radius 1/2 and center (0,1/2), exactly as it looks.
The moving "picture" of this curve is quite different. Between 0 and π it spins once around the circle but then from π to 2π it goes around the circle another time! So this is really somehow two circles, even though it looks like only one geometrically.

More information about these curves is available here

Length of polar curves
The formula is ∫_θ=α^θ=βsqrt(r²+(dr/dθ)²)dθ. This formula is gotten from the parametric curve formula on p.652 of the textbook. I used it to find the length of the cardioid (the double angle formula from trig is needed). Then I used it to find the length of a circle (!), but here the novelty is that we actually trace the circle r=cos(θ) twice from 0 to 2π, so some care is needed if we only wanted to find the length of one circumference.

"Sketching" roses
Here are dynamic pictures of two roses. The first is the one I sketched in class r=cos(3θ). It is covered twice and has 3 "petals". The second is r=cos(4θ). It is only covered once, and it has 8 petals! Wow, polar coordinates can be annoying!

r=cos(3) This is a three-leafed rose. Please note that the graph shows one sweep, as θ goes from 0 to 2Π. The rectangular graph, shown here, has three pairs of ups and downs. The polar trace covers the leaves twice. The six up-and-downs of cos(3θ) (magically?) reduce to retracings of half of the loops. I hope I made this evident. I introduced some deliberate distortion in the second tracing. (!) Without the distortion, the second tracing could not be seen at all. The imaginary "point" travels over the computer screen's pixels and colors them (from white to black in this case). The second trip could not be noticed since the pixels had already been flipped to black. So I put in a small perturbation so that the second layer of travel could be seen.

r=cos(4θ) This "rose" has 8 leaves or petals, and the dynamic way it is traced is weird and wonderful to me. The rectangular graph. to the right, shows four bumps up and four bumps down. There are no retracings of already colored points, so that the wiggles up and down of cos(4θ) all result in 8 leaves.

Area inside one petal of r=cos(3θ)
Well, cos(3θ) "first" (going from 0 to 2π) is 0 when 3θ=π/2. So we get half a petal by integrating from 0 to π/6. The formula is ∫_α^β(1/2)r²dθ for area in polar coordinates (see the discussion on p.649 of the textbook), so this becomes (for the whole petal, we need to double):
2·(1/2)∫₀^π/6 cos(3θ)²dθ. This can be computed using a trig identity.

Another kind of spiral
I didn't talk about Exponentials and snails, darn it! Curves of the form r=a e^bθ are spirals of a different kind than what we've drawn (for example, different from the spiral drawn for the QotD last time). All of these spirals have a strange and wonderful geometric property. If a ray is drawn from the pole (the origin) then the angle the ray makes with the tangent line of the spiral at any intersection is the same. I attempted to illustrate this with the first picture to the right. It turns out that this silly geometric property has natural consequences in terms of the energy efficiency of its construction. A snail shell when considered transversal (perpendicular) to the axis of symmetry usually is one of these curves. You can read a wikipedia article about these spirals, which have a number of different names (of course!).
To the right is a machine-drawn picture of r=e^.25θ as θ goes from –Π to 5Π. The .25 was put in to make the exponential not grow too fast so the picture would be tolerable.

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 r²=3r+r·sin(θ). I multiplied by r so that I would get some stuff I'd recognize from the polar/rectangular conversion equations. r² is x²+y² and r·sin(θ) is y. So I have x²+y²=3r+y, or x²+y²–y=3r. I would rather avoid square roots so I will square this, and get (x²+y²–y)²=9r²=9(x²+y²). 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. Finally, r increases (as θ increases in the counterclockwise fashion) to r=3 again when θ=2Π. Here the "deviation" from circularity in the curve is certainly visible. The bottom seems especially dented.
r=1+sin(θ) We decrease the constant a bit more, and look at r=1+sin(θ). The values of sine are all between –1 and 1, so r will be between 0 and 2. The (red) inner circle has shrunk to a point. This curve will be inside a circle of radius 2 centered at the origin. We begin our sweep of the curve at 0, when r is 1. Then r increases to 2, and the curve goes through the point (0,2). In the θ interval from Π/2 to Π, sin(θ) decreases from 1 to 0, and the curves moves closer to the origin as r decreases from 2 to 1. Something rather interesting now happens as θ travels from Π to 3Π/2 and then from 3Π/2 to 2Π. The rectangular graph of 1+sine, shown here, decreases down to 0 and then increases to +1. The polar graph dips to 0 and then goes back up to 1. The dip to 0 in polar form is geometrically a sharp point! I used "!" here because I don't believe this behavior is easily anticipated. The technical name for the behavior when r=3Π/2 is cusp. This curve is called a cardioid from the Latin for "heart" because if it is turned upside down, and if you squint a bit, maybe it sort of looks like the symbolic representation of a heart. Maybe.
r=1/2+sin(θ) Let's consider r=1/2+sin(θ). The values of sine are all between –1 and 1, so r will be between –1/2 and 3/2. The (red) inner circle actually had "radius" –1/2, and it consists, of course, of points whose distance to the pole, (0,0), is 1/2. When θ is 0, r is 1/2. In the first two quadrants, 1/2+sin(θ) increases from 1/2 to 3/2 and then backs down to 1/2. In the second two quadrants, when θ is between Π and 2Π, more interesting things happen. The rectangular graph on the interval [0,2Π] of sine moved up by 1/2 shows that this function is 0 at two values, and is negative between two values. The values are where 1/2+sin(θ)=0 or sin(θ)=–1/2. The values of θ satisfying that equation in the interval of interest are Π+Π/6 and 2Π–Π/6. The curves goes down to 0 distance from the origin at Π+Π/6, and then r is negative until 2Π–Π/6. The natural continuation of the curve does allow negative r's, and the curve moves "behind" the pole, making a little loop inside the big loop. Finally, at 2Π–Π/6, the values of r become positive, and the curve links up to the start of the big loop. This curve is called a limaçon. The blue lines are lines with θ=Π+Π/2 and θ=2Π–Π/6. These lines, for the θ values which cross the pole, are actually tangent to the curve at the crossing points.
r=0+sin(θ) Let's try a last curve in this family, with the constant equal to 0. What does r=sin(θ) look like? A graph is shown to the right. There are several interesting features of this graph. First, this is a polar curve which does have a nice rectangular (xy) description. If we multiply r=sin(θ) by r, we get r²=r·sin(θ), so that x²+y²=y. This is x²+y²–y=0 or, completing the square, x²+y²–2(1/2)y+(1/2)²–(1/2)²=0 so that (x–0)²+(y–1/2)²=(1/2)². This is a circle of radius 1/2 and center (0,1/2), exactly as it looks. The moving "picture" of this curve is quite different. Between 0 and π it spins once around the circle but then from π to 2π it goes around the circle another time! So this is really somehow two circles, even though it looks like only one geometrically.