Students, A couple of students have noted that there are several questions on this week's homework which we didn't really cover well in class, related to calculus-type questions we could ask about curves given in polar coordinates. I'm sorry about that: different coordinate questions are a favorite topic of mine and I got kind of distracted. (Ask me some time about shower controls and Moebius strips!) I have advised the TA of this situation. Tomorrow (Thursday) he will give you more help with these topics. I have asked him not to include this newest material on his quiz. Here, in brief, is what you need to know to do the homework questions. 1. We can give parameterizations ("driving instructions") in polar coordinates just as well as we give them in Cartesian coordinates. Just tell me "where" to be at each time t ; the location can be given in polar coordinates ( r(t), theta(t) ) but that can also be converted to the corresponding Cartesian coordinates in the usual way: x(t) = r(t) cos(theta(t)) and y(t) = r(t) sin(theta(t)) so there is really nothing new here. 2. We can compute slopes of these curves (and thus get Cartesian-coordinate equations for the tangent lines). The slope is always dy/dx, which we can compute as before as (dy/dt)/(dx/dt) . 3. We can also compute areas and arc lengths by the same process: polar parameterization leads to Cartesian parameterization which leads to the formulas we gave on Monday. But there are simpler formulas in some cases. In summary: you can always convert to Cartesian coordinates and then do calculus Let me expand on these three comments. 1. Try some simple examples of polar parameterizations, e.g. what if my position at time t is (2t, 3t)? If you plot a few points, you will see that this path traces out a spiral. Sometimes the curve can be expressed implicitly in polar coordinates -- in this case, for example, you can eliminate t from the equations { r = 2t, theta = 3t } to see that 3r = 2 theta; that equation describes a curve (implicitly, in polar coordinates). You can also describe this curve explicitly in polar coordinates; obviously r = (2/3) theta . Since we have formulas to convert between Cartesian and polar coordinates, you can also express this curve implicitly in Cartesian coordinates, for example as 3 sqrt(x^2+y^2) = 2 arctan(y/x). Or you can always use the conversions to give a Cartesian parameterization of the same curve, in this case (x,y) = ( (2t) cos(3t), (2t) sin(3t) ). So there you have many different expressions for the same curve, each of them deducible from the parameterization in polar coordinates. What you see here is typical: the parameterization in polar coordinates is often the simplest of these many ways to describe the curve. Keep this in mind in particular if the curve is given explicitly in polar coordinates, e.g. as r = 2cos(theta). First rewrite as a polar parameterization: (r, theta) = ( 2cos(t), t ). Then use this to get a Cartesian parameterization: (x,y) = ( 2cos(t) cos(t), 2cos(t) sin(t) ). (BTW this is a very simple curve to write implicitly in Cartesian coordinates if you first multiply the defining equation by r : the resulting equation r^2=2r cos(theta) simply says (in Cartesian coordinates) that x^2+y^2 = 2x, i.e. (x-1)^2 + y^2 = 1. That's a circle.) 2. Are you good at memorizing formulas? If so, you can memorize the general form of the (Cartesian) slope of a line given by a polar parameterization. Suppose r and theta are parameterized as functions of t. Then the Cartesian coordinates of your location at time t are, as usual, (x,y) = ( r(t) cos(theta(t)), r(t) sin(theta(t)) ). So the line tangent to the curve right there will have slope equal to dy/dx = (dy/dt)/(dx/dt) ; the numerator is, by the chain rule, r'(t) sin(theta(t)) + r(t) cos(theta(t)) theta'(t) and the denominator is r'(t) cos(theta(t)) - r(t) sin(theta(t)) theta'(t) so the slope can be computed as the ratio of these. Memorize that if you wish, or simply go through the intermediate stage of computing ( x(t), y(t) ) first. 3. I have given above a process that can help you find the area inside a curve given parametrically in polar coordinates: convert to Cartesian coordinates and then integrate y dx around the curve (clockwise). It turns out there is another formula that you can use in the special case that the curve is given explicitly in polar coordinates in the form r = f(theta). Most students find this special-case formula to be simpler and so they memorize it (and use it) when applicable, but you can just continue to use the general procedure if you wish. You can see that formula in the book, and they justify it by computing the areas of little pie-slice-shaped pieces; add those and take a limit, then recognize the whole thing as a limit of Riemann sums, i.e. an integral. But allow me to convince you that you will get exactly the same area if you do what I proposed, integrating y dx. Indeed, we can first give a (polar) parametrization of the same curve: (r, theta) = ( f(t), t ) and then the corresponding Cartesian parametrization (x,y) = ( f(t) cos(t), f(t) sin(t) ) so that what you must integrate along the boundary of the region is y dx = y(t) x'(t) dt = ( f(t) sin(t) )( f'(t) cos(t) - f(t) sin(t) ) dt = f(t) f'(t) sin(t) cos(t) dt - f(t)^2 sin(t)^2 dt (Note that taking increasing t will trace out the curve COUNTER clockwise.) Use Integration By Parts on the first term, with u = sin(t) cos(t) and dv = f(t) f'(t) dt . Then v = (1/2) (f(t))^2 and du = (cos(t)^2 - sin(t)^2)dt =(1 - 2 sin(t)^2) dt . Thus the " - v du " term in Integration By Parts is the integral of - (1/2) (f(t))^2 dt + (f(t))^2 sin(t)^2) dt and the second term here cancels the second term in the last line of the previous paragraph. Also note that uv = (1/2) ( f(t) cos(t) )( f(t) sin(t) ) = (1/2) x(t) y(t); so if your curve is closed, then at the beginning and ending values of t you are at the same point (x(t1)=x(t2) and y(t1)=y(t2)) so when you evaluate "uv" at these two values of t and subtract, you get zero. So all that will remain of our integral of y dx isn the integral of -(1/2) f(t)^2 dt . Since we are traversing the curve counter-clockwise instead of clockwise, we are computing the negative of the enclosed area; in other words we conclude that the enclosed area is exactly A = integral from t=t1 to t=t2 of (1/2) ( f(t) )^2 dt or equivalently it's the integral of (1/2) r^2 d theta, exactly the formula given in the book. You may use this formula whenever you are in the special situation of a curve given explicitly in polar coordinates (i.e. r = f(t)), and it's usually a little faster than working out what y dx is and then integrating that. Your choice!