We will have Friday quizzes for a few weeks to help you prepare for the midterm exam on 10 March. The score will count in the homework pool. So, the important thing is to prepare for it and doit, even if it does not "count" very much. On Monday, 22 February, we discussed problems 5,7 and 11 on p 192. I was pleased to see that some of you worked on them over the weekend. Rather than spoil the game (Kaplan offers very few examples) by just doing the problems in the book, here is a sample problem. Problem A. Show that curl _v = 0. Then solve grad f = _v for f. Consider f = z-xy. grad f = [-y,-x,1] whose curl really is 0. The "partial integration" that allows you to recover f from grad f has to be done in one of 6 = 3! ways of ordering x,y,z. But once you have made the choice, proceed thus Step 1. -y = f_x => f = -xy + g(y,z) . (Note how the "constant of integration" now becomes a function independent of the variable being integrated with respect to.) Step 2 -x = f_y = -x + g_y => g_y = 0 => g=g(z) Setp 3 1 = f_z = g'(z) => g = z + k Check that grad( -xy + z + k) really is [-y,-z,1]. Comment, two solutions of the problem will differ (locally) by a constant, just as in the baby calculus. But this is not the case for the next version of this problem: Problem B. Show that div _v = 0. Then solve curl _u = _v for _u. If we name the components of the unknown vector field _u=[a,b,c] and let's call the known components of _v = [A,B,C], we have three partial differential equations to solve, for which there is no systematic method. So we utilize two facts: Theorem. Two solutions will differ by a gradient vector field. Proof: curl _u = curl _w => curl (_u-_w) = _0. Hence, by Problem A, we can solve for grad f = _u - _w. Assumption. Let's try to find a solution for which c = 0. This changes the three differential equations to these: A(x,y,z) = c_y - b_z = - b_z => b = - \int A dz + g(x,y) B = a_z - c_x = a_z => a = \int B dz + h(x,y) C = b_x - a_y = known(x,y,z) + h_x - g_y . Now you get a choice for integrating C(x,y,z)-known(x,y,z) = h_x by assuming g=0 or = -g_y by assuming h=0. The reason you can make all these choices is that you have all the gradient fields in the world to make up the difference. So given this scheme we integrated _v = [ 42, xz, ye^x ] which is divergence free. b = -42z + g a = xz^2/2 (letting h=0) ye^x = g_x - 0 => g = ye^x. \d_x \d_y \d_z Check: curl [xz^2/2, -42z+ye^x, 0] = [ 42, xz, ye^x ] . Finally, we discussed some useful ways of approaching 11. The importance of Kaplan's advice, to use the identities already found'' cannot be overestimated. Here's the point of this exercise. Each of these examples come up somewhere in the application of the vectore calculuse to the engineering and physical sciences. So one must be prepared to solve such monstrosities. However, one can, in the time honored way of mathematics, try to reduce the problem to simpler components, and then try to remember these (think back to your first trig course!). These in turn can be remembered by the clever tricks built into the Gibbs notation. To verify these elementary identities, however, we will use the Cartan notation because it makes more sense. Later .... not now! To appreciate the Gibbs notation I need a type symbol # to stand for the upside down triangle. (When you copy this lesson into your notebooks, be sure to use the correct notation for hand writing.) We will also let * stand for any one or all of the three products in the vector calculus: scalar product, dot product, cross product, and decide by "dimensional analysis" which it must be. For example For example, additivity of grad curl and div is summarized by saying # * (A + B) = # * A + # * B . The multiplicative rule is # * (A * B) = (# * A) * B + A * ( # * B) 1 2 3 4 5 6 Now, if A is a scalar and B a vector, so that *2 is scalar product, and, suppose *1 is cross (i.e. you are computing the curl(f _v ), then the LHS must end up being a vector. Therefore *3 must be scalar, producing a vector (the gradient), and *4 must be cross (a dot here would reduce two vectors to a scalar.) Since *5 has to be scalar product, *6 must again by cross, so the identinty is curl(f _v) = grad(f) x _v + f curl(_v). Etc.