Dr. Allen Back. Nov. 5, 2014
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1 Dr. Allen Back Nov. 5, 2014
2 12 lectures, 4 recitations left including today.
3 a Most of what remains is vector integration and the integral theorems. b We ll start 7.1, 7.2,4.2 on Friday. c If you are not in physics, please bear with us on this material. There are good reasons why it is always in multivariable calculus courses, though it will not be that evident at an elementary level.
4 Right inverses and existence of solutions:s Set theoretically, for f : A B, a function g : B A satisfying f g = Id B (i.e. f (g(q)) = q for all q B) is EQUIVALENT to the equation f (p) = q having a solution p A for any q B.
5 Left inverses and uniqueness of solutions: Set theoretically, for f : A B, a function g : B A satisfying g f = Id A (i.e. g(f (p)) = p for all p A) is EQUIVALENT to the equation f (p) = q whenever (i.e. for which q) it has a solution, having that solution be unique among elements of A.
6 Of course in linear algebra, we learn that for linear transformations f : R n R n given by f (x) = Ax, either of the above are equivalent to det(a) 0.
7 Recall also for f : R n R n and f (x) = Ax: a Linear transformations take parallelograms to parallelograms. (Similarly in higher dim.) b If e 1, e 2,..., e n are the standard basis vectors, then the i th column of the n n matrix A is given by Ae i. c So, e.g. it is easy to map the unit square with sides e 1, e 2 in R 2 (and lower left corner at the origin) to any parallelogram with sides v 1, v 2 (and lower left corner at the origin. (A = [ v 1 v 2 ].) d So A 1 can take us from an arbitrary parallelogram with sides v 1, v 2 to the unit square.
8 Recall also for f : R n R n and f (x) = Ax: e By composition of two of the above, we can map any parallelogram with sides w 1, w 2 in R 2 (and lower left corner at the origin) to any parallelogram with sides v 1, v 2 (and lower left corner at the origin. f Throwing in a translation lets us deal with parallelograms for which the origin is not a vertex.
9 A r r θ by geometry
10 The Polar Coordinate Transformation
11 A x u y u x v y v u v
12 For a C 1 1:1 onto map F (u, v) = (x(u, v), y(u, v) taking D R 2 in the uv-plane to D R 2 in the xy plane, da = dx dy = x u x v dudv lets us move between xy and uv integrals. (Similarly for more than 2 variables.) y u y v
13 Polar: da = r dr dθ. Cylindrical: dv = r dr dθ dz. Spherical: dv = ρ 2 sin φ dρ dθ dφ.
14 The Basic Spherical Coordinate Picture
15 Polar/Spherical/Cylindrical Problem: D (x 2 + y 2 ) 3 2 dx dy for D the disk x 2 + y 2 1.
16 Polar/Spherical/Cylindrical Problem: e x2 dx.
17 Polar/Spherical/Cylindrical Problem: Volume of a right circular cone of height H and largest radius R.
18 Polar/Spherical/Cylindrical Problem: Volume of the portion of the Earth above latitude 45. Ambiguous as written; could mean all points with colatitude φ in spherical coordinates less than π 4. Or could refer to all points whose z value is above the z value at this latitude. Either integral could be set up....
19 Polar/Spherical/Cylindrical Problem: Volume, using spherical coordinates, of a ball of radius R with a hole of radius a (centered on a diameter) drilled out of it.
20 Right inverses and existence of solutions
21 Left inverses and uniqueness of solutions
22 Consider the map F : D R 2 D R 2 defined by F (x, y) = (x 2 y 2, x + y) Label the components of image points F (x, y) as (u, v); i.e. we think of the above transformation as u = x 2 y 2 v = x + y
23 To study one-to-oneness and ontoness of F, consider the algebra: u = x 2 y 2 v x + y v = x + y = x y
24 Adding and subtracting the above two equations: x = 1 ( v + u ) 2 v y = 1 ( v u ) 2 v
25 You may take as given the fact that these formulae check showing that G(u, v) = ( 1 2 ( v + u ), 1 ( v u ) ) v 2 v gives the inverse of F where everything is defined; i.e. F (G(u, v)) = (u, v) G(F (x, y) = (x, y)
26 What is the natural domain of the function G? In other words, describe the largest subset (call it U) of the uv plane on which G is defined.
27 Find the largest set V in the xy plane so that for all (x, y) V, F (x, y) belongs to the domain U of the function G which you found above.
28 Find a point (x, y) so that F (x, y) = (4, 2). More generally, briefly explain why the equation F (G(u, v)) = (u, v) shows that F : V U is onto.
29 Note F (1, 1) = (0, 2). Can there be a different point (x, y) besides (1, 1) with F (x, y) = (0, 2)? More generally, briefly explain why the equation G(F (x, y)) = (x, y) shows that F : V U is one-to-one.
30 Two Definitions Continuity at each Point of a set S: x S and ɛ > 0 δ > 0 so that x y < δ f (x) f (y) < ɛ. Uniform Continuity on a Set S: ɛ > 0 δ > 0 so that x, y S, x y < δ f (x) f (y) < ɛ.
31 An Important Theorem Bolzano Weierstrass: Every bounded sequence in R n has a convergent subsequence. This theorem is also at the heart of the proof that continuous functions with closed and bounded domains are automatically bounded and attain their extrema.
32 The idea of why Bolzano Weirrstrass holds:
33 Based on Bolzano Weierstrass, one can show that every continuous function on a closed and bounded set is uniformly continuous there.
34 Suppose one is considering the integrability of a continuous function f over a rectangle of area A.
35
36 Suppose one is considering the integrability of a continuous function f over a rectangle of area A. Consider any regular partition each of whose constituent rectangles is smaller in diameter than the δ given in the definition of uniform continuity.
37 Suppose one is considering the integrability of a continuous function f over a rectangle of area A. Consider any regular partition each of whose constituent rectangles is smaller in diameter than the δ given in the definition of uniform continuity. Then any two Riemann sums for this partition will differ by at most ɛa. This allows one to show that all Riemann sums settle down to a single limit as the partitions becomes sufficiently fine.
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