Dr. Allen Back. Nov. 19, 2014

Transcription

1 Why of Dr. Allen Back Nov. 19, 2014 Graph

2 Picture of T u, T v for a Lat/Long Param. of the Sphere. Why of Graph

3 Basic Picture Why of Graph

4 Why Φ(u, v) = (x(u, v), y(u, v), z(u, v)) Tangents T u = (x u, y u, z u ) T v = (x v, y v, z v ) Area Element ds = T u T v du dv Normal N = T u T v Unit normal ˆn = ± T u T v T u T v (Choosing the ± sign corresponds to an orientation of the surface.) of Graph

5 Why Two Kinds of Integral of a scalar function f (x, y, z) : f (x, y, z) ds Integral of a vector field F (x, y, z) : F (x, y, z) ˆn ds. S S of Graph

6 Why of Integral of a scalar function f (x, y, z) calculated by f (x, y, z) ds = f (Φ(u, v)) T u T v du dv S D where D is the domain of the parametrization Φ. Integral of a vector field F (x, y, z) calculated by F (x, y, z) ˆn ds S ( Tu = ± F (Φ(u, v)) ) T v D T u T T u T v du dv v where D is the domain of the parametrization Φ. Graph

7 3d Flux Picture Why of Graph

8 Why of The preceding picture can be used to argue that if F (x, y, z) is the velocity vector field, e.g. of a fluid of density ρ(x, y, z), then the surface integral S ρ F ˆn ds (with associated Riemann Sum ρ(x i, yj, zk ) F (xi, yj, zk ) ˆn(x i, yj, zk ) S ijk) represents the rate at which material (e.g. grams per second) crosses the surface. Graph

9 Why From this point of view the orientation of a surface simple tells us which side is accumulatiing mass, in the case where the value of the integral is positive. of Graph

10 2d Flux Picture Why of Graph There s an analagous 2d Riemann sum and interp of F ˆn ds. C

11 Why of Graph

12 Why Problem: Calculate S F (x, y, z) ˆn ds for the vector field F (x, y, z) = (x, y, z) and S the part of the paraboloid z = 1 x 2 y 2 above the xy-plane. Choose the positive orientation of the paraboloid to be the one with normal pointing downward. of Graph

13 Why Problem: Calculate the surface area of the above paraboloid. of Graph

14 Why Why of The Vector 1 Electromagnetism. (Two of Maxwell s equations involve surface integrals.) 2 Partial differential equations. (Integration by parts is an important technique here, and is based on the chapter 8 integral theorems.) 3 Heat, fluid flow and difffusion. 4 (Related to Markov chains in probability.) 5 At an advanced level, financial mathematics. (e.g. Black-Scholes is similar to the heat equation.) Graph

15 Why Why of Graph

16 Why Why The Scalar 1 As with multiple integrals, going from a density of something (this time per unit surface area) to a total. 2 Vector surface integrals are surface integrals of the scalar function normal component. of Graph

17 Why Why 1d-Heat Equation T t = c 2 T x 2 (T (x, t) is the temperature at position x and time t along a rod.) Black-Scholes V t + cs 2 2 V V + rs rv = 0. S 2 S (V (S, t) is the price of a derivative as a function of an underlying stock price S at time t.) (See e.g. Black-Scholes model in Wikipedia.) of Graph

18 Why Problem: Find S F ˆn ds for F (x, y, z) = (0, yz, z 2 ) and S the portion of the cylinder y 2 + z 2 = 1 with 0 x 1, z 0, and the positive orientation chosen to be a radial outward (from the axis of the cylinder) normal. of Graph

19 Why Problem: Letting r denote the vector field (x, y, z), find the value of the surface integral r ˆn ds r 3 S where S is the sphere of radius R about the origin, oriented with an outward normal. of Graph

20 Why Spherical coordinates (letting ρ = R be constant) gives a natural parametrization of the sphere of radius R centered at the origin. It is geometrically to be expected (and you ve done some verifications like this at least in look at problems) that the Jacobian result for spherical coordinates translates to the area element on such a sphere to be ds = R 2 sin φ dφ dθ. of Graph

21 Why Green s theorem says that for simple closed (piecewise smooth) curve C whose inside is a region R, we have Q P(x, y) dx + Q(x, y) dy = x P dx dy y C as long as the vector field F (x, y) = (P(x, y), Q(x, y)) is C 1 on the set R and C is given its usual inside to the left orientation. R of Graph

22 Why For a y-simple region P y dy dx = R is fairly easily justified. C= R P dx. of Graph

23 Why For an x-simple region Q x dx dy = R C= R Q dy. of Graph

24 Why So for a region that is both y-simple and x-simple we have Green: P(x, y) dx + Q(x, y) dy = Q y P x dx dy. R R of Graph

25 Why Intuitively, why the different signs? +Q x yet P y. And why this combination of Q x P y? of Graph

26 Why Intuitively, why the different signs? +Q x yet P y. And why this combination of Q x P y? Think about the line integral around a small rectangle with sides x and y. If you assume (or justify) that evaluating the vector field in the middle of each edge gives a good approximation in the line integral, then Q x P y emerges quite naturally. of Graph

27 Why If you can cut a region into two pieces where we know Green s holds on each piece (e.g. a ring shaped region), then Green also holds for the entire region. (Because the line integrals over the cuts show up twice with opposite signs (think inside to the left ) and cancel. of Graph

28 Why So in the end, Green s theorem holds for regions whose boundaries include several closed curves (multiply-connected regions) as long as we orient each boundar curve according to the inside to the left rule. of Graph

29 Why 1 2 C x dy y dx oriented counterclockwise. for C the boundary of the ellipse x y = 1 of Graph

30 Why Let F = 1 x 2 ( y, x). + y 2 If C 1 and C 2 are two simple closed curves enclosing the origin (and oriented with the usual inside to the left), can you say whether one of C 1 F d s and C 2 F d s is bigger than the other? of Graph

31 Why Definition: A path c : [a, b] R 3 is said to be a flow line (or integral curve) of a vector field F (x, y, z) if for all t [a, b]. c (t) = F (c(t)) of Graph

32 Why Flow lines always exist at least locally because of existence and uniqueness theorems for ordinary differential equations. of Graph

33 Why At the end of Math 2210, you studied how to use eigenvalues/eigenvectors/diagonalization/(jordan Normal form) to solve linear differential equations. of Graph

34 Why Show that c(t) = (x 0 e t, y 0 e t ) is a flow line of F (x, y) = (x, y). of Graph

35 Why Show that c(t) = [ ] x(t) = y(t) is a flow line of F (x, y) = ( y, x). [ cos t ] [ ] sin t x0 sin t cos t y 0 of Graph

36 Why We will discuss the meaning of these when we get to the 3d integral theorems (8.2 and 8.4.) For a vector field F (x, y, z) = (P, Q, R) div( F ) is a (scalar) function; i.e. a number at each point. curl( F ) is another vector field; i.e. a vector at each point. of Graph

37 Why for F (x, y, z) = (P, Q, R). div( F ) = P x + Q y + R z ( F ). of Graph

38 Why for F (x, y, z) = (P, Q, R). î ĵ ˆk curl( F ) = x y z P Q R ( F ). of Graph

39 Why Please take a look at the table of vector analysis identities on page 255. Many are just the 1-variable product or chain rules. 1 (fg) 2 div(f F ) 3 curl(f F ) 4 curl( G F ) would be more complicated. of Graph

40 of Why This is not worth memorizing! If one rotates about the z-axis the path (curve) z = f (x) in the xz-plane for 0 a x b, one obtains a surface of revolution with a parametrization and ds =? Φ(u, v) = (u cos v, u sin v, f (u)) of Graph

41 of Why ds = u 1 + (f (u)) 2 du dv. of Graph

42 Graph Why This is not worth memorizing! For the graph parametrization of z = f (x, y), and ds =? Φ(u, v) = (u, v, f (u, v)) of Graph

43 Graph Why ds = 1 + fu 2 + fv 2 du dv. of Graph

44 Graph Why of Graph For such a graph, the normal to the surface at a point (x, y, f (x, y)) (this is the level set z f (x, y) = 0) is so we can see that ( f x, f y, 1) 1 cos γ = 1 + fx 2 + fy 2 determines the angle γ of the normal with the z-axis. And so at the point (u, v, f (u, v)) on a graph, ds = 1 cos γ du dv. (Note that du dv is essentially the same as dx dy here.)

45 Paraboloid z = x 2 + 4y 2 Why The graph z = F (x, y) can always be parameterized by Φ(u, v) =< u, v, F (u, v) >. of Graph

46 Paraboloid z = x 2 + 4y 2 Why The graph z = F (x, y) can always be parameterized by Φ(u, v) =< u, v, F (u, v) >. Parameters u and v just different names for x and y resp. of Graph

47 Paraboloid z = x 2 + 4y 2 Why The graph z = F (x, y) can always be parameterized by Φ(u, v) =< u, v, F (u, v) >. Use this idea if you can t think of something better. of Graph

48 Paraboloid z = x 2 + 4y 2 The graph z = F (x, y) can always be parameterized by Φ(u, v) =< u, v, F (u, v) >. Why of Graph

49 Paraboloid z = x 2 + 4y 2 The graph z = F (x, y) can always be parameterized by Φ(u, v) =< u, v, F (u, v) >. Why of Graph Note the curves where u and v are constant are visible in the wireframe.

50 Paraboloid z = x 2 + 4y 2 Why A trigonometric parametrization will often be better if you have to calculate a surface integral. of Graph

51 Paraboloid z = x 2 + 4y 2 Why A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u 2 >. of Graph

52 Paraboloid z = x 2 + 4y 2 Why A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u 2 >. of Graph

53 Paraboloid z = x 2 + 4y 2 Why A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u 2 >. of Graph Algebraically, we are rescaling the algebra behind polar coordinates where x = r cos θ y = r sin θ leads to r 2 = x 2 + y 2.

54 Paraboloid z = x 2 + 4y 2 Why A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u 2 >. of Graph Here we want x 2 + 4y 2 to be simple. So x = 2r cos θ y = r sin θ will do better.

55 Paraboloid z = x 2 + 4y 2 Why A trigonometric parametrization will often be better if you have to calculate a surface integral. Φ(u, v) =< 2u cos v, u sin v, 4u 2 >. of Graph Here we want x 2 + 4y 2 to be simple. So x = y = 2r cos θ r sin θ will do better. Plug x and y into z = x 2 + 4y 2 to get the z-component.

56 Parabolic Cylinder z = x 2 Why Graph parametrizations are often optimal for parabolic cylinders. of Graph

57 Parabolic Cylinder z = x 2 Why Φ(u, v) =< u, v, u 2 > of Graph

58 Parabolic Cylinder z = x 2 Φ(u, v) =< u, v, u 2 > Why of Graph

59 Parabolic Cylinder z = x 2 Φ(u, v) =< u, v, u 2 > Why of Graph One of the parameters (v) is giving us the extrusion direction. The parameter u is just being used to describe the curve z = x 2 in the zx plane.

60 Elliptic Cylinder x 2 + 2z 2 = 6 Why The trigonometric trick is often good for elliptic cylinders of Graph

61 Elliptic Cylinder x 2 + 2z 2 = 6 Why Φ(u, v) =< 3 2 cos v, u, 3 sin v >=< 6 cos v, u, 3 sin v > of Graph

62 Elliptic Cylinder x 2 + 2z 2 = 6 Why Φ(u, v) =< 6 cos v, u, 3 sin v > of Graph

63 Elliptic Cylinder x 2 + 2z 2 = 6 Φ(u, v) =< 6 cos v, u, 3 sin v > Why of Graph

64 Elliptic Cylinder x 2 + 2z 2 = 6 Why Φ(u, v) =< 6 cos v, u, 3 sin v > of Graph

65 Elliptic Cylinder x 2 + 2z 2 = 6 Φ(u, v) =< 6 cos v, u, 3 sin v > Why of Graph What happened here is we started with the polar coordinate idea x = r cos θ z = r sin θ but noted that the algebra wasn t right for x 2 + 2z 2 so shifted to x = 2r cos θ z = r sin θ

66 Elliptic Cylinder x 2 + 2z 2 = 6 Φ(u, v) =< 6 cos v, u, 3 sin v > Why of Graph x = z = 2r cos θ r sin θ makes the left hand side work out to 2r 2 which will be 6 when r = 3.

67 Ellipsoid x 2 + 2y 2 + 3z 2 = 4 Why A similar trick occurs for using spherical coordinate ideas in parameterizing ellipsoids. of Graph

68 Ellipsoid x 2 + 2y 2 + 3z 2 = 4 Why A similar trick occurs for using spherical coordinate ideas in parameterizing ellipsoids. Φ(u, v) =< 2 sin u cos v, 4 2 sin u sin v, 3 cos u > of Graph

69 Ellipsoid x 2 + 2y 2 + 3z 2 = 4 Φ(u, v) =< 2 sin u cos v, 2 sin u sin v, 4 3 cos u > Why of Graph

70 Hyperbolic Cylinder x 2 z 2 = 4 Why You may have run into the hyperbolic functions cosh x = ex + e x 2 sinh x = ex e x 2 of Graph

71 Hyperbolic Cylinder x 2 z 2 = 4 Why of You may have run into the hyperbolic functions cosh x = sinh x = ex + e x 2 ex e x Just as cos 2 θ + sin 2 θ = 1 helps with ellipses, the hyperbolic version cosh 2 θ sinh 2 θ = 1 leads to the nicest hyperbola parameterizations. 2 Graph

72 Hyperbolic Cylinder x 2 z 2 = 4 Why Just as cos 2 θ + sin 2 θ = 1 helps with ellipses, the hyperbolic version cosh 2 θ sinh 2 θ = 1 leads to the nicest hyperbola parameterizations. Φ(u, v) =< 2 sinh v, u, 2 cosh v > of Graph

73 Hyperbolic Cylinder x 2 z 2 = 4 Φ(u, v) =< 2 sinh v, u, 2 cosh v > Why of Graph

74 Saddle z = x 2 y 2 Why The hyperbolic trick also works with saddles of Graph

75 Saddle z = x 2 y 2 Why Φ(u, v) =< u cosh v, u sinh v, u 2 > of Graph

76 Saddle z = x 2 y 2 Φ(u, v) =< u cosh v, u sinh v, u 2 > Why of Graph

77 Hyperboloid of 1 Sheet x 2 + y 2 z 2 = 1 Why The spherical coordinate idea for ellipsoids with sin φ replaced by cosh u works well here. of Graph

78 Hyperboloid of 1 Sheet x 2 + y 2 z 2 = 1 Why Φ(u, v) =< cosh u cos v, cosh u sin v, sinh u > of Graph

79 Hyperboloid of 1 Sheet x 2 + y 2 z 2 = 1 Why of Graph

80 Hyperboloid of 2-Sheets x 2 + y 2 z 2 = 1 Why Φ(u, v) =< sinh u cos v, sinh u sin v, cosh u > of Graph

81 Hyperboloid of 2-Sheets x 2 + y 2 z 2 = 1 Why of Graph

82 Top Part of Cone z 2 = x 2 + y 2 Why So z = x 2 + y 2. of Graph

83 Top Part of Cone z 2 = x 2 + y 2 Why So z = x 2 + y 2. The polar coordinate idea leads to Φ(u, v) =< u cos v, u sin v, u > of Graph

84 Top Part of Cone z 2 = x 2 + y 2 So z = x 2 + y 2. The polar coordinate idea leads to Φ(u, v) =< u cos v, u sin v, u > Why of Graph

85 Mercator of the Sphere Why For 0 v, 0 u 2π Φ(u, v) = (sech(v) cos u, sech(v) sin u, tanh(v)). (Note tanh 2 (v) + sech 2 (v) = 1) of Graph

86 ctionn Why of Graph