STUDY GUIDE: VECTOR CALCULUS 1. Glossary/Notation × = vector cross product. If x1 = (x1; y1; z1); x2 = (x2; y2; z2), then the cross product is computed as the symbolic determinant 2 3 6 i j k 7 6 7 x1 × x2 = det 6 x1 y1 z1 7 : 46 57 x2 y2 z2 The cross product produces a vector perpendicular to both x1 and x2: ———————————————————————————————— · = scalar (“dot”) product. If x1 = (x1; y1; z1); x2 = (x2; y2; z2), then x1 · x2 = x1 x2 + y1y2 + z1z2: The dot product measures the angle θ between the two vectors: x · x 1 2 = cos θ: jx1j jx2j In particular, if the vectors are orthogonal, then their dot product is 0. ————————————————————————————————- coordinates and components/projections. If u is a unit vector, and x is an arbitrary vector, then the coordinate (aka “scalar component” or “scalar projection”) of x in the direction of u is compux = x · u: The projection of x in the direction of u, aka the vector projection of x onto u is projux = (x · u) u: The vector projection of x perpendicular to u (aka “orthogonal projec- tion”) is orthux = x − (x · u) u: ————————————————————————————————- 1 2 STUDY GUIDE: VECTOR CALCULUS magnitude and distance The magnitude (aka “length”) of a vector x = (x; y; z) is q jxj = x2 + y2 + z2; similarly in 2-D. The distance between two vectors x1 = (x1; y1; z1) and x2 = (x2; y2; z2) is the magnitude of the vector x1 − x2, i.e., q 2 2 2 dist(x1; x2) = jx1 − x2j = (x1 − x2) + (y1 − y2) + (z1 − z2) : ———————————————————————————————— @ @ @ r = @x ; @y ; @z = gradient. For a function f (x; y; z), we have ! @ f @ f @ f r f = ; ; = f ; f ; f ; @x @y @z x y z similarly in 2-D. ———————————————————————————————— r · = divergence. For a vector-valued function F = (P; Q; R), we have @P @Q @R r · F = div F = + + = P + Q + R ; @x @y @z x y z similarly in 2-D. ———————————————————————————————— r × = curl. For a vector valued function in 3-D, F = (P; Q; R)), we have 2 i j k 3 6 7 curl F = r × F = det 6 @ @ @ 7 : 6 @x @y @z 7 4 PQR 5 For a 2-D vector field F = (P; Q); the curl is simpler: @Q @P curl F = − : @x @y ———————————————————————————————— directional derivative. If u is a unit vector, then the directional derivative in the direction of u equals @ f = u · r f: @u Remark: the unit vector u may be constant, or it may be a variable vector field (i.e., which changes from point to point). For example, if n is the outer unit normal to a closed orientable surface S , then the “outer normal derivative” is @ f = n · r f: @n ———————————————————————————————— STUDY GUIDE: VECTOR CALCULUS 3 Chain Rule. Let f (x; y; z) 2 C1. Set r(u; v) = (x(u; v); y(u; v); z(u; v). Then @ f (r(u; v)) @r @ f (r(u; v)) @r = r f (r(u; v)) · ; = r f (r(u; v)) · : @u @u @v @v If r(t) = (x(t); y(t); z(t)is a parametrized curve, then d f (r(t)) dr = r f (r(t)) · : dt dt ———————————————————————————————— Derivative matrix and Jacobian • If f (x1; x2; :::; xn) = f1(x1; x2; :::; xn); f2(x1; x2; :::; xn); :::; fm(x1; x2; :::; xn) is a vector valued function from Rn ! Rm, then its derivative matrix is 2 @ f1 @ f1 @ f1 3 6 ::: 7 6 @x1 @x2 @xn 7 6 7 6 @ f @ f @ f 7 0 6 2 2 ::: 2 7 f (x) = 6 @x1 @x2 @xn 7 : 6 : : : 7 6 : : ::: : 7 6 7 4 @ fm @ fm ::: @ fm 5 @x1 @x2 @xn • If m = 1, then ! @ f @ f @ f f 0 = r f = ; ; :::; @x1 @x2 @xn is the usual gradient. In this case we say that f is differentiable at a given point x0 if f (x) − f (x ) − r f (x ) · (x − x ) lim 0 0 0 = 0: x!x0 jx − x0j • If n = m, then the derivative matrix f 0 is sometimes called the Jacobian ma- trix; the absolute value of its determinant is called the Jacobian. For example, if f (r; θ) = (r cos θ; r sin θ) is the co-ordinate transformation for polar co-ordinates, which we may also write as a column vector " # " # x r cos θ = ; y r sin θ then the Jacobian equals 2 @x @x 3 " # 6 7 cos θ −r sin θ det 6 @r ∂θ 7 = det = r: 46 @y @y 57 sin θ r cos θ @r ∂θ Thus when changing to polar co-ordinates in 2-D integrals, we have dx dy = r dr dθ: Similarly for 3-D spherical co-ordinates 2 x 3 2 ρ cos θ sin φ 3 6 7 6 7 6 y 7 = 6 ρ sin θ sin φ 7 ; 6 7 6 7 4 z 5 4 ρ cos φ 5 we may compute the Jacobian as ρ2 sin φ, so dx dy dz = ρ2 sin φ dρ dφ dθ: 4 STUDY GUIDE: VECTOR CALCULUS ds = arclength differential. For a parametric curve x(t) = (x(t); y(t); z(t)), we have q ds = (x0(t))2 + (y0(t))2 + (z0(t))2 dt q (or ds = (x0(t))2 + (y0(t))2 dt for a curve in 2-D). In the special case that the curve is a graph y = f (x), in 2-D, then the parametrization is x(x) = (x; f (x)) (or, if you prefer, x(t) = (t; f (t))) and q q ds = 1 + ( f 0(x))2 dx or ds = 1 + ( f 0(t))2 dt: ———————————————————————————————— dσ = surface area differential. For a parametrized surface r(u; v) = (x(u; v); y(u; v); z(u; v), we have @r @r dσ = × dudv: @u @v In the special case that the surface is a graph z = f (x; y), then we have the parametrization r(x; y) = (x; y; f (x; y)), and in this case q q 2 2 2 dσ = 1 + jr f (x; y)j dxdy = 1 + ( fx) + ( fy) dxdy: Remark: The textbook uses the notation dS in place of dσ. ———————————————————————————————— dS = n dσ = vector surface differerential Let r(u; v) = (x(u; v); y(u; v); z(u; v) be a parametrization of an orientable surface S . Then the vectors @r=@u and @r=@v are tangent to the surface, so @r @r that the cross product @u × @v is perpendicular to the surface. The vector surface differential is ! @r @r dS = n dσ = ± × dudv ; @u @v where n denotes a fixed choice of unit normal to the surface S . Here, ± is chosen depending on orientation. For example, if S is a closed surface @r @r (i.e., it encloses some 3-D region), and if @u × @v points outward, then @r × @r n = @u @v @r @r @u × @v is the outward unit normal, so if S has positive (i.e., outward) orientation, then @r @r ! @u × @v @r @r @r @r n dσ = × dudv = × dudv = dS: @r @r @u @v @u @v @u × @v ———————————————————————————————— @R = boundary of R. E.g., if R is the ball x2 + y2 + z2 ≤ 9, then @R is the sphere x2 + y2 + z2 = 9. ———————————————————————————————— STUDY GUIDE: VECTOR CALCULUS 5 2. Max-Min 2.1. Unconstrained Max-Min’s in 2D. Suppose that f (x; y) 2 C2 has a critical point at (x0; y0); i.e., r f (x0; y0) = 0: Consider the quantity 2 D = fxx fyy − fxy : The following “2nd derivative test” holds: • If D(x0; y0) > 0 and fxx(x0; y0) > 0, then f has a local min at (x0; y0): • If D(x0; y0) > 0 and fxx(x0; y0) < 0, then f has a local max at (x0; y0): • If D(x0; y0) < 0, then f has a saddle point at (x0; y0): • If D(x0; y0) = 0, the test is inconclusive. ———————————————————————————————— 2.2. Constrained Max-Min’s - Lagrange Multipliers. Consider f (x; y; z) subject to the constraint g(x; y; z) = k. The constrained extreme values of f can only occur where (x; y; z; λ) solves the system of equations (1) r f (x; y; z) = λrg(x; y; z); g(x; y; z) = k: In the case of multiple constraints g1(x; y; z) = k1; :::; gm(x; y; z) = km, one should solve the system (2) r f = λ1rg1 + ::: + λmrgm; g1(x; y; z) = k1; :::; gm(x; y; z) = km: NOTE: In principle, the Lagrange method only produces critical points, but often some ad hoc reasoning will allow one to determine whether the solution of the systems (1) or (2) does in fact yield a max or min. ———————————————————————————————— 3. Planes, Tangent Planes 3.1. Planes. The equation of the plane through the point x0, and perpendicular to a vector P is given by (3) P · (x − x0) = 0: Notice that this can also be written as P · x − c0 = 0; or P · x = c0; where c0 = P · x0: For example, 3x − 2y + 4z = 10 is the equation of a plane with perpendicular vector P = (3; −2; 4), because it can be re- written as (3; −2; 4) · (x; y; z) = 10: Sometimes it is convenient to normalize so that the perpendicular to the plane is a unit normal; e.g., for the previous plane, the magnitude of p p p (3; −2; 4) = 32 + (−2)2 + 42 = 9 + 4 + 16 = 30: 6 STUDY GUIDE: VECTOR CALCULUS Therefore, the unit normal n = p3 ; p−2 ; p4 ; so the plane can be re-written in normal- 30 30 30 ized form as ! 3 −2 4 10 (4) p ; p ; p · (x; y; z) = p : 30 30 30 30 ———————————————————————————————— 3.2.