STUDY GUIDE: VECTOR CALCULUS 1. Glossary/Notation × = Vector

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STUDY GUIDE: VECTOR CALCULUS

1. Glossary/Notation

If x1 = (x1, y1, z1), x2 = (x2, y2, z2), then the cross product is computed as the symbolic determinant    i j k    x1 × x2 = det  x1 y1 z1  .   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. |x1| |x2| 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 projection”) 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 |x| = 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) = |x1 − x2| = (x1 − x2) + (y1 − y2) + (z1 − z2) . ————————————————————————————————

∂ ∂ ∂ ∇ = ∂x , ∂y , ∂z = gradient. For a function f (x, y, z), we have ! ∂ f ∂ f ∂ f ∇ f = , , = f , f , f ; ∂x ∂y ∂z x y z similarly in 2-D. ————————————————————————————————

∇ · = divergence. For a vector-valued function F = (P, Q, R), we have ∂P ∂Q ∂R ∇ · F = div F = + + = P + Q + R ; ∂x ∂y ∂z x y z similarly in 2-D. ————————————————————————————————

∇ × = curl. For a vector valued function in 3-D, F = (P, Q, R)), we have  i j k    curl F = ∇ × F = det  ∂ ∂ ∂  .  ∂x ∂y ∂z   PQR  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 · ∇ 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 · ∇ f. ∂n ———————————————————————————————— STUDY GUIDE: VECTOR CALCULUS 3

Chain Rule. Let f (x, y, z) ∈ 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 = ∇ f (r(u, v)) · , = ∇ 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 = ∇ 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

 ∂ f1 ∂ f1 ∂ f1   ...   ∂x1 ∂x2 ∂xn     ∂ f ∂ f ∂ f  0  2 2 ... 2  f (x) =  ∂x1 ∂x2 ∂xn  .  . . .   ......     ∂ fm ∂ fm ... ∂ fm  ∂x1 ∂x2 ∂xn • If m = 1, then ! ∂ f ∂ f ∂ f f 0 = ∇ 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 ) − ∇ f (x ) · (x − x ) lim 0 0 0 = 0. x→x0 |x − x0| • If n = m, then the derivative matrix f 0 is sometimes called the Jacobian matrix; 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  ∂x ∂x  " #   cos θ −r sin θ det  ∂r ∂θ  = det = r.  ∂y ∂y  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  x   ρ cos θ sin φ       y  =  ρ sin θ sin φ  ,      z   ρ cos φ  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 + |∇ f (x, y)| 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) ∈ C2 has a critical point at (x0, y0), i.e., ∇ 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.

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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) ∇ f (x, y, z) = λ∇g(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) ∇ f = λ1∇g1 + ... + λm∇gm, 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 √ √ (3, −2, 4) = 32 + (−2)2 + 42 = 9 + 4 + 16 = 30. 6 STUDY GUIDE: VECTOR CALCULUS Therefore, the unit normal n = √3 , √−2 , √4 , so the plane can be re-written in normal- 30 30 30 ized form as ! 3 −2 4 10 (4) √ , √ , √ · (x, y, z) = √ . 30 30 30 30 ———————————————————————————————— 3.2. Tangent Planes. If a surface S is given parametrically as r(u, v) = (x(u, v), y(u, v), z(u, v), then the vectors ∂r/∂u and ∂r/∂v are tangent to the surface, so that the cross product ∂r ∂r P = ∂u × ∂v is perpendicular to the surface. So, to ﬁnd the equation of the tangent plane to a surface at the parametric value (u0, v0), just plug into the formula (3) above, with ! ! ∂r ∂r P = (u , v ) × (u , v ) ∂u 0 0 ∂v 0 0 and x0 = r((u0, v0). For surfaces given as graphs z = f (x, y), we can view the parameters as (x, y) = (u, v), so in this case r(x, y) = (x, y, f (x, y)), and    i j k  ∂r ∂r   P = × = det  1 0 fx  = (− fx, − fy, 1). ∂x ∂y   0 1 fy

Thus, the tangent plane at the point x0, y0, f (x0, y0) is given as − fx(x0, y0), − fy(x0, y0), 1 · (x, y, z) − (x0, y0, f (x0, y0)) = 0.

Notice that the second factor can be re-written as (x − x0, y − y0, z − f (x0, y0)). Thus, com- puting the dot product and simplifying, we get

z = f (x0, y0) + ∇ f (x0, y0) · (x − x0, y − y0).

For a level surface F(x, y, z) = k, we have ∇F = P is perpendicular to the surface. Consequently, to compute the equation of the tangent plane to the level surface F(x, y, z) = k, at the point (x0, y0, z0), where F(x0, y0, z0) = k, just plug into the formula (3) above, with P = ∇F(x0, y0, z0), x0 = (x0, y0, z0), and x = (x, y, z).

4. Path Integrals, “Fundamental Theorem” and Green’s Theorem In this section all curves will be piecewise smooth.

4.1. Arclength. • If γ is a curve, then its length is Z `(γ) = ds, γ where ds is the arclength diﬀerential deﬁned above; e.g., if the curve has the parametrization x(t) = (x(t), y(t)), then the arclength of the curve between the parametric values t = a and t = b is Z b q (x0(t))2 + (y0(t))2 dt. a STUDY GUIDE: VECTOR CALCULUS 7

Remark: 1) the analogous formula holds for 3-D curves (x(t), y(t), z(t)); 2) in 2-D, if the γ is given as a graph y = f (x), then use the graph parametrization for ds. ———————————————————————————————— R · 4.2. “Work Integrals”: γ F dr. • If the curve γ has the parametrization r(t) = (x(t), y(t)), a ≤ t ≤ b, then for the vector field F = (P(x, y), Q(x, y)), we have Z Z F · dr = (P, Q) · (dx, dy) γ γ Z Z b = Pdx + Qdy = P x(t), y(t) x0(t) + Q x(t), y(t) y0(t) dt. γ a ———————————————————————————————— 4.3. Green’s Theorem. Let γ be a simple, closed curve which forms the boundary of a region D ⊂ R2, and suppose that γ is parametrized with a “positive orientation”, i.e., the region D lies to the left of a bug walking along the boundary curve in the direction induced by the parametrization. Moreover, we suppose that the curve is parametrized so that γ is traversed exactly once (so that the curve is simple with this parametrization). Let F = (P, Q) be a C1 vector field. Then ∂Q ∂P Z − dxdy = F · dr. D ∂x ∂y γ R " · Remark: Notice that if F is conservative, then γ F dr = 0. Indeed, “conservative” means ∂Q ∂P that there is a potential function f such that F = ∇ f , which is equivalent to ∂x − ∂y = 0. ———————————————————————————————— 4.4. Fundamental Theorem. If γ is a curve with initial point a and terminal point b then Z ∇ f · dr = f (b) − f (a). γ Note that for such integrals, the path doesn’t matter, only the initial and final points. Also, notice that if γ is a closed curve, then the initial and final points are the same, so the integrals is zero. This is another way to see that for conservative fields, the closed path R · ∇ integral γ F dr = 0, since a conservative field F equals f , for some potential function f . ————————————————————————————————

5. Surface Integrals and Gauss’s Theorem 5.1. Surface Area. • If S is a parametrized surface, then its surface area is

σ(S ) = dσ, S where dσ is the surface area diﬀerential" as above (NOTE: the book uses the notation dS in place of dσ). E.g., if S is parametrized as r(u, v) = x(u, v), y(u, v), z(u, v) , then the surface area of the portion of the surface with (u, v) ∈ D is given by ∂r ∂r × dudv. D ∂u ∂v " 8 STUDY GUIDE: VECTOR CALCULUS

Here is a specific example: we will set up an integral which represents the surface area of the upper hemisphere of the unit sphere x2 + y2 + z2 = 1. The unit sphere has the standard parametrization  x   cos θ sin φ       y  =  sin θ sin φ  , 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π      z   cos φ  (this is just standard spherical co-ordinates with ρ = 1); i.e., r(θ, φ) = (cos θ sin φ, sin θ sin φ, cos φ). A computation yields ∂r ∂r × = sin φ. ∂θ ∂φ For the upper hemisphere, the parametric region D is given by 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2. Thus, Z 2πZ π/2 σ(upper hemisphere) = sin φ dφ dθ. 0 0 ———————————————————————————————— · · 5.2. “Flux Integrals”: S F dS = S F n dσ. • Recall the definition of the vector surface differential dS. Then, if the surface S is parametrized by!r(u, v) = (x!(u, v), y(u, v), z(u, v)), (u, v) ∈ D, and if F = (P, Q, R), we have ! ∂r ∂r F · dS = F · n dσ = ± (P, Q, R) · × dudv , S S D ∂u ∂v where"± depends on" orientation. For" example, if S is a closed surface with outward ∂r ∂r orientation, and if the normal vector ∂u × ∂v points outward, then we choose +.

Example of a flux integral computation. We set up an integral to compute the downward flux of F(x, y, z) = (z, y, x) across the upper hemisphere x2 + y2 + z2 = 1, z ≥ 0. The flux is

F · n dσ, S where the unit normal should be oriented" downward. We use the standard spherical coordinate parametrization for the unit upper hemisphere: r(φ, θ) = (sin φ cos θ, sin φ sin θ, cos φ), 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2. Then  i j k    r × r = det  cos φ cos θ cos φ sin θ − sin φ  φ θ    − sin φ sin θ sin φ cos θ 0 

= (sin2 φ cos θ, sin2 φ sin θ, sin φ cos φ). Consequently dS = ± (sin2 φ cos θ, sin2 φ sin θ, sin φ cos φ) dφdθ, and we need to determine the correct choice of “+” or “-”. To do this, let’s ﬁx one point that is easy to compute, say φ = π/4, θ = 0. Then

(sin2 φ cos θ, sin2 φ sin θ, sin φ cos φ) = (1/2, 0, 1/2), φ=π/4,θ=0 STUDY GUIDE: VECTOR CALCULUS 9 which points upward, since the z-component of the vector is positive. Thus we choose “-” to get the downward orientation. Therefore,

F · n dσ S Z 2πZ π/2 " = − F · (sin2 φ cos θ, sin2 φ sin θ, sin φ cos φ) dφdθ 0 0 Z 2πZ π/2 = − (cos φ, sin φ sin θ sin φ cos θ) · (sin2 φ cos θ, sin2 φ sin θ, sin φ cos φ) dφdθ 0 0 Z 2πZ π/2 2π = − 2 sin2 cos φ cos θ + sin3 φ sin2 θ dφdθ = − . 0 0 3 ———————————————————————————————— 5.3. Gauss’s Theorem, aka “Divergence Theorem”. Let R denote a 3-D region whose boundary ∂R = S is a piecewise smooth orientable surface, with positive orientation (i.e., where we choose the unit normal n that points outward). Then, if F is a C1 vector field, we have div F dV = div F dxdydz = F · dS = F · n dσ. R R ∂R S Remark: notice$ that if S is a closed$ surface that completely" encloses" a 3-D region R, then it is often easier to use Gauss’s Theorem when evaluating flux integrals, especially if div F is R · pretty simple. As a specific example, we compute the flux integral S (x, 0, z) dS, where S is the sphere x2 +y2 +z2 = 4. Note that this surface is the boundary of the ball x2 +y2 +z2 ≤ 4, and that div (x, 0, z) = 2. Thus, ! Z Z 4 64 (x, 0, z) · dS = 2 dV = 2 π 23 = π S x2+y2+z2≤4 3 3 ————————————————————————————————

5.4. Stokes’s Theorem. Let γ be a simple, closed curve which forms the boundary of S0, 3 which is a region contained in a surface S ⊂ R , and suppose that γ and S0 are parametrized with a “positive orientation”, i.e., we imagine a bug walking along the boundary curve on the same side as the unit normal points, so that the surface region S0 lies to his left as he traverses the curve in the direction induced by the parametrization; e.g., if the unit normals point “upward”, then the bug should walk on the “top” side, with S0 to his left. Moreover, we suppose that the curve is parametrized so that γ is traversed exactly once. Let F = (P, Q, R) be a C1 vector ﬁeld. Then Z curl F · dS = F · dr. S0 γ

Remarks: one can also denote γ"= ∂S0. 10 STUDY GUIDE: VECTOR CALCULUS

6. Course Outline and Practice Problems Ch. 12 • vectors, components/projections, magnitude/length • Dot product, angle between lines, curves, planes • Cross product • Eq’ns of lines and planes • Tangent lines Ch. 13 • Arclength • Velocity and acceleration Ch. 14 • Tangent planes and Differentiability • Chain rule • Gradient and Directional derivatives • Max-Min • Lagrange multipliers • Level curves and surfaces Ch. 15 • Double and triple integrals • Polar, cylindrical and spherical co-ordinates • Jacobians and change of variables Ch. 16 • Vector fields • Line (aka “Path”) integrals; “Work” integrals • Curl and divergence • Conservative fields, Fundamental Theorem • Green’s Thm • Parametric surfaces: Tangent Planes, Surface area, surface and “Flux” integrals • Stokes Thm • Gauss’s Thm (“aka” Divergence Thm) Some Practice Problems for Final Exam • 12.5: 26, 50 • 13.4: 43, 45 • 14.4: 4 • 14.6: 45 • 14.7: 11 • 14.8: 10 • 15.10: 26 • 16.2: 22 • 16.3: 19 • 16.4: 11 • 16.6: 35, 50 • 16.7: 24, 25 (compute directly, not using Gauss’s Theorem) • 16.8: 10, 16 • 16.9: 12, 14, 24, 29 • 16 review: 29, 33, 34, 37, 39