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MA
Multivariable Calculus
Basic Vectors
Thm 1: ‖cu‖ = |c| ‖u‖
Thm 2: (unit vector in direction of a) = (^) ‖aa‖
Thm 3 [Dot product properties]: a · b = b · a a · (b + c) = a · b + a · c (a + b) · c = a · c + b · c
(da) · b = d (a · b) = a · (db) 0 · a = 0 a · a = ‖a‖^2
Thm 4 [Dot product & angle]: a · b = ‖a‖ ‖b‖ cos θ
Thm 5 [Orthogonality]: a ⊥ b ⇐⇒ a · b = 0
Component (signed scalar): compa b = ‖b‖ cos θ = (^) ‖aa·b‖
Projection (vector): proja b = compa b × (^) ‖aa‖ = aa··ba a
Cross product: 〈a 1 , a 2 , a 3 〉 × 〈b 1 , b 2 , b 3 〉 :=
i j k a 1 a 2 a 3 b 1 b 2 b 3
〈a 2 b 3 − a 3 b 2 , a 3 b 1 − a 1 b 3 , a 1 b 2 − a 2 b 1 〉
Thm 6: (a × b) ⊥ a and (a × b) ⊥ b
Thm 7 [Cross prod. & angle]: ‖a × b‖ = ‖a‖ ‖b‖ sin θ
Thm 8 [Cross product properties]: a × b = − (b × a) a × (b + c) = a × b + a × c
(a + b) × c = a × c + b × c (da)×b = d (a × b) = a×(db)
Scalar triple product (= signed vol. of parallelepiped):
a · (b × c) :=
a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3
a 1 (b 2 c 3 − b 3 c 2 ) , a 2 (b 3 c 1 − b 1 c 3 ) , a 3 (b 1 c 2 − b 2 c 1 )
Thm 10 & 11 [Plane]: n · r = n · r 0 ⇐⇒ ax + by + cz = ax 0 + by 0 + cz 0 = d
Thm 13 [Derivative properties for vectors]: d dt (r^ (t) +^ s^ (t)) =^ r
′ (^) (t) + s′ (^) (t) d dt (c^ r^ (t)) =^ c^ r
′ (^) (t) d dt (f^ (t)^ r^ (t)) =^ f^
′ (^) (t) r (t) + f (t) r′ (^) (t) d dt (r^ (t)^ ·^ s^ (t)) =^ r
′ (^) (t) · s (t) + r (t) · s′ (^) (t) d dt (r^ (t)^ ×^ s^ (t)) =^ r
′ (^) (t) × s (t) + r (t) × s′ (^) (t)
Thm 14 [Arc length]: (length from a to b) =
∫ (^) b a ‖r
′(t)‖ dt
Vector rotation: 90 ° anticlockwise: 〈x, y〉 → 〈−y, x〉 90 ° clockwise: 〈x, y〉 → 〈y, −x〉
Surfaces
Level curve of f (x, y) = horizontal trace (for functions in two vars) = 2-D graph of f (x, y) = k for some constant k Contour plot = numerous level curves on the same graph
Level surface of f (x, y, z) = 3-D graph of f (x, y, z) = k for some constant k.
Quadric surfaces
Cylinder = infinite prism
Elliptic paraboloid: x
2 a^2 +^
y^2 b^2 =^
z c
Hyperbolic paraboloid: x
2 a^2 −^
y^2 b^2 =^
z c
Ellipsoid: x
2 a^2 +^
y^2 b^2 +^
z^2 c^2 = 1
Elliptic cone: x
2 a^2 +^
y^2 b^2 −^
z^2 c^2 = 0
Hyperboloid of one sheet
: x
2 a^2 +^
y^2 b^2 −^
z^2 c^2 = 1
Hyperboloid of two sheets
: x
2 a^2 +^
y^2 b^2 −^
z^2 c^2 =^ −^1
Limits
Limit: lim (x,y)→(a,b)
f (x, y) = L iff for any > 0 there exists δ > 0 such that |f (x, y) − L| < whenever 0 <
(x − a)^2 + (y − b)^2 < δ
Thm 15: To show limit does not exist, take the limit via two different paths that have different limits
Thm 16 & 17 [Limit theorems]: Limits may be taken into addition, subtraction, multiplication, division
Thm 18 [Squeeze theorem]: If |f (x, y) − L| ≤ g(x, y) ∀(x, y) close to (a, b) and lim (x,y)→(a,b)
g(x, y) = 0 then lim (x,y)→(a,b)
f (x, y) = L
Continuity: f is continuous at (a, b) ⇐⇒ lim (x,y)→(a,b)
f (x, y) = f (a, b)
i.e. the limit exists and the f is valid at (a, b)
Thm 20 & 21 [Continuity theorems]: If two functions are continuous (at (a, b)), then their sum, difference, product, quotient, and composition are continuous too (quotient requires denominator 6 = 0)
All polynomials, trigonometric, exponential, and rational functions are continuous
Partial Derivatives
Thm 2 [Clairaut’s theorem]: If fxy and fyx are both continuous on disk containing (a, b) then fxy (a, b) = fyx(a, b) Thm 3 [Tangent plane eqn]: Given surface z = f (x, y) with point (a, b):
- normal vector: 〈fx(a, b), fy (a, b), − 1 〉
- tangent plane: z = f (a, b) + fx(a, b)(x−a) + fy (a, b)(y−b) Multivariable differentiability: z = f (x, y) is differentiable at (a, b) if 4 z = fx(a, b) 4 x + fy (a, b) 4 y + 14 x + 24 y (with vanishing 1 and 2 ) i.e. zooming in to (a, b) will make surface approximate tangent plane fx & fy are continuous at (a, b) =⇒ f is diff.able at (a, b) f is differentiable at (a, b) =⇒ f is continuous at (a, b)
Differentiation Techniques
Chain rule: For z = f (x, y) and x = x(t), y = y(t): dz dt
∂f ∂x
dx dt
∂f ∂y
dy dt For z = f (x, y) and x = x(s, t), y = y(s, t): ∂z ∂s
∂f ∂x
∂x ∂s
∂f ∂y
∂y ∂s
Thm 11 [Implicit differentiation]: Given F (x, y, z) = 0 that defines z implicitly as a function of x and y, then: ∂z ∂x
Fx(x, y, z) Fz (x, y, z) ∂z ∂y
Fy (x, y, z) Fz (x, y, z) provided Fz (x, y, z) 6 = 0 Quotient rule: f (x) =
g(x) h(x)
=⇒ f ′(x) =
g′(x)h(x) − g(x)h′(x) [h(x)]^2 F (x, y, z) = 0 =⇒ normal vector = 〈Fx, Fy , Fz 〉
Gradient Vectors
Thm 13 [Dir. derivatives]: Duf (x, y) = Of (x, y) · u where Of (x, y) := 〈fx, fy 〉 = gradient vector at (x, y) and u := direction (as unit vector) Direction of Of (x, y) = steepest upward direction ‖Of (x, y)‖ = steepest upward gradient Thm 1 [Level curve ⊥ Of ]: 0 6 = Of (x 0 , y 0 ) is normal to the level curve f (x, y) = k that contains (x 0 , y 0 )
Thm 2 [Level surface ⊥ OF ]: 0 6 = OF (x 0 , y 0 , z 0 ) is normal to the level surface F (x, y, z) = k that contains (x 0 , y 0 , z 0 )
Critical Points, Minimum, Maximum
given f (x, y) : D → R
Local maximum: (a, b) is a local maximum if f (x, y) ≤ f (a, b) for all points (x, y) near (a, b)
Local minimum: (a, b) is a local minimum if f (x, y) ≥ f (a, b) for all points (x, y) near (a, b)
Saddle point: (a, b) is a saddle point if fx(a, b) = fy (a, b) = 0 and every neighbourhood at (a, b) contains points (x, y) ∈ D for which f (x, y) < f (a, b) and points (x, y) ∈ D for which f (x, y) > f (a, b)
Critical point: (a, b) is a critical point if fx(a, b) = fy (a, b) = 0 (If point P is a local maximum/minimum then: fx(P ) and fy (P ) both exist =⇒ P is a critical point)
Local maximum/minimum and critical points cannot be boundary points
Absolute maximum: f has an absolute max. at (a, b) if ∀(x, y) ∈ D, f (x, y) ≤ f (a, b)
Absolute minimum: f has an absolute min. at (a, b) if ∀(x, y) ∈ D, f (x, y) ≥ f (a, b)
Boundary point of R: point (a, b) such that every disk with center (a, b) both contains points in R and not in R
Closed set: Set that contains all its boundary points
Bounded set: Set that is contained in some (finite) disk
Thm 14 [Extreme Value Theorem]: If f (x, y) is continuous on a closed & bounded set D, then the absolute maximum & minimum must exist
To find absolute maximum/minimum of f with domain D:
- Find the values of f at all critical points in D
- Find the extreme values of f on the boundary of D
- Take largest/smallest of the values of Steps 1 & 2
Lagrange Multipliers
Suppose f (x, y) and g(x, y) are differentiable functions such that Og(x, y) 6 = 0 on the constraint curve g(x, y) = k. If (x 0 , y 0 ) is a (local) maximum/minimum of f (x, y) constrained by g(x, y) = k, then Of (x 0 , y 0 ) = λOg(x 0 , y 0 ) for some constant λ (the Lagrange multiplier).
To find the maximum/minimum points of f (x, y) constrained by g(x, y) = k, we solve { Of (x 0 , y 0 ) = λOg(x 0 , y 0 ) g(x 0 , y 0 ) = k
for x 0 , y 0 , λ.
Integration Techniques
Integration by parts: ∫ u
dv dx
dx = u v −
du dx
v dx
Area & Volume Integrals
Thm 4 [Fubini’s theorem]: If f is continuous on rectangle R = [a, b] × [c, d] then: ∫ ∫
R
f (x, y) dA =
∫ (^) b
a
∫ (^) d
c
f (x, y) dy dx =
∫ (^) d
c
∫ (^) b
a
f (x, y) dx dy
Region types (double integration): Type I: D = {(x, y) : a ≤ x ≤ b, g 1 (x) ≤ y ≤ g 2 (x)} Type II: D = {(x, y) : c ≤ y ≤ d, h 1 (y) ≤ x ≤ h 2 (y)}
Polar coords. ←→ rectangular coords.: x = r cos θ y = r sin θ
r =
x^2 + y^2 θ = atan2(y, x) Integrating over a polar rectangle: If R = {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β} then: ∫ ∫
R
f (x, y) dA =
∫ (^) β
α
∫ (^) b
a
f (r cos θ, r sin θ) r dr dθ
Region types (polar): Type I: D = {(r, θ) : 0 ≤ a ≤ r ≤ b, g 1 (r) ≤ θ ≤ g 2 (r)} Type II: D = {(r, θ) : α ≤ θ ≤ β, h 1 (θ) ≤ r ≤ h 2 (θ)}
Region types (triple integration): Type I: E = {(x, y, z) : (x, y) ∈ D, u 1 (x, y) ≤ z ≤ u 2 (x, y)} Type II: E = {(x, y, z) : (y, z) ∈ D, u 1 (y, z) ≤ x ≤ u 2 (y, z)} Type III: E = {(x, y, z) : (x, z) ∈ D, u 1 (x, z) ≤ y ≤ u 2 (x, z)}
Spherical coords. ←→ rectangular coords.: x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ
ρ =
x^2 + y^2 + z^2 θ = atan2(y, x) φ = cos−^1
z ρ
= cos−^1
z √ x^2 + y^2 + z^2
Integrating over a spherical wedge: If E = {(ρ, θ, φ) : 0 ≤ a ≤ ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d} then: ∫ ∫ ∫
E
f (x, y, z) dV = ∫ (^) d
c
∫ (^) β
α
∫ (^) b
a
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ^2 sin φ dρ dθ dφ
Jacobian (2D) of transformation (u, v) 7 → (x, y):
∂(x, y) ∂(u, v)
∂x ∂u
∂x ∂y ∂v ∂u
∂y ∂v
∣∣ = ∂x ∂u
∂y ∂v
∂x ∂v
∂y ∂u ∫ ∫
R
f (x, y) dA =
S
f (x(u, v), y(u, v))
∂(x, y) ∂(u, v)
∣ du dv
Jacobian (3D) of transformation (u, v, w) 7 → (x, y, z):
∂(x, y, z) ∂(u, v, w)
∂x ∂u
∂x ∂v
∂x ∂y ∂w ∂u
∂y ∂v
∂y ∂z ∂w ∂u
∂z ∂v
∂z ∂w
R
f (x, y, z) dV = ∫ ∫ ∫
S
f (x(u, v, w), y(u, v, w), z(u, v, w))
∂(x, y, z) ∂(u, v, w)
∣ du dv dw
Consider choosing transformation to make bounds constants
Using
∣ ∂ ∂((x,yu,v))
∣ ∂ ∂((u,vx,y))
− 1 and appropriate f (x, y) may avoid needing to express x, y in terms of u, v
Line Integrals
Line integral for scalar field: If curve C is given by r(t) = 〈x(t), y(t), z(t)〉, a ≤ t ≤ b then: ∫
C
f (x, y, z) ds =
∫ (^) b
a
f (x(t), y(t), z(t)) ‖r′(t)‖ dt
Answer is indep. of orientation and parameterization of r(t) Line integral for vector field: If curve C is given by r(t) = 〈x(t), y(t), z(t)〉, a ≤ t ≤ b then: ∫
C
F(x, y, z) · dr =
∫ (^) b
a
F(x(t), y(t), z(t)) · r′(t) dt = ∫
C
(P dx + Q dy + R dz) =
∫ (^) b
a
P x′(t)dt +
∫ (^) b
a
Qy′(t)dt +
∫ (^) b
a
Rz′(t)dt
Answer is its negation when r(t) has opposite orientation
Conservative vector fields
A vector field F is conservative on D iff F = Of for some scalar function f on D f is called the potential function of F To recover f from F = 〈fx, fy 〉, do partial integration of fx to get g(x, y) + h(y) [= f (x, y)] (where h(y) is the unknown integration constant), then differentiate g(x, y) + h(y) w.r.t. y and compare with fy to determine h(y) Test for conservative field (2D): If F = 〈P, Q〉 is a vector field in an open (excludes all boundary points) and simply-connected (has no “holes”) region D and both P and Q have continuous first-order partial derivatives on D then: ∂Q ∂x
∂P
∂y
⇐⇒ F is conservative on D
Test for conservative field (3D): F = 〈P, Q, R〉 (similar requirements as 2D case): ∂Q ∂x
∂P
∂y
∂R
∂y
∂Q
∂z
∂P
∂z
∂R
∂x
F is conservative on D
Fundamental theorem for line integrals: If F is conservative with potential function f , and C is a smooth curve from point A to point B, then: ∫
C
F · dr =
C
Of · dr = f (B) − f (A)
=⇒ line integral for conservative field is path-independent
Two paths with different line integrals but same initial and terminal points =⇒ vector field is not conservative
Green’s Theorem
If C is a positively oriented (anticlockwise), piecewise smooth, simple closed curve in the plane, and D is the region bounded by C, and F = 〈P, Q〉 then: ∫
C
F · dr =
C
P dx + Q dy =
D
∂Q
∂x
∂P
∂y
dA
(useful when ∂Q∂x − ∂P∂y is simpler than P and Q)
!!! Look out for holes (÷ 0) – use extended Green’s theorem
!!! When borrowing other question result, check orientation
Reverse application of Green’s theorem: If A is the area of D, then (choose whichever is convenient):
A =
C
x dy = −
C
y dx =
C
(x dy − y dx)
Parameterize the boundary curve in terms of t (a ≤ t ≤ b)
(e.g.
∫ (^) b
a
x(t)
dy dt
− y(t)
dx dt
dt )
Surface Integrals
Parametric form of a surface in R^3 : r(u, v) = 〈x(u, v), y(u, v), z(u, v)〉 , (u, v) ∈ D
Smooth surface: A surface that is parameterized by r(u, v) where (u, v) ∈ D, such that ru and rv are continuous and ru × rv 6 = 0 ∀ (u, v) ∈ D
Thm 6 [Normal vector of parametric surface]: If a smooth surface S has parameterization r(u, v) = 〈x(u, v), y(u, v), z(u, v)〉 , (u, v) ∈ D then: ru(a, b) × rv (a, b) is normal to S at (x(a, b), y(a, b), z(a, b))
Thm 7 [Surface integral for scalar field]: If a smooth surface S has parameterization r(u, v) = 〈x(u, v), y(u, v), z(u, v)〉 , (u, v) ∈ D then: ∫ ∫
S
f (x, y, z) dS =
D
f (x(u, v), y(u, v), z(u, v)) ‖ru ×rv ‖ dA
Thm 7a [Surface integral special case z = g(x, y)]: If S is the surface z = g(x, y) where (x, y) ∈ D then:
∫ ∫
S
f (x, y, z) dS =
D
f (x, y, g(x, y))
∂g ∂x
∂g ∂y
dA
Orientable surface: two-sided surface Positive orientation: outward from enclosed region
Thm 6 [Surface integral for vector field]: If a smooth surface S has parameterization r(u, v) = 〈x(u, v), y(u, v), z(u, v)〉 , (u, v) ∈ D then: ∫ ∫
S
F · dS =
S
F · ˆn dS =
D
F · (ru × rv ) dA
Thm 6a [Surface integral special case z = g(x, y)]: If F = 〈P, Q, R〉, and S is the surface z = g(x, y) where (x, y) ∈ D, then the flux in the upward orientation: ∫ ∫
S
F · dS =
D
−P
∂g ∂x
− Q
∂g ∂y
+ R
dA
Vector Differential Operator
O =
∂x
∂y
∂z
Divergence
If F = 〈P, Q, R〉 then:
div F = O · F =
∂P
∂x
∂Q
∂y
∂R
∂z
Gauss’ theorem
If E is a solid region with piecewise smooth boundary surface S with positive (outward) orientation then: ∫ ∫
S
F · dS =
E
div F dV
Curl
If F = 〈P, Q, R〉 then:
curl F = O × F =
∂R
∂y
∂Q
∂z
∂P
∂z
∂R
∂x
∂Q
∂x
∂P
∂y
Stokes’ theorem
If S is a surface with a boundary curve C (positively oriented w.r.t. S) then: ∫
C
F · dr =
S
curl F · dS
Positive orientation of boundary curve: If surface S has unit normals pointing towards you, then the positive orientation of boundary curve C goes anti-clockwise
Trigonometric Formulae
Double angle sin 2x = 2 sin x cos x cos 2x = cos^2 x − sin^2 x = 2 cos^2 x − 1 = 1 − 2 sin^2 x tan 2x =
2 tan x 1 − tan^2 x Triple angle sin 3x = 3 sin x − 4 sin^3 x cos 3x = 4 cos^3 x − 3 cos x Pythagorean sin^2 x + cos^2 x = 1 tan^2 x + 1 = sec^2 x cot^2 x + 1 = csc^2 x
Integrals∫
sin^2 x dx =
(2x − sin 2x) ∫ cos^2 x dx =
(2x + sin 2x) ∫ tan^2 x dx = tan x − x ∫ sin^3 x dx =
(cos 3x − 9 cos x) ∫ cos^3 x dx =
(sin 3x + 9 sin x) ∫ sin x cos x dx = −
cos^2 x + C 1
=
sin^2 x + C 2
Sum of angles sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β ∓ sin α sin β
tan(α ± β) =
tan α ± tan β 1 ∓ tan α tan β cot(α ± β) =
cot α cot β ∓ 1 cot β ± cot α
