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Quadric surfaces Partial Derivatives Critical Points, Minimum, Maximum
MA2104
Multivariable Calculus ❼ Cylinder = infinite prism ❼ Thm 2 [Clairaut’s theorem]: If f and f are both given f(x,y): D → R
xy yx
continuous on disk containing (a,b) then fxy(a,b)=fyx(a,b) ❼ Local maximum: (a,b) is a local maximum if
Basic Vectors ❼ Elliptic paraboloid: x2 + y2 = z ❼ Thm 3 [Tangent plane eqn]: f(x,y) ≤ f(a,b) for all points (x,y) near (a,b)
2 2
a b c Given surface z = f(x,y) with point (a,b):
❼ Thm 1: kcuk = |c|kuk - normal vector: hf (a,b),f (a,b),−1i ❼ Local minimum: (a,b) is a local minimum if
x y f(x,y) ≥ f(a,b) for all points (x,y) near (a,b)
❼ Thm 2: (unit vector in direction of a) = a - tangent plane: z = f(a,b) +fx(a,b)(x−a)+fy(a,b)(y−b)
kak ❼ Multivariable differentiability: ❼ Saddle point: (a,b) is a saddle point if
2 y2
x z f (a,b) = f (a,b) = 0 and every neighbourhood at (a,b)
❼ Hyperbolic paraboloid: 2 − 2 = z = f(x,y) is differentiable at (a,b) if x y
❼ Thm 3 [Dot product properties]: a b c contains points (x,y) ∈ D for which f(x,y) < f(a,b) and
△z=f (a,b)△x+f (a,b)△y+ǫ △x+ǫ △y
a·b=b·a (da)·b = d(a·b) = a·(db) x y 1 2 points (x,y) ∈ D for which f(x,y) > f(a,b)
(with vanishing ǫ and ǫ ) i.e. zooming in to (a,b) will
a·(b+c)=a·b+a·c 0·a=0 1 2
2 make surface approximate tangent plane ❼ Critical point: (a,b) is a critical point if
(a+b)·c=a·c+b·c a·a=kak f (a,b) = f (a,b) = 0
2 2 2 ❼ fx & fy are continuous at (a,b) =⇒ f is diff.able at (a,b) x y
❼ Thm 4 [Dot product & angle]: a·b = kakkbkcosθ ❼ Ellipsoid: x + y + z = 1 (If point P is a local maximum/minimum then:
2 2 2 ❼ f is differentiable at (a,b) =⇒ f is continuous at (a,b)
a b c
f (P) and f (P) both exist =⇒ P is a critical point)
x y
❼ Thm 5 [Orthogonality]: a ⊥ b ⇐⇒ a·b = 0 Differentiation Techniques ❼ Local maximum/minimum and critical points cannot be
❼ Component (signed scalar): comp b = kbkcosθ = a·b boundary points
a kak
❼ Elliptic cone: x2 + y2 − z2 = 0 ❼ Chain rule: For z = f(x,y) and x = x(t), y = y(t): ❼ Absolute maximum: f has an absolute max. at (a,b) if
a a·b 2 2 2
❼ Projection (vector): proj b = comp b× = a a b c
a a kak a·a dz ∂f dx ∂f dy ∀(x,y) ∈ D,f(x,y) ≤ f(a,b)
dt = ∂x dt + ∂y dt
i j k ❼ Absolute minimum: f has an absolute min. at (a,b) if
For z = f(x,y) and x = x(s,t), y = y(s,t):
: ∀(x,y) ∈ D,f(x,y) ≥ f(a,b)
❼ Cross product: ha ,a ,a i×hb ,b ,b i = a a a =
1 2 3 1 2 3 1 2 3
b b b ∂z ∂f ∂x ∂f ∂y
1 2 3 Hyperboloid 2 y2 2 ❼ Boundary point of R: point (a,b) such that every disk
ha b −a b ,a b −a b ,a b −a b i ❼ : x + −z =1 = +
2 3 3 2 3 1 1 3 1 2 2 1 2 2 2
of one sheet a b c ∂s ∂x ∂s ∂y ∂s with center (a,b) both contains points in R and not in R
❼ Thm 6: (a×b)⊥a and (a×b)⊥b ❼ Thm 11 [Implicit differentiation]: Given F(x,y,z) = 0 ❼ Closed set: Set that contains all its boundary points
that defines z implicitly as a function of x and y, then:
❼ Thm 7 [Cross prod. & angle]: ka×bk = kakkbksinθ ∂z F (x,y,z) ❼ Bounded set: Set that is contained in some (finite) disk
=− x
Hyperboloid 2 y2 2
❼ : x + −z =−1 ∂x F (x,y,z) ❼ Thm 14 [Extreme Value Theorem]:
❼ Thm 8 [Cross product properties]: 2 2 2 z
of two sheets a b c
a×b=−(b×a) (a+b)×c=a×c+b×c ∂z F (x,y,z) If f(x,y) is continuous on a closed & bounded set D, then
=− y the absolute maximum & minimum must exist
a×(b+c)=a×b+a×c (da)×b=d(a×b)=a×(db) ∂y F (x,y,z)
z
provided F (x,y,z) 6= 0 ❼ To find absolute maximum/minimum of f with domain D:
❼ Scalar triple product (= signed vol. of parallelepiped): z
Limits 1) Find the values of f at all critical points in D
a a a ❼ Quotient rule:
1 2 3 2) Find the extreme values of f on the boundary of D
′ ′
: b b b g(x) g (x)h(x)−g(x)h (x)
a·(b×c) = 1 2 3 = ❼ Limit: lim f(x,y) = L f(x) = =⇒ f′(x) = 2 3) Take largest/smallest of the values of Steps 1 & 2
c c c (x,y)→(a,b) h(x)
1 2 3 [h(x)]
a (b c −b c ),a (b c −b c ),a (b c −b c ) iff for any ǫ > 0 there exists δ > 0 such that
1 2 3 3 2 2 3 1 1 3 3 1 2 2 1 q ❼ F(x,y,z) = 0 =⇒ normal vector = hF ,F ,F i
2 2 x y z Lagrange Multipliers
❼ Thm 10 & 11 [Plane]: |f(x,y)−L| < ǫ whenever 0 < (x−a) +(y−b) <δ
n·r=n·r ⇐⇒ ax+by+cz=ax +by +cz =d
0 0 0 0 ❼ Thm 15: To show limit does not exist, take the limit via Gradient Vectors ❼ Suppose f(x,y) and g(x,y) are differentiable functions such
❼ Thm 13 [Derivative properties for vectors]: two different paths that have different limits that ▽g(x,y) 6= 0 on the constraint curve g(x,y) = k.
d ′ ′ ❼ Thm 13 [Dir. derivatives]: Duf(x,y) = ▽f(x,y)·u If (x ,y ) is a (local) maximum/minimum of f(x,y)
dt (r(t) + s(t)) = r (t) + s (t) ❼ Thm 16 & 17 [Limit theorems]: Limits may be taken : 0 0
where ▽f(x,y) = hf ,f i = gradient vector at (x,y)
d ′ x y constrained by g(x,y) = k, then ▽f(x ,y ) = λ▽g(x ,y )
(cr(t)) = cr (t) into addition, subtraction, multiplication, division : 0 0 0 0
dt and u = direction (as unit vector) for some constant λ (the Lagrange multiplier).
d (f (t)r(t)) = f′ (t)r(t) + f (t)r′ (t)
dt ❼ Direction of ▽f(x,y) = steepest upward direction
d (r(t) · s(t)) = r′ (t) · s(t) + r(t) · s′ (t) ❼ Thm 18 [Squeeze theorem]:
dt k▽f(x,y)k = steepest upward gradient
d (r(t) ×s(t)) = r′(t)×s(t)+r(t)×s′(t) If |f(x,y) − L| ≤ g(x,y) ∀(x,y) close to (a,b)
dt and lim g(x,y) = 0 ❼ Thm 1 [Level curve ⊥ ▽f]: 0 6= ▽f(x ,y ) is normal to
R (x,y)→(a,b) 0 0
b ′ the level curve f(x,y) = k that contains (x ,y )
❼ Thm 14 [Arc length]: (length from a to b) = a kr (t)kdt then lim f(x,y) = L 0 0
(x,y)→(a,b)
❼ Vector rotation: 90➦ anticlockwise: hx,yi → h−y,xi
90➦ clockwise: hx,yi → hy,−xi ❼ Continuity: f is continuous at (a,b)
⇐⇒ lim f(x,y) = f(a,b)
(x,y)→(a,b)
i.e. the limit exists and the f is valid at (a,b)
Surfaces
❼ Thm 20 & 21 [Continuity theorems]: ❼ To find the maximum/minimum points of f(x,y)
❼ Level curve of f(x,y) = horizontal trace (for functions in If two functions are continuous (at (a,b)), then their sum, constrained by g(x,y) = k, we solve
two vars) = 2-D graph of f(x,y) = k for some constant k difference, product, quotient, and composition are
Contour plot = numerous level curves on the same graph continuous too (quotient requires denominator 6= 0) ▽f(x ,y ) = λ▽g(x ,y )
❼ Thm 2 [Level surface ⊥ ▽F]: 0 6= ▽F(x ,y ,z ) is 0 0 0 0
0 0 0 g(x ,y ) = k
❼ Level surface of f(x,y,z) = 3-D graph of f(x,y,z) = k ❼ All polynomials, trigonometric, exponential, and rational normal to the level surface F(x,y,z) = k that contains 0 0
for some constant k. functions are continuous (x ,y ,z ) for x , y , λ.
0 0 0 0 0
Integration Techniques ❼ Jacobian (3D) of transformation (u,v,w) 7→ (x,y,z): Green’s Theorem Vector Differential Operator
∂x ∂x ∂x
∂u ∂v ∂w ❼ If C is a positively oriented (anticlockwise), piecewise
❼ Integration by parts: ∂(x,y,z) ∂y ∂y ∂y ∂ ∂ ∂
= smooth, simple closed curve in the plane, and D is the ▽= , ,
Z Z ∂(u,v,w) ∂u ∂v ∂w ∂x ∂y ∂z
dv du ∂z ∂z ∂z region bounded by C, and F = hP,Qi then:
u dx = uv− vdx ZZZ ∂u ∂v ∂w
dx dx Z Z ZZ
f(x,y,z)dV = ∂Q ∂P Divergence
ZZZ R F·dr= Pdx+Qdy= − dA
∂(x,y,z) C C D ∂x ∂y
Area & Volume Integrals f(x(u,v,w),y(u,v,w),z(u,v,w)) dudvdw
S ∂(u,v,w) (useful when ∂Q − ∂P is simpler than P and Q) If F = hP,Q,Ri then:
❼ Consider choosing transformation to make bounds constants ∂x ∂y
❼ Thm 4 [Fubini’s theorem]: ∂P ∂Q ∂R
❼ !!! Look out for holes (÷0) – use extended Green’s theorem divF=▽·F= + +
If f is continuous on rectangle R = [a,b] × [c,d] then: −1
❼ Using ∂(x,y) = ∂(u,v) and appropriate f(x,y) may ∂x ∂y ∂z
ZZ Z bZ d Z dZ b ∂(u,v) ∂(x,y) ❼ !!! When borrowing other question result, check orientation
f(x,y)dA = f(x,y)dydx = f(x,y)dxdy avoid needing to express x,y in terms of u,v Gauss’ theorem
R a c c a ❼ Reverse application of Green’s theorem:
Line Integrals If A is the area of D, then (choose whichever is convenient): If E is a solid region with piecewise smooth boundary surface
❼ Region types (double integration): Z Z Z S with positive (outward) orientation then:
Type I: D = {(x,y) : a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)} A= xdy=− ydx=1 (xdy−ydx) ZZ ZZZ
Type II:D = {(x,y) : c ≤ y ≤ d, h (y) ≤ x ≤ h (y)} ❼ Line integral for scalar field: C C 2 C
1 2 If curve C is given by r(t) = hx(t),y(t),z(t)i, a ≤ t ≤ b then: F·dS= divFdV
Z Z Parameterize the boundary curve in terms of t (a ≤ t ≤ b) S E
❼ Polar coords. ←→ rectangular coords.: b Z
p 2 2 f(x,y,z)ds = f(x(t),y(t),z(t))kr′(t)k dt 1 b dy dx
x=rcosθ r = x +y (e.g. x(t) −y(t) dt ) Curl
y = rsinθ θ = atan2(y,x) C a 2 a dt dt
❼ Integrating over a polar rectangle: Answer is indep. of orientation and parameterization of r(t) If F = hP,Q,Ri then:
If R = {(r,θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β} then: ❼ Line integral for vector field: Surface Integrals
ZZ Z Z If curve C is given by r(t) = hx(t),y(t),z(t)i, a ≤ t ≤ b then: curlF = ▽×F= ∂R − ∂Q,∂P − ∂R,∂Q − ∂P
β b Z Z 3 ∂y ∂z ∂z ∂x ∂x ∂y
f(x,y)dA = f(rcosθ,rsinθ)rdrdθ b ❼ Parametric form of a surface in R :
R α a F(x,y,z)·dr = F(x(t),y(t),z(t)) · r′(t)dt = r(u,v) = hx(u,v),y(u,v),z(u,v)i, (u,v) ∈ D
C a Stokes’ theorem
❼ Region types (polar): Z Z b ′ Z b ′ Z b ′ ❼ Smooth surface: A surface that is parameterized by
Type I: D = {(r,θ) : 0 ≤ a ≤ r ≤ b, g (r) ≤ θ ≤ g (r)} (P dx+Qdy+Rdz)= Px(t)dt+ Qy(t)dt+ Rz(t)dt r(u,v) where (u,v) ∈ D, such that ru and rv are continuous If S is a surface with a boundary curve C (positively oriented
1 2 C a a a w.r.t. S) then:
Type II:D = {(r,θ) : α ≤ θ ≤ β, h (θ) ≤ r ≤ h (θ)} and ru ×rv 6= 0 ∀ (u,v) ∈ D
1 2 Answer is its negation when r(t) has opposite orientation Z ZZ
❼ Region types (triple integration): ❼ Thm 6 [Normal vector of parametric surface]: F·dr= curlF·dS
Type I: E={(x,y,z):(x,y)∈D, u (x,y)≤z ≤u (x,y)} Conservative vector fields If a smooth surface S has parameterization C S
1 2 r(u,v) = hx(u,v),y(u,v),z(u,v)i,(u,v) ∈ D then:
Type II: E = {(x,y,z) : (y,z) ∈ D, u (y,z) ≤ x ≤ u (y,z)}
1 2 ❼ A vector field F is conservative on D iff F = ▽f
Type III:E = {(x,y,z) : (x,z) ∈ D, u (x,z) ≤ y ≤ u (x,z)} r (a,b)×r (a,b) is normal to S at (x(a,b),y(a,b),z(a,b)) Positive orientation of boundary curve:
1 2 for some scalar function f on D u v
If surface S has unit normals pointing towards you, then the
❼ Spherical coords. ←→ rectangular coords.: f is called the potential function of F ❼ Thm 7 [Surface integral for scalar field]: positive orientation of boundary curve C goes anti-clockwise
x=ρsinφcosθ ❼ To recover f from F = hf ,f i, do partial integration of f If a smooth surface S has parameterization
x y x r(u,v) = hx(u,v),y(u,v),z(u,v)i,(u,v) ∈ D then:
y = ρsinφsinθ to get g(x,y) +h(y) [= f(x,y)] (where h(y) is the unknown Trigonometric Formulae
z = ρcosφ integration constant), then differentiate g(x,y) + h(y) w.r.t. ZZ f(x,y,z)dS = ZZ f(x(u,v),y(u,v),z(u,v))kr ×r kdA
y and compare with fy to determine h(y) u v Double angle Integrals
S D Z
p 2 2 2 ❼ Test for conservative field (2D): sin2x = 2sinxcosx 2 1
ρ = x +y +z If F = hP,Qi is a vector field in an open (excludes all ❼ Thm 7a [Surface integral special case z = g(x,y)]: 2 2 sin xdx = 4 (2x−sin2x)
θ = atan2(y,x) boundary points) and simply-connected (has no “holes”) If S is the surface z = g(x,y) where (x,y) ∈ D then: cos2x = cos x−sin x Z 1
z =2cos2x−1 cos2xdx = (2x+sin2x)
−1
φ=cos region D and both P and Q have continuous first-order s 2 Z 4
ZZ ZZ =1−2sin x
ρ 2 2
! partial derivatives on D then: ∂g ∂g 2
−1 p z f(x,y,z)dS = f(x,y,g(x,y)) ∂x + ∂y +1 dA tan2x = 2tanx tan xdx = tanx−x
=cos ∂Q ∂P S D 2 Z
2 2 2 ≡ ⇐⇒ Fisconservative on D 1−tan x 1
x +y +z Triple angle 3
∂x ∂y 3 Z sin xdx = 12 (cos3x−9cosx)
❼ Test for conservative field (3D): ❼ Orientable surface: two-sided surface sin3x = 3sinx−4sin x 3 1
❼ Integrating over a spherical wedge: Positive orientation: outward from enclosed region cos3x = 4cos3x−3cosx cos xdx = (sin3x+9sinx)
F=hP,Q,Ri (similar requirements as 2D case): Z 12
If E = {(ρ,θ,φ) : 0 ≤ a ≤ ρ ≤ b,α ≤ θ ≤ β,c ≤ φ ≤ d} then: ❼ Thm 6 [Surface integral for vector field]: Pythagorean 1
∂Q ∂P ∂R ∂Q ∂P ∂R Fis conservative 2 2 sinxcosxdx = − cos2x+C
ZZZ ≡ , ≡ , ≡ ⇐⇒ If a smooth surface S has parameterization sin x+cos x=1 2 1
∂x ∂y ∂y ∂z ∂z ∂x on D 2 2
f(x,y,z)dV = r(u,v) = hx(u,v),y(u,v),z(u,v)i,(u,v) ∈ D then: tan x+1=sec x 1 2
Z Z Z E 2 2 = sin x+C2
d β b ❼ Fundamental theorem for line integrals: ZZ ZZ ZZ cot x+1=csc x 2
f(ρsinφcosθ,ρsinφsinθ,ρcosφ)ρ2sinφdρdθdφ
If F is conservative with potential function f, and C is a ˆ
c α a smooth curve from point A to point B, then: S F·dS= S F·ndS = DF·(ru×rv) dA Sum of angles
❼ Jacobian (2D) of transformation (u,v) 7→ (x,y): Z F·dr=Z ▽f·dr=f(B)−f(A) ❼ Thm 6a [Surface integral special case z = g(x,y)]: sin(α ±β) = sinαcosβ ±cosαsinβ
∂x ∂x If F = hP,Q,Ri, and S is the surface z = g(x,y) where cos(α±β)=cosαcosβ∓sinαsinβ
∂(x,y) ∂u ∂v ∂x∂y ∂x∂y C C
= ∂y ∂y = − (x,y) ∈ D, then the flux in the upward orientation: tanα±tanβ
∂(u,v) ∂u ∂v ∂u∂v ∂v ∂u =⇒ line integral for conservative field is path-independent tan(α±β)= 1∓tanαtanβ
ZZ ZZ ZZ ZZ
∂(x,y) ❼ Two paths with different line integrals but same initial and ∂g ∂g cotαcotβ ∓1
f(x,y)dA = f(x(u,v),y(u,v)) dudv F·dS= −P −Q +R dA cot(α±β)=
R S ∂(u,v) terminal points =⇒ vector field is not conservative S D ∂x ∂y cotβ ±cotα
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