1. Review of first 1/3 of course
Parameterized curves
x = f(t) 1D parameterized function
Mapping from 1D to 1D
Gives x coordinate of object moving in 1D at time t. x
(t=0)
(t=1)
(t=2)
(t=3) y
x = f(t), y=g(t) 1D parameterized function
Mapping from 1D to 2D
Gives (x.y) coordinate of object moving in 2D at time t.
(t=2)
(t=3)
(t=1)
(t=0) x z
(t=3) y
x = f(t), y=g(t), z=h(t) 1D parameterized function
Mapping from 1D to 3D
Gives (x.y,z) coordinate of object moving in 3D at time t.
(t=2)
(t=1)
(t=0) x
J.A. Sethian/UC Berkeley

2. Derivatives of parameterized curvesy
R(t1) = (x(t1), y(t1))
R(t) = (x(t), y(t)) is vector description of curve in 2D.
R(t0) = (x(t0), y(t0)) x y
dR(t)/dt=(dx(t)/dt , dy(t)/dt, dz(t)/dt) is vector
derivative of particle trajectory
It is tangent to the curve at the point x(t=t0), y(t=t0)
(t=t0)
dR(t=t0)/dt
R(t=t0) = (x(t=t0), y(t=t0)) x
J.A. Sethian/UC Berkeley

3. Arc-length of parameterized curvesy B
(At t=t1)
(x(t), y(t)) describes curve in 2D.
What is the total arc-length S from A to B? A
At t=t0 S x
dS(t)/dt=[ (dx(t)/dt)2 + (dy(t)/dt)2 ](1/2) using “bob’s”
theorem.
Add them all up to get total arc-length
S = [ds(t)/dt] dt
S = = [ (dx(t)/dt)2 + (dy(t)/dt)2 ](1/2) dt t1 y t0 t1
ds(t)
dy(t)/dt
dx(t)/dt t0 x
J.A. Sethian/UC Berkeley

4. Vectors Dot product Cross product Two vectors in 3 dimensions
V=(v1,v2,v3)
Definition of dot product:
U dot V = (u1*v1 + u2*v2 + u3*v3) z Ѳ y
Angle between vectors is Ѳ
U=(u1,u2,u3)
Proved: U dot V = |U| |V| cos Ѳ
Proved:
Two vectors are orthogonal U dot V=0 x
Cross product
W=(w1,w2,w3) z
Given two vectors in 3 dimensions
Can define their cross product as U x V =
W=(u2*v3-u3*v2,u3*v1-u1*v3,u1*v2-u2*v1) y
V=(v1,v2,v3) Ѳ
Proved: W=U x V is orthogonal to U and to V
U=(u1,u2,u3)
Proved: |U x V| = |U||V| sin Ѳ x
Proved: |U x V| = area of parallelogram
J.A. Sethian/UC Berkeley

5. Defining planes using vectors
A plane is completely described if you are given a point P=(x0,y0,z0) and normal vector N
For any point (x,y,z) in the plane, the vector that connects (x,y,z) to P is given by V = (x-x0,y-y0,z-z0)
N = (a,b,c) normal vector
We can write an equation for any point (x,y,z) in the plane by realizing that the any such vector V in the plane must be orthogonal to N: that is, N dot V = 0 R
P=(xo,yo,zo) Q
V = (x-x0,y-y0,z-z0)
So any point (x,y,z) in the plane must satisfy
N dot V = 0: which is the same as
(a,b,c) dot (x-x0,y-y0,z-z0) = 0 which becomes
a(x-x0) + b(x-y0) + x(z-z0) = 0
(x,y,z)
If we are not given a normal vector N , but instead have three points P, Q, and R in the plane, we can find a normal vector N by taking the cross product
N = (Q-P) x (R-P)
J.A. Sethian/UC Berkeley

6. Functions of more than one variable
Input space
Need an extra dimension to graph it
f(x): 1D input
to 1D output
z=f(x0) y
y=f(x) x
(Red arrow is output space)
Function of one variable x x0
z=f(x0,y0) z y
f(x,y): 2D input to 1D output
z=f(x,y)
(x,y) y x
(x0,y0)
Function of two variables x z w
w=f(x0,y0,z0)
f(x,y,z): 3D input to 1D output
(x,y,z) z
w=f(x,y,z) y y
(I don’t know how to draw the graph, but it’s there somewhere!)
(x0,y0,z0) x
Function of three variables x

7. Derivatives of functions of more than one variable
Start with a function f(x,y)
The graph is 2D surface in R3
z=f(x0,y0+h)
Slope=fy(x0,y0) z
z=f(x0+h,y0)
At an input point (x0,y0) the function has an output f(x0,y0)
z=f(x,y)
We can ask “how does the output change as x increases, holding y fixed?”
z=f(x0,y0) y
Slope=fx(x0,y0)
(x0,y0+h)
(x0,y0)
The slope of the line connecting these changes is 𝐟𝐱 𝐱𝟎,𝐲𝟎 x
(x0+h,y0)
We can also ask “how does the output change as y increases, holding x fixed?” Answer= 𝐟𝐲 𝐱𝟎,𝐲𝟎
J.A. Sethian/UC Berkeley

8. The Multi-Dimensional Chain Rule
Of one variable: t
f(t) x
g(x) y
𝒅𝒚 𝒅𝒚 = 𝒅𝒚 𝒅𝒙 𝒅𝒙 𝒅𝒕
Of many variables: r
f(r,s,t) u
a(u,v) x z s
g(r,s,t) v
c(x,y)
b(v,w) y
h(r,s,t) w r
To find 𝝏𝒛 𝒅𝒓 follow all pathways through the graph:
𝝏𝒛 𝒅𝒓 = 𝝏𝒛 𝝏𝒙 𝝏𝒙 𝝏𝒖 𝛛𝒖 𝝏𝒓 𝝏𝒛 𝝏𝒙 𝝏𝒙 𝝏𝒗 𝛛𝒗 𝝏𝒓 𝝏𝒛 𝝏𝒚 𝝏𝒚 𝝏𝒗 𝛛𝒗 𝝏𝒓 𝝏𝒛 𝝏𝒚 𝛛𝒚 𝝏𝒘 𝝏𝒘 𝝏𝒓
J.A. Sethian/UC Berkeley

9. Critical Values of Multivariable functions
One dimension y=f(x) y y
Critical points: Places where df/dx =0
If, in addition, d2f/dx2 <0 (>0) then max (min) x x a b c d e
x=a: global maximum ||| x=b: global min ||| x=c: local maximum |||x=d: local min |||x=e: (inflection)
Two dimensions z=f(x,y)
Critical points: Places where both df/dx =0 and df/dy=0
f(a2,b2)
At the input point (a,b) define D as
D(a,b)=fxx(a,b) fyy(a,b) – [fxy(a,b)]2
Suppose (a,b) is a critical point, then
(a) if D(a,b)>0 and fxx(a,b)>0, then local max
(b) if D(a,b)>0 and fxx(a,b)<0, then local min
(c) if D(a,b)<0, then neither max nor min
f(a0,b0)
f(a1,b1)
(a0,b0) = local max. (a1,b1) = neither (saddle) (a2,b2)= global max
J.A. Sethian/UC Berkeley

10. Direction Derivatives
We can also ask: Standing at the input point (x0,y0), how does the output change if we move in the direction u=(a,b), where u is a unit vector? z
z=f(x,y)
z=f(x0,y0)
Slope=Du(x0,y0)= direction derivative y
u=(a,b)
(x0+ah,y0+bh)
(x0,y0) x
And we proved that:
Du(x0,y0)=a*fx(x0,y0) + b * fy(x0,y0)
J.A. Sethian/UC Berkeley

11. The gradient Start with a function f(x,y)
The graph is 2D surface in R3
z=f(x0,y0+h)
Slope=fy(x0,y0) z
z=f(x0+h,y0)
At an input point (x0,y0) the function has an output f(x0,y0)
z=f(x,y)
We can ask “how does the output change as x increases, holding y fixed?”
z=f(x0,y0) y
Slope=fx(x0,y0)
(x0,y0+h)
𝜵𝐟 𝐱𝟎,𝐲𝟎 =<𝐟𝐱 𝐱𝟎,𝐲𝟎 ,𝐟𝐲 𝐱𝟎,𝐲𝟎 >
(x0,y0)
The slope of the line connecting these changes is 𝐟𝐱 𝐱𝟎,𝐲𝟎 x
(x0+h,y0)
We can also ask “how does the output change as y increases, holding x fixed?” Answer= 𝐟𝐲 𝐱𝟎,𝐲𝟎
Finally,we can ask:
“Standing at input point (x0,y0), what input direction gives the maximal change in f(x,y)?"
Why? Because
Duf(x0,y0) = 𝛁𝐟 𝐱𝟎,𝐲𝟎 𝐝𝐨𝐭 𝐮
=|𝛁𝐟| |u| cos
The answer is to move in the gradient direction
𝜵𝐟 𝐱𝟎,𝐲𝟎 =<𝐟𝐱 𝐱𝟎,𝐲𝟎 ,𝐟𝐲 𝐱𝟎,𝐲𝟎 > Ѳ
J.A. Sethian/UC Berkeley

12. Level Sets z Start with a function f(x,y)
The graph is 2D surface in R3
z=f(x,y)=x2+y2
Level set with value k is the set of all input points sent to the value k.
Height k=f(x0,y0)
Moving along the k level set in input space, the function f(x,y) doesn’t change y
(x0,y0)
𝛁𝐟(𝐱𝟎,𝐲𝟎)
So, if u is a direction along the k level set, then Duf(x0,y0)=0
Direction of maximal change
“k level set”: all input points sent to the same output k u x
The gradient at f(x0,y0) is the direction of maximal change
Duf(x0,y0)=0
Since Duf(x0,y0)=0 and since 0=Duf(x0,y0)= 𝛁𝐟(𝐱𝟎,𝐲𝟎) dot u
Then u(x0,y0) is perpendicular to 𝛁𝐟(𝐱𝟎,𝐲𝟎)
So the gradient at 𝛁𝐟(𝐱𝟎,𝐲𝟎) is normal to the tangent to the level set going through (𝒙𝟎,𝒚𝟎)
J.A. Sethian/UC Berkeley

13. Level Sets in Higher Dimensions
Again. At input (x0,y0) the gradient is normal to the line tangent to the k=f(x0,y0) level curve passing through (x0,y0).
This same idea is true in higher dimensions. For a function w=f(x,y,z), the input space is three-dimensional.
Input space for f(x,y,z) z
Tangent plane
𝛁𝐟(𝐱𝟎,𝐲𝟎,𝐳𝟎)
Gradient
At input (x0,y0,z0), the gradient is normal to the *plane tangent* to the k=f(x0,y0,z0) level set passing through (x0,y0,z0). y w
(x0,y0,z0)
w0=f(x0,y0,z0) z
Input space
(x,y,z) y
So we have an equation for the tangent plane at the point (x0,y0,z0), since we know a point and the normal
(x0,y0,z0) x
Every (x,y,z) input point on this k-level surface is sent to the same output k=f(x0,y0,z0) x
Graph of output vs. input
𝛁𝐟(𝐱𝟎,𝐲𝟎,𝐳𝟎)
J.A. Sethian/UC Berkeley

14. Lagrange Multipliers Want to maximize f(x,y) = - [x^2 +y^2]
Let’s plot the level curves:
Constraint g(x,y) = 2
Level set k=-9: f(x,y)=-9
Seems clear that the maximum occurs at (0,0). And the maximum f(0,0) is 0
Level set k=-4: f(x,y)=-4
Level set k=-1: f(x,y)=-1
Level set k=0: f(x,y)=0
Input space
By adding this constraint, the maximum now occurs at (0,2), and the maximum value f(x,y)=-4
Let’s now add a constraint. We require that (x,y) satisfy y-x2=2.
So if g(x,y)=y-x2, we are requiring that g(x,y)=2
This maximum occurs where the constraint curve touches a level set exactly once!
Extreme value occurs where normal to the constraint curve points in the direction as the normal to the level

15. Chapter 15: Integration y-axis y=f(x) 1D Integration: f(x) f(x) x-axisdx
x=a
x=b
x=a
x=b
x=b
𝑓 𝑥 𝑑𝑥 = amount of f(x) between x=a and x=b
= area under curve y=f(x) between x=a and x=b
x=a
z-axis
z=f(x,y)
f(x,y)
y-axis
y=d
2D Integration: f(x,y)
y-axis
y=c
dxdy
x=a
x-axis
y=d
x=b
x=b
𝑓 𝑥,𝑦 𝑑𝑥𝑑𝑦 = amount of f(x,y) between x=a and x=b and y=c and y=d
= volume under surface z=f(x,y) between x=a and x=b and y=c and y=d
x-axis
y=c
x=a
f(x,y,z)
w-axis
w=f(x,y,z)
3D Integration: f(x,y,z)
z-axis
y-axis
x-axis
z=f
y=d
x=b
𝑓 𝑥,𝑦 𝑑𝑥𝑑𝑦𝑑𝑧 = amount of f(x,y,z) between x=a and x=b and y=c and y=d and z=e and z=f
= volume under graph z=f(x,y,z) between x=a and x=b and y=c and y=d and z=e and z=f
z=e
y=c
x=a

16. Integration over weird regions
h(x)
Integration over weird regions
y-axis
f(x,y)
g(x)
x-axis
x=a
x=b
Choose outer limits of integration first:
y-axis
𝑥=𝑎 ? 𝑥=𝑏 ? 𝑑𝑥
f(x,y)
x-axis
x=a
x=b
For each value of the outer limit of integration, find the inner limits of integration:
y-axis
h(x)
f(x,y)
𝑥=𝑎 𝑔(𝑥) 𝑥=𝑏 ℎ(𝑥) 𝑓 𝑥,𝑦 𝑑𝑦 𝑑𝑥
g(x)
x-axis x

17. Other coordinate systems:
Polar coordinates:
x=r cos (Ѳ)
y=r sin (Ѳ)
rdѲ r
(x,y) Ѳ r
dxdy=rdrdѲ
Cylindrical coordinates:
x=r cos (Ѳ)
y=r sin (Ѳ)
z=z
(r,Ѳ,z)=(x,y,z)
dxdydz=rdrdѲdz
(ρ,ϕ,Ѳ)=(x,y,z)
Spherical coordinates:
x=ρsin (ϕ) cos (Ѳ)
y=ρsin (ϕ) sin (Ѳ)
z=ρcos (ϕ)
dxdydz=ρ2sin(∅)dρdϕdѲ

18. Change of Variable: v y x=x(u,v) y=y(u,v) dv du u x
What is dxdy? (What is the magic factor?)
𝝏𝒙 𝒅𝒖 𝝏𝒙 𝒅𝒗
𝝏𝒚 𝒅𝒖 𝝏𝒚 𝒅𝒗
dxdy = dudv = J(u,v) dudv
𝑥1.𝑦1 𝑥2.𝑦2 𝑓 𝑥,𝑦 𝑑𝑥𝑑𝑦= 𝑢1,𝑣1 𝑢2,𝑣2 𝑓 𝑥 𝑢,𝑣 ,𝑦 𝑢,𝑣 𝐽 𝑢,𝑣 𝑑𝑢 𝑑𝑣

19. Chapter 16: Integration and the fundamental theorem of calculusy
Line integrals of a scalar function f:R2 to R
Curve C=(x(t),y(t))
t=b
t=a
f(x,y) defined in
input space x
𝐶 𝑓 𝑥,𝑦 𝑑𝑠=
𝑓 𝑥,𝑦 𝝏𝒙 𝒅𝒕 2+ 𝝏𝒚 𝒅𝒕 2 (1/2)𝑑𝑡 z
Line integrals of a scalar function f:R3 to R
Curve C=(x(t),y(t),z(t))
t=b y
t=a
f(x,y,z) defined in
input space x
𝐶 𝑓 𝑥,𝑦,𝑧 𝑑𝑠=
𝑓 𝑥,𝑦,𝑧 𝝏𝒙 𝒅𝒕 2+ 𝝏𝒚 𝒅𝒕 2+ 𝝏𝒛 𝒅𝒕 2 (1/2)𝑑𝑡

20. Chapter 16: Integration and the fundamental theorem of calculusy
F(x,y)=(P(x,y),Q(x,y))
Line integrals over a vector field from R2 to R2
Curve C=(x(t),y(t))
t=b
t=a x y
Collect all the tangential components 𝝉 of F along C
t=b
t=a
𝑐 𝑷𝒅𝒙+𝑸𝒅𝒚
𝐶 𝐹 𝑑τ= . x
3D: Collect all the tangential components 𝝉 of F along C
𝑐 𝑷𝒅𝒙+𝑸𝒅𝒚+𝑹𝒅𝒛
𝐶 𝐹 𝑑τ= .

21. Chapter 16: Integration and the fundamental theorem of calculus
Suppose that the vector field F comes from a gradient of a scalar field f(x,y,z): F =
𝛁𝐟(𝐱,𝐲,𝐳)
The fundamental theorem of calculus for line integrals: z
Curve C=(x(t),y(t),z(t))
t=b y
t=a
F =𝛁𝐟(𝐱,𝐲,𝐳) x
𝐶 𝑑𝒓= 𝒇(𝒙 𝒃 ,𝒚 𝒃 ,𝒛 𝒃) −𝒇(𝒙 𝒂 ,𝒚 𝒂 ,𝒛 𝒂 ) .
𝛁𝐟(𝐱,𝐲,𝐳)
f(x(b),y(b),z(b)) – f(x(a),y(a),z(a))
= f(end) – f (beginning)

22. Curl and Divergence
Curl
Let F(x,y,z) be a vector field =(P(x,y,z),Q(x,y,z),R(x,y,z))
We can define the “curl” of F:
𝛁 𝐗 𝐅 which represents the amount of "twist"
𝝏𝑹 𝒅𝒚
𝝏𝑸 𝒅𝒛
𝝏𝑷 𝒅𝒛
𝝏𝑹 𝒅𝒙
𝝏𝑸 𝒅𝒙
𝝏𝑷 𝒅𝒚
𝛁 𝐗 𝐅= - , - , -
Example: Imagine 2D: F(x,y)=(P(x,y),Q(x,y))
Then the only non-zero term is the last one:
Which represents the “twist” on a box
𝝏𝑸 𝒅𝒙
𝝏𝑷 𝒅𝒚 -
𝝏𝑸 𝒅𝒙
𝝏𝑷 𝒅𝒚
= minus
= minus

23. Curl and Divergence
Divergence
Let F(x,y,z) be a vector field =(P(x,y,z),Q(x,y,z),R(x,y,z))
We can define the “div” of F:
𝛁 𝐅 which represents the amount of expansion" . .
𝝏𝑷 𝒅𝒙
𝝏𝑸 𝒅𝒚
𝝏𝑹 𝒅𝒛
𝛁 𝐅= . + +
Example: Imagine 2D: F(x,y)=(P(x,y),Q(x,y))
Then the divergence is the expansion (or contraction):
𝝏𝑷 𝒅𝒙
𝝏𝑸 𝒅𝒚 +
𝝏𝑷 𝒅𝒙
𝝏𝑸 𝒅𝒚
= minus
= minus

24. The Equivalence Loop:
(Caution: there are some caveats having to do with smoothness, connectivity of regions, etc. Check them!)
(1) The vector field F =P(x,y,z),Q(x,y,z),R(x,y,z) is the gradient of a scalar field f(x,y,z)
(2) The vector field F is conservative
𝐶 𝐹 𝑑𝑟=0 .
(3)
around any closed path C
𝐶 𝐹 𝑑𝑟 .
(4)
is path independent
(5)
𝛁 X F = 0

25. Green’s Theorem: Two versions
Curve C
(1) Collecting all the tangential components of F=(P,Q)
D=interior
𝐶 𝐹 𝑑τ=
𝑐 𝑷𝒅𝒙+𝑸𝒅𝒚 = 𝐷
𝝏𝑸 𝒅𝒙
𝝏𝑷 𝒅𝒚 . - dA
= 𝐷
𝛁 𝐗 𝐅 .
k dA
(2) Collecting all the normal components of F=(P,Q)
𝑐 −𝑸𝒅𝒙+𝑷𝒅𝒚
𝐶 𝐹 𝑑𝒏= .
= 𝐷
Div 𝐅 dA