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

Derivatives of parameterized curves y 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) 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

Arc-length of parameterized curves y 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 t0 y t1 ds(t) dy(t)/dt dx(t)/dt t0 x J.A. Sethian/UC Berkeley

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

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

Functions of more than one variable Input space Need an extra dimension to graph it f(x): 1D input to 1D output y=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

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

The Multi-Dimensional Chain Rule Of one variable: t f(t) x g(x) y 𝒅𝒚 𝒅𝒚 = 𝒅𝒚 𝒅𝒙 𝒅𝒙 𝒅𝒕 Of many variables: t 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 𝝏𝒓 = 𝝏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

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

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

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

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

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