Multivariable Calculus, 14.4 Curvature, Math 14.2-14.5, Math new material 15.5-15.8

Kepler's Laws of Planetary Motion r ''(t) =

( - k / ‖r(t)‖² ) e(r)

Unit vector v

( 1 / ‖v‖ ) v

tangential component aT(T) =

( a · v / v · v ) v

aT(T) =

( a⋅v / v⋅v )v

Component of u along v

( u · v / ‖v‖) = ‖u‖cosθ gives length of the proj

The curvature without κ(s) is if r(t) is an arc length parameterization

(1 / v(t)) ‖ T'(t) ‖

∫ du / (a² + u²)

(1/a) tan⁻¹ (u/a) + K

Sphere of radius R and center (a,b,c)

(x-a)² + (y-b)² + (z-c)² = R²

Cylinder of radius R with vertical axis (a,b,0)

(x-a)² + (y-b)² = R²

Pulling out scalars with dot product

(λv) · w = v · (λw) = λ(v · w)

d/dt arccos(x)

- 1 / √(1 - x²)

∂z/∂x =

- Fx / Fz

∂z/∂y =

- Fy / Fz

∫ csc²(x)

- cotx + c

d/dt csc(x)

- csc(x)cot(x)

d/dt cot(x)

- csc²(x)

Is f(x,y) differentiable at (a,b)?

- partial derivatives exist and are continous - f(x,y) = L(x,y) + e(x,y) where e(x,y) is defined by: lim (a,b --> x,y) e(x,y) / √(x-a)² + (y-b)² = 0 (local linearity)

v x w =

- w x v

d/dt cos(x)

-sin(x)

d/dx cos(x) =

-sin(x)

i x i = j x j = k x k =

0

i · j =

0

i · j = i · k = j · k =

0

v x v =

0

i · i =

1

i · i = j · j = k · k =

1

sec(x) =

1 / cos(x)

csc(x) =

1 / sin(x)

d/dt arctan(x)

1 / x² + 1

Radius =

1 / κ(x) --- The radius of curvature is the reciprocal of the curvature

d/dt arcsin(x)

1 / √(1 - x²)

n = <1,3,2> P₀=(4,-1,1) What is the scalar equation of the plane?

1(x-4) + 3(y+1) + 2(z-1) = 0

find the intersection of two planes x + y + z + 1 = 0 and x + 2y + 3z + 4 = 0

1. set equal to each other y = -2z - 3 set z = t find others. Need a parameterization ! equation for a line in 3D is a plane, not a line.

Analyze the critical points of this function.

1. set the partial derivatives equal to 0 2. calculate fxx, fyy, fyx 3. plug in, find D

Find global extremum

1. set the partial derivatives equal to 0 2. check the boundary

Kepler's A =

1/2 JT (π√p)a^(3/2)

(v · v)'

2v' · v

Airplane ‖u‖ = ‖v + w‖

40<cos0, sin0> + 200<cos(π/4), sin(π/4)> ‖u‖=√

a(t) =

< 0, -g>

The gradient of a function f is the vector of partial derivatives

< ∂f / ∂x , ∂f / ∂y > < ∂f / ∂x , ∂f / ∂y , ∂f / ∂z >

Chain rule ∂f / ∂θ

= x ∂f/∂y - y ∂f/∂x

Area under a curve

A = ₀∫¹ y(t) x'(t) dt

Do these four points lie on the same plane? P(2,4,4) Q(3,1,6) R(2,8,0) S(7,2,1)

Check if PS · (PR x PQ) = 0 ?

Prove : v · w = ‖v‖ ‖w‖ cosθ

Cosines: ‖v-w‖² = ‖v‖² + ‖w‖² - 2cosθ‖v‖‖w‖ ‖v-w‖² = (v · v) - 2(v · w) + (w · w) = ‖v‖² + ‖w‖² - 2(v · w) ‖v‖² + ‖w‖² - 2(v · w) = ‖v‖² + ‖w‖² - 2cosθ ‖v‖ ‖w‖ - 2(v · w) = - 2cosθ ‖v‖ ‖w‖ v · w = cosθ ‖v‖ ‖w‖

Calculate the directional derivative in the direction of v at the given point. Du f(P) "as we move one unit the direction v, ___ will increase by Du f(P)"

Du f(P) = ∇f · u 1. normalize the directional vector

Parameterization of tangent line at r(t₀) in terms of x,y

Find: L(t) = r(t₀) + t r'(t₀) Set: x = ___ t y = ___ t Eliminate t to get an equation

T, B, and N make up the

Frenet frame

If a function of two variables f is continuous at (a, b) and a function of one variable G is continuous at c = f(a, b), then the composite function ____ is continuous at (a, b).

G(f(x, y))

Kepler's k =

GM

How many different direction vectors does a line have?

Infinitely many direction vectors

Kepler's p =

J² / k

Parameterization of tangent line at or r(t) at t=___

L(t) = r(t₀) + t r'(t₀)

Tangent line at r(t₀)

L(t) = r(t₀) + t r'(t₀)

normal vector

N(t) = T'(t)

Unit normal vector

N(t) = T'(t) / ‖T'(t)‖

Normal vector:

N(t) = T'(t) / ‖T'(t)‖ T(t)* = r'(t) / ‖r'(t)‖

Is ‖u + v‖ ≥ ‖u‖ + ‖v‖ ?

No, ‖u + v‖ ≤ ‖u‖ + ‖v‖

Compute the curvature of a circle of radius R. A circle has a parameterization of : Arc length parameterization of:

Parameterization: r(θ) = 〈R cos θ, R sin θ〉 Arc length parameterization: s(θ) = ∫(θ, 0) ‖r'(u)‖ du = Rθ s = Rθ θ = s/R r₁(s) = 〈R cos (s/R), R sin (s/R)〉 T(s) = dr₁/ds = 〈-sin(s/R), cos(s/R)〉 dT/ds = -(1/R)〈cos(s/R), sin(s/R)〉 κ(s) = ‖dT/ds‖ = 1/R

Determine the radius, center, and plane containing the circle. r(t) = 〈sin t, 0, 4 + cos t〉

Radius 1, center (0, 0, 4), xz-plane

If R₁(t) and R₂(t) are differentiable and R₁'(t) = R₂'(t)

R₁(t) = R₂(t) + c

Binormal vector = B(t) =

T(t) x N(t)

Show that the lines r(t) = <-1,2,2> + t<4,-2,1> and r2(t) = <0,1,1> + t<2,0,1> do not intersect.

They intersect only if: <-1,2,2> + t₁<4,-2,1> = <0,1,1> + t₂<2,0,1> x = -1+4t₁ = 2t₂ t₁ and t₂ don't work for x,y,z

Do these equations parameterize the same line? r1(t) = (3, −1, 4) + t<8, 12, −6> r2(t) = (11, 11, −2) + t<4, 6, −3>

Two lines in R3 coincide (are the same) if they are parallel and pass through a common point. thus, v = λ(w) and same t must work in parametric equations

projection of u onto v (written projv U)

U‖v = ( u · v / ‖v‖²) * v or ( u · v / v · v ) v

A domain D is bounded if it is contained in some way

You can put a fence around a bounded function

(d/dt)(r₁(t) x r₂(t))

[r₁'(t) x r₂(t)] + [r₁(t) x r₂'(t)]

aN(N) =

a - aT(T) a - ( a⋅v / v⋅v )v

normal component aN(N) =

a - aT(T) = a - ( a · v / v · v ) v

tangential component aT =

a · T = a · v / ‖v‖

Acceleration :

a(t) = v''(t)

Equation of a plane through P₀ = (x₀, y₀, z₀) and n=<a,b,c> scalar form

a(x-x₀) + b(y-y₀) + c(z-z₀) = 0

aN(N) --> N

aN = ‖aN(N)‖ N = aN(N) / aN

a =

aT(T) + aN(N) tangential component + normal component

a^x

a^x * ln(a)

A domain D is called closed if D contains all its boundary points

all boundary points

(c r(t) )'

c r'(t)

Find parametric equations for a circle of radius 4 and center (3,9)

c(t) = (3 + 4cost, 9 + 4sint)

d/dt sin(x)

cos(x)

d/dx sin(x) =

cos(x)

d/dt sinh(x)

cosh(x)

hyperbolic functions

cosh² - sinh² = 1

∫ - 1 / √(a² - u²)

cos⁻¹ (u/a) + K

Chain rule for Paths Rate of change in temperate along a path c(t)

d/dt f(r(t)) = ∇f r(t) · r'(t) 1. x and y in terms of temperature from r(t) 2. Calculate ∇f and make all t ^ 3. Calculate r'(t) 4. Plug given t into all 5. Take dot product

Kepler's 2nd Law. Covers equal area in equal times. Area dA/dt =

dA/dt = dA/dθ dθ/dt = (1/2) r(t)² θ'(t) = ‖ J ‖ by fundamental theorem of calculus: dA/dθ = 1/2 r²

Speed at time t

ds/dt = ‖ r'(t) ‖ or √x '(t)² + y '(t)²

Find the maximum height of the rocket.

dy / dt, solve for t plug t into original y

If D < 0 then

f has a saddle point at f(a,b)

(d/dt)(f(t) * r(t)) =

f'(t)r(t) + f(t)r'(t)

(d/dt)(f(t) r(t) ) =

f'(t)r(t) + f(t)r'(t)

Linear approximation / linearization

f(a + ∆x, b + ∆y) = f(a,b) + fx(a,b)∆x + fy(a,b)∆y

fx =

f(a+h, b+k) - (a,b) / .1 (or whatever distance between x's are

fy =

f(a+h, b+k) - (a,b) / .1 (or whatever distance between y's are

If D > 0 and fxx(a,b) < 0 then

f(a,b) is a local maximum

If D > 0 and fxx(a,b) > 0 then

f(a,b) is a local minimum

u x (v x w) = (u x v) x w ?

false

is √x²+y² differentiable?

fx and fy continuous at all except 0,0 so it is differentiable everywhere except at the origin.

We say that P = (a, b) is a critical point of f(x, y) if

fx(a, b) = 0 or fx(a, b) does not exist, and fy(a, b) = 0 or fy(a, b) does not exist.

Fermat's Theorem: If f(x, y) has a local minimum or maximum at P = (a, b), then (a, b) is a critical point of f(x, y). A point is a critical point if

fx(a,b) = 0 and fy(a,b) = 0

∆f =

fx(a,b)∆x + fy(a,b)∆y

lagrange -- subject to a constraint (or two!) g(x, y) = 0 occur at critical points P when

fx(x,y) = λgx(x,y) and fy(x,y) = λgy(x,y) two constraints! ∇f = λ∇g + µ∇h curves g(x, y, z) = 0, h(x, y, z) = 0

Show that f(x, y) = 5x + 4y² is differentiable in one way

fx(x,y) and fy(x,y) exist and are continuous

D = D(a,b) = D is for discriminant of f(x,y) at P=(a,b)

fxx(a,b)fyy(a,b) - f²xy (a,b)

slope of tangent line

g'(t) / f'(t) y = g'(t) x = f'(t)

Product rule derivatives D(f(x) * g(x) )

g(x) * f'(x) + g'(x) * f(x)

Quotient rule derivatives D(f(x) / g(x) )

g(x) f'(x) - g'(x)f(x) / (g(x))^2 f'(x)*g(x) - f(x)*g'(x) / (g(x))^2

i x j = j x k = k x i =

i x j = k , j x i = -k j x k = i k x i = j

because i =〈1, 0, 0〉, j =〈0, 1, 0〉, and k =〈0, 0, 1〉

i · j = i · k = j · k and i · i = j · j = k · k

When is v x w = 0?

if v and w are parallel

If an object's acceleration is given, we can solve for v(t) and r(t) by

integrating twice v(t) = ∫ a(t) --> + c₀ r(t) = ∫₀t v(t) --> + c₁

lim(x,y → P) k f(x,y) =

k lim(x,y → P) f(x,y)

aN =

k(t) u(t)² a ⋅ N √‖a‖² - |aT|²

Prove that v · w = ‖ v ‖‖ w ‖cosθ

law of cosines: c² = a² - b² - 2abcosθ → ‖ v-w ‖² = ‖v‖² + ‖w‖² - 2‖v‖‖w‖θ

Local linearity f(x,y) = L(x,y) + e(x,y) where e(x,y) is defined by:

lim (a,b --> x,y) e(x,y) / √(x-a)² + (y-b)² = 0

limit r(t) = <x,y,z> as t→t₀

lim(t→t₀) r(t) = <lim(t→t₀) x(t) , lim(t→t₀) y(t) , lim(t→t₀) z(t) >

A vector valued function r(t) approaches the limit u (a vector) as t approaches t₀ if, In this case,

lim(t→t₀) r(t) = u

lim(x,y → P) f(x,y) * g(x,y) ) =

lim(x,y → P) f(x,y) * lim(x,y → P) g(x,y) ) =

If lim(x,y → P) g(x,y) ≠ 0, then lim(x,y → P) ( f(x,y) / g(x,y) )

lim(x,y → P) f(x,y) / lim(x,y → P) g(x,y)

lim(x,y → P) ( f(x,y) + g(x,y) ) =

lim(x,y → P)f(x,y) + lim(x,y --> P)g(x,y)

A function ƒ of two variables is continuous at P = (a, b) if

lim(x,y → a,b) f(x,y) = f(a,b)

∫ (1 / x)

ln |x|

Equation of a plane through three points. P=(1,0,-1) Q=(2,2,1) R=(4,1,2)

n = PQ x PR d = n · OP 4x + 3y - 5z = 9

Equation of a plane through P₀ = (x₀, y₀, z₀) and n=<a,b,c> vector form

n · <x,y,z> = d where d = ax₀ + by₀ + cz₀

The *osculating plane* at a point P on a curve is the plane through P determined by the vectors TP and NP. It has normal vector BP. It is defined only if the curvature κP at P is ____.

nonzero

T′(t) and T(t) are ___

orthogonal

natural log (ln) must be ____ and cannot be

positive, 0 or negative

Acceleration of a particle along path r(t) a(t) =

r ''(t)

Velocity of a particle along path r(t) v(t) =

r '(t)

Compute the tangent vector at t = 1 r(t) = <1 - t², t>

r '(t) = < -2t, 1 > r '(t) = <-2, 1> --> draw from y = 1 to here

Find the velocity of the particle at time t =4

r'( 4) = r'(t) = v(t) is velocity at time t

(d/dt)(r(g(t)))

r'(g(t))g'(t)

Velocity :

r'(t)

Unit tangent vector T(t)

r'(t) / ‖r'(t)‖

Show that a line has no curvature r(t) = 〈x₀, y₀, z₀〉 + tu and ‖u‖ = 1

r'(t) = u , then , ‖r'(t)‖ = ‖u‖ = 1 T(t) = r'(t) / ‖r'(t)‖ → r'(t) T′(t) = r″(t) = 0 because r'(t) = u is a constant!

e(r) = e(r) is the unit vector that points to the planet as it moves around the sun (at origin)

r(t) / ‖ r(t) ‖

Parameterization of line that passes through two points Midpoint occurs at: Or midpoint formula:

r(t) = (1-t)OP + t OQ t = 1/2 (x + x0 / 2, y + y0 / 2, z + z0 / 2) where x and x0 are x values of each end point

Parameterization of the line perpendicular to the xz-plane through P=(4,9,8)

r(t) = <4,9,8> + t<0,1,0>

Vector parameterization equation

r(t) = P₀ + vt

vector parameterization Given a position vector (direction) or P and a point, find the equation of the line passing through that point, parallel to position vector.

r(t) = r₀ + tv = <x₀, y₀, z₀> + t <a, b, c> r(t) describes a vector whose terminal point traces out a line from -∞ to ∞

Kepler's 2nd Law. Covers equal area in equal times. J =

r(t) x r'(t) where J is a constant: AKA dJ/dt = 0

Equation of tangent line at t₀

r(t₀) + t r'(t₀)

Sum rule (r₁(t) + r₂(t))' =

r₁'(t) + r₂'(t)

(d/dt)(r₁(t) · r₂(t))

r₁'(t) · r₂(t) + r₁(t) · r₂'(t)

an arc length parametrization, where t = g⁻¹(s) is the inverse of the arc length function s = g(t)

r₁(s) = r(g⁻¹(s))

Find the arc length parameterization:

s = ∫ ‖r'(t)‖ <- arc length function s = __ t turn into t = __ s plug into r(t)

d/dt sec(x)

sec(x)tan(x)

d/dt tan(x)

sec²(x)

Find the points of collision

set equal to each other

Find the points of intersection

set equal to each other but use t for one and s for the other

Express cos in terms of sin and cos

sin(t) = √(1-cos²t)

d/dt cosh(x)

sinh(x)

1 - cost / 2 =

sin²t

∫ 1 / √(a² - u²)

sin⁻¹ (u/a) + K

a =

tangential component of acceleration + normal component of acceleration aT(T) + aN(N)

∫ sec²(x)

tanx + c

By theorem, r(t) is only continuous at t₀ only if

the components x(t), y(t), z(t) are continuous at t₀

Multiply the vectors with

the dot product

What is J? The center that r and r'(t) rotate around. Orthogonal to both! Meaning:

the motion of a planet around the Sun takes place on a plane

aN is

the normal component of acceleration. aN = k(t)*u(t)² describes the change in v due to a change in direction.

aT is

the tangential component of acceleration. aT = v'(t) is the rate at which the speed v(t) changes

If D = 0 then

the test is inconclusive

u x v is perpendicular to 10u + 22v

true

v · w = w · v ?

true

Decomposition of u

u = projvU + u _|_ v

If u =λv then

u and v are parallel

Distributive law with dot product

u · (v + w) = u · v + u · w

component of u along v

u · v / ‖v‖

aT =

u'(t) a ⋅ T a⋅v / ‖v‖

v(t) =

u(t) * <cos_ , sin_ >,

Speed :

u(t) = ‖v(t)‖

projvV

v

Normal vector for the curve between intersecting objects F and G

v = ∇Fp x ∇Gp

Angle between v and u is obtuse if

v · u < 0

Two vectors v and w are perpendicular if

v · w = 0

v · w = ‖v‖ ‖w‖ cosθ = 0

v · w = 0 , the orthogonal

Given N(t) = T'(t) / ‖T'(t)‖ unit normal vector: find ||T′(t)|| = T′(t) =

v(t)κ(t) v(t)κ(t)N(t)

v x w = 0 if

w = λv

Find the parameterization c(t) x² + y² = 4 c(0) = (1, √3)

x = 2cos(θ + w) → 1 = 2cos(0 + w) y = 2sin(θ + w) → √3 = 2sin(0 + w)

cycloid

x = Rt - R sin t y = R - Rcost

parameterize the line through P=(a,b) with slope m

x = a + t y = b + mt

Find the parameterization of the line through P=(a,b) Q=(c,d)

x = a + t(c-a) y = b + t(d-b)

quadratic formula

x = b ± √(b² - 4ac) / 2a

Parametric equations

x = x₀ + at y = y₀ + at z = z₀ + at

∫ ln(x)dx

x ∙ (ln(x) - 1) + C

parameterizing a cycloid with radius R

x(t) = R ( t - sin(t)) y(t) = R ( 1 - cos(t))

Find the parameterization c(t) y = x² c(0) = (3,9)

x(t) = a + t → x(0) = a + 0 → 3 = a + 0 → a=3 x(t) = 3+t y = (3 + t)² (3+t, 9+6t+t^2)

Find a parameterization c(t) for y = 3x-4 c(0) = (2,2)

x(t) = t + a x = t + 2 y = 3(t+2) - 4 (t+2, 3t+2)

Ellipsoid

x² + y² + z² = 1 horizontal trace: ellipse vertical traces: ellipses

Elliptic cylinder

x² + y² = 1 horizontal trace: ellipse vertical traces: vertical lines

Elliptic cone

x² + y² = z² horizontal trace: ellipse vertical traces: hyperbolas

One sheeted hyperboloid

x² + y² = z² + 1 horizontal trace: ellipse vertical traces: hyperbolas

Two sheeted hyperboloid

x² + y² = z² - 1 horizontal trace: ellipse vertical traces: hyperbolas

Equation for tangent vector c(t) = <x, y>

y - y(t) = slope (x - x(t) ) slope = dy/dx

Parabolic cylinder

y = ax² horizontal trace: parabola vertical traces: 1-2 vertical lines

Hyperbolic cylinder

y² - x² = 1 horizontal trace: hyberboloid vertical traces: vertical lines

Equation of tangent plane

z = f(a,b) + fx(a,b)(x-a) + fy(a,b)(y-b)

Elliptic paraboloid

z = x² + y² horizontal trace: ellipse vertical traces: parabolas

Hyperbolic paraboloid

z = x² - y² or z = y² - x² POSITIVE determines which it "covered-wagons" horizontal trace: hyperbola vertical traces: parabolas

distance between a point and a plane *

| PQ x n | / ‖n‖ or? | d | / √(a² + b² + c²) (a(xo) + b(yo) + c(zo) + d) PQ is vector between point and point on line n is normal vector to the plane

Distance from P to the plane

| d | / √(a² + b² + c²) | d | = (a(xo) + b(yo) + c(zo) + d) (d is -d when it is on left side) P=(xo,yo,zo) Ex. P=(4,-4,3) 2x-2y+5z = -8 d = | 2(4) -2(-4) + 5(3) + 8 | / √(2² + (-2)² + 5²) = 6.8

distance between two planes *

| d | / √(a² + b² + c²) | d | = (a(xo) + b(yo) + c(zo) + d) (d is -d when it is on left side) P=(xo,yo,zo) on one plane <a,b,c> and d on the other

Assuming f(x, y) is defined near P = (a,b), then lim(x,y → P) f(x,y) = L if for any ε > 0 there exists δ > 0 such that if (x,y) satisfies 0 < d((x,y), (a,b)) < δ, then

| f(x, y) - L | < ε

Volume of a parallelepiped spanned by u, v and w

| u · (v x w)|

Another equation for curvature

| x'(t)y''(t) x x''(t)y'(t) | / (x'(t)² + y'(t)²)^(3/2)

Curvature of a Graph in the Plane κ(x) =

| ƒ ''(x) | / (1 + ƒ '(x)² )^(3/2)

∇fP points in the direction of maximum rate of increase. The maximum rate of increase is

||∇fP||

λv has a length of

|λ| ‖v‖ v if λ>0 , -v if λ<0

Curvature if r(t) is NOT an arc length parameterization

κ(s) = ( 1 / ‖r'(t)‖ )* ‖dT/dt‖ where u(t) = ‖r'(t)‖

The curvature at κ(s) is if r(s) is an arc length parameterization

κ(s) = ‖ dT / ds ‖

normal component aN =

κ(t)v(t)² ‖ a x v ‖ / ‖v‖ !!!!!

The curvature at a point on a graph y = f(x) in the plane is

κ(x) = |ƒ ''(x)| / (1 + ƒ '(x)² )^(3/2)

Two nonzero vectors v and w are called perpendicular or orthogonal if the angle between them is

π/2

distance between a line and a line (maybe not needed since they would be parallel

‖ PQ x b ‖ / ‖b‖ PQ is vector between point on each line b is the cross product between the two direction vectors

distance between a point and a line *

‖ PQ x n ‖ / ‖n‖ PQ is vector between point on each line n is direction vector of the line

Curvature: r(s) is an an *arc length parametrization* and T the unit tangent vector. k(s) =

‖ dT / ds ‖ T(t) = r'(t) / ‖r'(t)‖ arc length parameterization at r(s) if ‖r'(s)‖ = 1

Arc length parameterization

‖ r '(s) ‖ = 1

In practice, we compute curvature using the following formula, which is valid for arbitrary regular parametrization: κ(t) =

‖ r'(t) x r''(t) ‖ / ‖r'(t)‖³

In practice, we compute curvature using the following formula, which is valid for arbitrary regular parametrizations:

‖ r'(t) x r''(t) ‖ / ‖r'(t)‖³

What are the components of u that makes an angle of 20 below the x-axis ?

‖ u ‖<cos(-20°), sin(-20°)>

What are the components of u that makes an angle of 45° with the x-axis?

‖ u ‖<cos(45°), sin(45°)>

Area of a parallelogram spanned by v and w

‖ v x w ‖

v x w using geometric properties

‖ v x w ‖ = ‖ v ‖‖ w ‖ sinθ

Speed of a particle along path r(t) u(t) =

‖ v(t) ‖

compvV

‖v‖

Proof for distance equation

‖v‖ = ( |d| / (n · n) )‖n‖ = |d|/‖n‖ = |d| / √(a² + b² + c²)

Unit vector

‖v‖ ev = ‖v‖<cosθ, sinθ>

v · w =

‖v‖ ‖w‖ cosθ

Second, the dot product of a vector with itself is: v · v:

‖v‖²

v · v =

‖v‖²

law of cosines ‖v-w‖² =

‖v‖² + ‖w‖² - 2cosθ‖v‖‖w‖

‖ v x w ‖² =

‖v‖² ‖w‖² - (v · w)² used to prove: ||v × w|| = ||v|| ||w|| sin θ

‖v x w‖ =

‖v‖‖w‖sinθ

Du f(P) = ∇f · u related to angle! ∇f and u

‖∇f ‖cosθ (cause ‖u‖ = 1)

Average rate of change from P to Q

∆ altitude / ∆horizontal

percentage increase in V

∆V / V * 100 = 2*∆r / r * 100 + ∆h/h * 100 ∆r / r = 2

Chain Rule for Paths

∇f r(t) · r'(t)

−∇fP points in the direction of maximum rate of decrease. The maximum rate of decrease is −||∇fP||

−||∇fP||

Length/magnitude of v = ‖PQ‖

√(a₂-a₁)² + (b₂-b₁)² + (c₂-c₁)²

‖v + w‖ =

√(v+w) · (v+w)

*Length of a Path* s =

∫(b, a) ‖ r '(t) ‖ dt = ∫(b, a) √x '(t)² + y '(t)² + z '(t)² dt

s(t) =

∫(b, a) ‖r'(t)‖ dt