Multi 2

graph of x²+y²=1

equation of a sphere with center (h,k,l) and radius r

r² = (x-h)² + (y-k)² + (z-l)²

a · b

|a| |b| cosθ (θ is angle between a and b and is always between 0 and pi)

a · a

|a|²

a · (b + c)

a · b + a · c

(ca) · b

c (a · b) = a · (cb)

a is perpendicular to b if a · b = 0

a · b = |a||b|cosθ

comp ab

scalar projection = signed magnitude of PS

comp ab =

(a · b) / |a| = (|b|cosθ)

vector projection of b onto a

projab = ( (a · b) / (a)² ) a = (|b|cosθ a) / |a|

a x b =

i (a₂b₃ - a₃b₂) - j(a₁b₃ - a₃b₁) + k(a₁b₂-a₂b₁)

a x a =

0 (area of a parllelogram is 0)

a x (ca) =

0

Cross product rules

Cross product it 1) orthogonal to a and b and 2) the length of the resulting vecto is equal to the area of the parallelogram generated by a and b.

a x b =

-(b x a) (right hand rule)

|a x b| =

|a||b|sinθ = area of the parallelogram

a || b iff

a x b = 0 (a,b x= 0)

scalar triple product:

let a, b_, c be vectors in R³. The scalar triple product of a,b,c is a · (b x c). To solve, replace ijk in cross product matrix with a₁, a₂, a₃

Volume of a box =

|a · (bxc)| (STP)

vector equation of a line

r = v₀ (vector from origin to point on a line (x₀, y₀, z₀) + tv (direction vector of a line), t ∈ R

Parametric equations

x = x₀ + at

y = y₀ + bt

z = z₀ + ct

P = (x₀, y₀, z₀),

v = <a,b,c>

Symmetric equation of a line

t = (x-x₀) / a = (y-y₀) / b = (z-z₀) / c. a,b,c x= 0

skew lines

2 different planes, not parallel but don’t intersect. set line coordinates equal to each other and see if they intersect

for a line segment…

give t parameters

how to describe a plane in R³

to describe a plane in R³ we need a point on it and a vector orthogonal to it

Equation of a plane

a(x-x0) + b(y-y0) + c(z-z0) =0 (right before 9/16)

equation of an ellipse

center = (h,k),

Semi-major axis (a): The distance from the center to a vertex along the longer axis. [1]

• Semi-minor axis (b): The distance from the center to a co-vertex along the shorter axis. [1]

• Orientation:

  • If a² is under x², the ellipse is horizontal.

  • If a² is under y², the ellipse is vertical.

Equation of a hyperbola

Horizontal Hyperbola:
(x-h)²/(a²) - (y-k)² / (b²)

Vertical Hyperbola:

(y-k)² / (a²) - (x-h)²/(b²)

a = distance from center to vertices

sphere

ax²+by²+cz² = 1 if a=b=c

r(t) is continuous at t=a if

lim r(t) = r(a)

t→ a

describe r(t) = <cost, sint, t>

unit tangent vector to a curve (T(t)) =

r’(t) / |r’(t)|

r(t) = tan^-1(t); r’(t) =

d/dt [u(t) + v(t)] =

d/dt [u(t) ]+ d/dt [v(t)] =

d/dt (c u(t))

c d/dt u(t)

d/dt [u (f(t))]

u’ (f(t)) * f’(t)

d/dt [u (t) · v(t) ]

u (t) · v(t) + u’(t) · v’(t)

d/dt [u (t) x v(t) ]

u (t) x v(t) + u’(t) x v’(t)

r(t) · r’(t)

= 0

D_{u = <a,b>}f(x,y) =

f_x (x,y)a + f_y(x,y)b

D_u f(x,y) =

gradient f · u = |gradient f| |u| cos ϴ. (ϴ is the angle between gradient and u)

2nd derivative test

fxxfyy-(fxy)²

In calc one local max/min has f’(x) = 0, but converse

isn’t always true.

In calc 1, if 2nd derivative is pos =

local min

In calc 1, if 2nd derivative is neg =

local max

In calc 1, if 2nd derivative is 0 =

test is inconclusive

If the cone's vertex is point (x₀, y₀, z₀), the equation becomes:

[(x-x₀)² / (a²) ] + [(y-y₀)² / b²] = [(z-z₀)² / c²]

• as c gets bigger, cone gets skinnier

• a stretches in x direction, b stretches in y direction

• If (a = b), it is a circular cone.

• If a x= b, it is an elliptic cone (its cross-sections are ellipses instead of circles).

[z/c] = sqrt ([(x-x₀)² / (a²) ] + [(y-y₀)² / b²] )

for 3d, local max has to be

the local max sideways f_y=0, and forward and backwards f_x=0

for 3d, saddle point means

the sideways path has a local max/min (f_y=0) and the forward and backwards path f_x=0 has the opposite

3d local min

the local min sideways (f_y=0), and forward and backwards (f_x=0)

Theorem: if f has a local max/min at (a,b), then

f_x(a,b) = 0 and f_y(a,b) = 0, if they exist. so, if f_x = 0 and f_y = 0, the ∇f = <f_x,f_y> = 0. ∇f = derivative in 2-d setting. Thus, ∇f(a,b) = 0.

f(x,y) is a 2-variable funtion if ∇f(a,b) = <0,0> or ∇f(a,b) DNE.

Then (a,b) is called a critical point, f(a,b) = critical value!

2nd derivative test:

Suppose the second partials of (x,y) are all continuous on a mixed partials are equal disk centered at (a,b) and that f has a critical point (a,b). Let D = |f_xx f_xy|

|f_yx f_{yy| =} f_xxf_{yy =} (f_xy)²

If D(a,b) = 0

Nothing, inconclusive

If D(a,b) < 0

We have a saddle point at (a,b)

If D > 0

a) f_xx (a,b) or f_yy (a,b) > 0 —> local min

b) f_xx (a,b) or f_yy(a,b) < 0 —> local max

When f(x,y) is subject to g(x,y) =k. The largest value that satisfies the requirement is when the level curves of f(x,y) and g(x,y)= k…

intersect tangentially

For Lagrange multiplier og equation

∇f = λ∇g. So, if ∇f = <a,b> and ∇g = <c,d>, then a = λc and b = λd

If you optimize (max/min) f(x,y) subject to a constraint g(x,y) = k, then

∇f = λ∇g, λ is ℝ, ∇g ≠ 0

Rectangle where a<=x<=b and c<=y<=d

R = [a,b] x [c,d]

volume of f(x,y) defined on a rectangle "R”

∫_Rf(x,y)dA

Fubini’s Theorem: If f is continuous on the rectange R = [a,b] x [c,d] then,

∫^b_a ∫_c^d f(x,y)dxdy = ∫∫_R f(x,y) dA

Type 1 region D:

{(x,y) , a ≤ x ≤ b, f₁(x) ≤ y ≤ f₂(x) }

Type 2 region:

{(x,y) , c ≤ y ≤ d, g₁(y) ≤ x ≤ g₂(y) }

D = D₁ U D_2. ∫∫_D f(x,y) dA =

∫∫_D f(x,y) dA = ∫∫_D1 f(x,y) dA + ∫∫_D2 f(x,y) dA

r and θ on a graph

x in polar coordinates

x = r cos θ

y in polar coordinates

y = r sin θ

θ =

tan⁻¹(y/x)

Polar rectangle:

{ (r,θ) | a ≤ r ≤ b, c ≤ θ ≤ d}

dA (polar coordinates) =

lim dr → 0 (dA) = rdθdr

r² in polar coordinates

r² = x²+y²

when do you use polar coordinates

circular regions

Paraboloid at center (h,k,j)

a = stretch along x-axis

b= stretch along y- axis

c = larger z = skinnier parabola

Hyperbolic paraboloid. b = stretch inner parabola towards y, a = gets skinnier, c= gets skinnier

Fubini’s theorem for triple integration

Let f(x,y,z) be a continuous function on B = [a,b] x [c,d] x [r,s]. Then ∭_Bf(x,y,z)dV ∫_c^d∫_r^s∫_a^b f(x,y,z)dxdzdy = …dxdydz = …dzdxdy = dzdydx = dydxdz = dydzdx (6 different ways)

∫∫_D(height)dA =

∫∫∫_E1dv

cyllindrical coordinates

x in cyllindrical

x = rcosθ

y in cylindrical

y = rsinθ

z in cylindrical coordinates

z = z

r = 1

cylinder

θ = pi/4

half of a plane

z = 2

plane

sin²θ

(-1/2) [cos(2θ)-1]

∫∫∫_Ef(x,y,z)dv in cylindrical coordinates

= ∫∫∫_Ef(x,y,z)dz(r)drdθ OR ∫∫∫_Ef(x,y,z) r dzdrdθ

cos(2θ) =

cos²θ + sin²θ

spherical coordinates

ρ = distance from the origin, θ = same θ as in cylindrical coordinates, ϕ = angle with the positive z-axis.

0 ≤ θ ≤ 2pi; 0 ≤ ϕ ≤ pi; 0 ≤ ρ ≤ infinity

z in spherical =

z = ρ cos (ϕ)

x in spherical

x = ρ sin (ϕ) cos θ

y in spherical

y = ρ sin (ϕ) sin θ

spherical dv =

dv = ρ² sin (ϕ) dρ dθ dϕ

ρ = 1 or ρ = c

sphere or radius c

ϕ = pi/4

cone

θ = pi/4

half plane

cylindrical rectangular box =

{(r,θ,z) | a≤ r ≤b, c≤θ≤d, r≤z≤s }

spherical coordinates rectangular box

{ (ρ,θ,ϕ) | a≤ρ≤b, c≤θ≤d, r≤ϕ≤s }

ball

sphere and everything inside

Use spherical if…

your shape looks like a cone, half-plane, or sphere