Calc 3

Distance formula in 3D

√(x₂-x₁)²+(y₂-y₁)²+(z₂-z₁)²

Equation of a sphere

(x-h)²+(y-k)²+(z-l)²=r²

ax²+bx+cy²+dy+ez²+fz=C is a…

sphere (complete the circle to transform into standard form)

y = f(x), z=f(y), or z=f(x) is a…

cylinder (any shape with no restriction on a parameter)

it will be centered around the parameter with no restriction

ax+by+cz=d is a…

plane

x²/a² + y²/b² + z²/c² = 1 is a…

elipsoid

z/c = x²/a + y²/b + d is a…

eliptic paraboloid

z/c = x²/a² - y²/b² is a…

hyperbolic paraboloid

x²/a² + y²/b² - z²/c² = 1 is a…

Hyperboloid of one sheet

-x²/a² -y²/b² + z²/c² = 1 is a…

Hyperboloid of two sheets

z²/c² = x²/a² + y²/b² is a…

cone

trace

curve on an intersection of a plane

3D shape where all traces are elipses

elipsoid

3D shape where horizontal traces are elipses and vertical traces are parabolas

eliptic paraboloid

3D shape where horizontal traces are hyperbolas and vertical traces are parabolas

hyperbolic paraboloid

3D shape where horizontal traces are elipses, vertical traces are hyperbolas, has a region in the middle that comes to a point/DNE

cone

3D shape where horizontal traces are elipses and vertical traces are hyperbolas

hyperbeloid of one sheet

3D shape where horizontal traces are elipses and vertical traces are hyperbolas but there is a region which DNE which splits the shape in two

hyperboloid of two sheets

Equation of a line

x = at = x₀

y = bt + y₀

z = ct + z₀

where (x₀,y₀,z₀) is a point on the line and

<a,b,c> is the direction vector of the line

Equation of a plane

a(x-x₀) + b(y-y₀) + c(z-z₀) = 0

where <a,b,c> is the normal vector (vector perpendicular to the plane)

distance from a point to a plane

let b = P₁P₀ = <x₁ - x₀, y₁ - y₀, z₁ - z₀>

and n be the normal vector to the plane

D = comp_nb = |n * b|/|n|

cross product gives

a vector which is orthogonal to both a and b

magnitude of a cross product |u x v| represents

the area of the parallelogram which is the result of adding vectors u and v

(can also represent torque)

(1/2)|u x v| represents…

the area of a triangle

triple product a * |b x c| represents

The volume of a parallelepiped

two vectors are coplanar (lie on the same plane) if and only if

their triple product a * | b x c | = 0

(1/6)( a * |b x c|) represents

the volume of a tetrahedron

dot product gives

a scalar which hints how similar the angles of two vectors are

0 is orthogonal, negative is obtuse, and positive is acute

component formula of cross product

a x b = <a₂b₃ - a₃b₂, a₃b₁ - a₁b₃, a₁b₂ - a₂b₁>

component formula of dot product

a₁b₁ + a₂b₂ + a₃b₃

magnitude/angle formula of dot product

a * b = |a||b|cosθ

magnitude/angle formula of cross product

|a x b| = |a||b|sinθ

two vectors are orthogonal if and only if

their dot product is 0

two vectors are parallel if

the ratios of their components are the same

two vectors are parallel if and only if

a x b = 0

projection of b onto a represents

the component of b that is going the same direction as a

scalar projection of b onto a

comp_ab = a*b/|a|

vector projection of b onto a

proj_ab = (a*b/|a|²)a

unit vector represents

vector going in the same direction as another but with magnitude 1

unit vector formula

a/|a|

direction angle

the angle between v = <α,β,γ> and the x,y, and z axes

vector value function

function whose domain is a set of real numbers and whose range is a vector

eg: r = <f(t), g(t), h(t)>

*domain is where f,g,h are all defined

space curve

the line C reated by parametric equations x = f(t), y = g(t), z = h(t)

derivative of a vector function r

r’(t) = <f’(t), g’(t), h’(t)>

unit Tangent vector T

T(t) = r’(t)/|r’(t)|

if |r’(t)| = c, then

r’(t) is orthogonal to r(t) for all t

vector derivative addition rule

d/dt (u + v) = u’ + v’

vector derivative constant multiple rule

d/dt (cu) = cu’

function vector product rule

d/dt (fu) = f’u + fu’

vector dot product derivative rule

d/dt (u * v) = u’ * v + u * v’

vector cross product derivative rule

d/dt(u x v) = u’ x v + u x v’

vector derivative chain rule

d/dt [u(f(t))] = f’(t)u’(f(t))

integral of a vector value function

∫r(t)dt = <∫f(t)dt, ∫g(t)dt, ∫h(t)dt>

and _a^b∫r(t)dt = R(t) |_a^b

Arc length of a parametric curve

L = _a^b∫|r’(t)|dt

AKA _a^b∫√( (dx/dt)² + (dy/dt)² + (dz/dt)² )dt

where the curve is traversed once and is continuous

Curvature K

curvature is measured by how fast the unit tangent vector changes

K = |T’(t)|/|r’(t)|

unit normal vector

the unit vector orthogonal to the tangent vector T(t)

N(t) = T’(t)/|T’(t)|

Binormal vector

the vector orthogonal to both T(t) and N(t)

B(t) = T(t) x N(t)

Particle motion equations in 3D

r(t) describes the motion of the position of a particle in 3D

v(t) = r’(t)

speed = |v’(t)|

a(t) = v’(t)

Projectile motion equations with initial speed v₀, initial angle α, and initial hight h

g = gravity = 9.8~

a(t) = <0, -g>

v(t) = <v₀cosα, v₀sinα - gt>

r(t) = <v₀cosαt, h + v₀sinαt - (1/2)gt²>

important values for projectile motion

time of impact: t where h + v₀sinαt - (1/2)gt² = 0

total distance: r(t) where t = time of impact (usually just relevant for x value)

speed at impact: |v(t)| for t = time of impact

max height: solve for t where velocity is 0, plug into y part of position vector

function of several variables

z = f(x,y)

{x,y, f(x,y) | (x,y) ∈ D}

where D is the domain where neither x nor y create a nonreal answer or discontinuity

Table of values

when there is no functional equation to describe a phenomenon of multiple variables, we can use a table to estimate values

(ex: windchill index takes in temperature and humidity and is expressed as a table)

contour map

way of representing higher dimensional functions in fewer dimensions by drawing their traces over the top of the domain;

looks like a topology map;

create one by setting k = f(x,y) where k is a constant and drawing it over a graph of the domain

Limits in 3D

let f(x,y) be a two variable function with (a,b) having points in D arbitrarily close

lim f(x,y) = L

.^{(x,y) → (a,b)}

if ∀ ε > 0 there is a corresponding δ > 0

if 0 < √((x-a)² - (y-b)²) < δ then |f(x,y) - L| < ε

How to prove that a limit exists in 3D

whereas in 2D where you had to check that the limit was the same in two directions, in 3D you have to check in an infinite number of directions

to do this you can use:

1) the epsilon-delta proof

2) the squeeze theorem

3) polar coordinates

4) prove that a limit for a generalized form of f(x,y) exists and then prove that any blind spots of that limit also exist (ie: pluggin y=mx into the limit and then also proving that the limit exists on the x axis)

notation for partial derivatives

with respect to x: f_x(x,y) = f_x = ∂f/∂x = ∂/∂x (f(x,y)) = Dx(f(x,y))

what do partial derivatives represent?

the instantaneous rate of change in only one direction for a multi variable (multi directional) function

tangent plane equation

z-z0 = ∂z/∂x(x-x0) + ∂z/∂y(y-y0)

limit defenition of a partial derivative

for x:

f_x = lim [f(a+h,b) - f(a,b)]/h

…^{h → 0}

to find the partial derivative with respect to one variable,

treat the other variables like a constant

for an implicitly defined function of 3 variables, you must use implicit differentiation to find ∂z/∂x by…

1) treating y like a constant

2) differentiating normally with respect to x

3) treating z as an undefined function of x (eg. the derivative of z is not 1, it’s ∂f/∂x)

for f_y and f_z , switch the roles

Clairaut’s theorem

if f is defined on a domain D containing (a,b), then f_xy(a,b) = f_yx(a,b)

AKA the order doesn’t matter for mixed derivatives (except for some special exceptions due to domain restrictions)

formula for implicit differentiation of a multivariable function

dz/dx = -F_x/F_z

Chain rule of a multivariable function

∂z/∂t = (∂z/∂x)(∂x/∂t) + (∂z/∂y)(∂y/∂t)

where z is defined as a function of x and y and x and y are defined by t (and possibly another variable)

Properties of the gradient vector

1) it represents <f_x,f_y>

2) it is always orthogonal to the curve because tangents are orthogonal

3) it is the way of steepest change of a curve

4) the largest possible rate of change for function f at point P is |∇f|

5) the biggest directional derivative is in the direction of u = ∇f/|∇f|

directional derivative formula

D_uf(x,y) = <f_x(x,y), f_y(x,y)> * <a,b>

where u is the unit vector <a,b>

the critical points of f(x,y) occur where…

both f_x = 0 or DNE and f_y = 0 or DNE

local max/min definition

max: f(x,y) >= f(a,b) for all points in some disc with center (a,b)

min: f(x,y) <= f(a,b) for all points in some disc with center (a,b)

saddle point

a critical point which is not an extrema (not a max/min)

second derivatives test for critical points

if the second partial derivatives of f are continous on a disc with center (a,b) and f_x(a,b) = 0 and f_y(a,b) = 0 such that (a,b) is a critical point of f,

let D = f_xx(a,b)f_yy(a,b) - [f_xy(a,b)]²

if:

1) D > 0 and f_xx > 0, then f(a,b) is a local min

2) D > 0 and f_xx < 0, then f(a,b) is a local max

3) D < 0 then f(a,b) is a saddle point

4) D = 0 then we don’t know anything

to find absolute max and min…

recall that if f is a closed bounded set D in ℝ², it has an absolute max and min

1) find the values of the CPs of f on D

2) find the extreme values of f on the boundary of D

3) the largest of these values is the max and the smallest is the min

lagrange multiplier

∇f(x₀,y₀,z₀) = λ∇g(x₀,y₀,z₀)

use this to set up a system of equations f_x = λg_x, f_y = λg_y, f_z = λg_z, and g(x,y,z) = k to find the max/min

*only useful for finding min/max on a set level curve, not everywhere

what do double integrals represent?

the volume under the surface z = f(x,y) over a region of ℝ² (which could be a variety of shapes)

AKA the infinite sum of boxes under a curve

Average value of f(x,y) over a region R=[a,b]x[c,d]

1/area(R) ∫_a^b∫_c^d f(x,y) dydx

Fubini’s Theorem

if f is a continous on a rectangle R = {(x,y) | a <= x <= b, c <= y <= d }

then ∫∫_Rf(x,y)dA = ∫_a^b∫_c^df(x,y)dydx = ∫_c^d∫_a^bf(x,y)dxdy

∫∫_Rf(x)g(y)dA =

∫f(x)dx * ∫g(y)dy

this applies to higher dimensions as well

Type I general region integrals

the region D is bounded below by y=g₁(x) and above by y=g₂(x)

the integral is ∫_x1^×2∫_g1(x)^g2(x)f(x,y)dydx

Type 1 general region integrals

the region D is bounded to the left by h₁(y) and to the right by h₂(y)

the integral is ∫_y1^y2∫_h1(x)^h2(x)f(x,y)dxdy

∫∫_DCdA is…

the area of D times the value of the constant C

One strategy for simplifying double integrals is…

Swap the order of dx and dy by reevaluating the bounds of the integral

It may be useful to swap a double integral in rectangular coordinates if…

1) the limits of integration are more easily expressed in polar (ex: circle, washer/ring, rose, cardeoid)

2) the integrand is more easily expressed in polar

∫∫_Df(x,y)dxdy =

where D is expressed in x and y

∫∫_Dr(θ) * rdrdθ

where D is expressed in r and θ

• if there is some sort of ring shape, r goes from the inner to the outer. Otherwise it goes from 0 to the radius.

useful polar conversions to know

y = rsinθ

x = rcosθ

x² + y² = r²

y/x = tanθ

surface area formula

∫∫_D√(1 + (f_x)² + (f_y)²) dA

the integrand of the surface area formula √(1 + (f_x)² (f_y)²) represents…

the magnitude of the gradient vector at any given point (AKA the stretch factor which needs to be taken into account)

notes on triple integral set up ∫∫∫_Ef(x,y,z)dV

dV can be any order of dz dy and dx. You can manually change the limits of integration if you can easily express them in a different order (not always possible bc it might not be x,y, and z simple)

the innermost integral’s limits can be of two variables

the middle integral’s limits can be of one variable

the outer integral’s limits must be constants

∫∫∫_E1dV represents…

the volume of the region E

z simple (can be extrapolated to y simple and x simple)

if you were to draw a line from anywhere on the z axis into the region E, it would always touch the same bounding functions z₁(x,y) and z₂(x,y)

• this are good functions to put as your innermost limits of integration in a triple integral

Cylindrical coordinates

x and y in terms of r and θ and z is the same

polar/cylindrical relationships

x = rcosθ

y = rsinθ

x² + y² = r²

y/x = tanθ

dV = r dzdrdθ