Multivariable Calc and DE 1A

Semester 1, Year 1

Define Gradients

The gradient measures the rate of change of the function. The gradient of a point on a curve is the gradient of the tangent line at that point.

Define Indefinite Integral

Let f:R→R. The indefinite integral of f(x) is a function g(x) s.t. g^I(x) = f(x). g is the antiderivative: ∫f(x) dx = g(x) + c where c is an arbitrary constant

Define Definite Integral

Let f:R→R. The definite integral [a,b] is the net signed area between f and the horizontal axis on [a,b]. Total area = ^b∫_af(x) dx

The Fundamental Theorem of Calculus

Let f be a continuous function on [a,b] ∈ R

A) f has antiderivative g given by g(x) = ^x∫_af(u) du ∀ x ∈ [a,b]

B) Let h be any antiderivative of f. Then: ^β∫_αf(u) du = h(β)-h(α) = g(β)-g(α) for all a≤α≤β≤b

It follows from A that this means g^I(x) = f(x)

Arc Length

Distance along the curve between a and b. S(a,b) = S(b)-S(a) = ^b∫_a√(1+(dy÷dx)²) dx

Surface area of revolution

Consider the surface created by rotating the curve around the x axis (exclude end faces).

A(a,b) = 2π ^b∫_a y√(1+(dy÷dx)²) dx

arsinhx

ln(x+√(x²+1))

arcoshx

±ln(x+√(x²-1))

For integrals containing (x²+a²)^½

try x = asinhθ as a substitution

For integrals containing (x²-a²)^½

try x = acoshθ as a substitution

Define trace

Curve produced when a surface intersects with a coordinate plane or another plane parallel to a coordinate plane

Cylindrical polar coordinates

P(r, θ, z)

(r, θ): polar coordinates of projection of P onto xy-plane

z: perpendicular distance from xy-plane to point P

Spherical polar coordinates

(ρ, θ, ∅)

ρ: distance from origin (radius)

θ: angle from positive x-axis (azimuth), 0≤θ≤2π

∅: angle from positive z axis to OP (inclination/polar angle), 0≤∅≤π

Converting to spherical

x = ρsin∅cosθ
y = ρsin∅sinθ
z = ρcos∅

ρ²=x²+y²+z²
tanθ=y÷x
cos∅=z÷(√(x²+y²+z²))

The partial derivative of f wrt x

F differentiated wrt x taking y as a constant

Interpretation of the partial derivative of f wrt x

The gradient of the surface in x-direction at (x₀, y₀)

Define f_xy

f partially differentiated first wrt x then wrt y

Clairaut’s theorem

For well-behaved functions, f_xy = f_yx

To find derivatives and partial derivatives, you can…

use chain rule

Define global maximum

A function f has a global maximum at (x₀, y₀) if f(x₀, y₀) ≥ f(x,y) ∀(x,y) ∈ Ω

Define local minimum

A function f has a local minimum at (x₀, y₀) if there is a disc B centred at (x₀, y₀) s.t. f(x₀, y₀) ≤ f(x,y) ∀ (x,y) ∈ B

Define extremum

If f has a local min or max, it has a local extremum

If f has a global min or max, it has a global extremum

Define critical point

(x₀, y₀) is a critical point of f(x,y) if both partial derivatives equal 0 at the point or one or both of them don’t exist at the point

Define saddle point

A critical point that is not an extremum (i.e. a pringle!)

What is the determinant of the Hessian matrix?

D = f_xx(x₀, y₀)f_yy(x₀, y₀) - f_xy(x₀, y₀)²

(x₀, y₀) is a local minimum if…

D > 0 and f_xx(x₀, y₀) > 0

(x₀, y₀) is a local maximum if…

D > 0 and 0 > f_xx(x₀, y₀)

(x₀, y₀) is a saddle point if…

0 > D

If D = 0

No conclusion can be drawn

To evaluate rectangular double integrals…

Integrate wrt x and then wrt y or vice versa, treating the other as a constant when you do

Fubini’s theorem

For double integrals over rectangular regions, the order of integration does not matter

For double integrals over non-rectangular regions…

Integrate wrt whichever variable is in the direction that the region is simple in (i.e. if vertically simple, integrate wrt y first)

Integrate first between two equations of the curves of the shape and then for the second integration, between two specific numbers

To change the variables…

You must change the limits and substitute in the new variables.

You must also find and multiply by the modulus of the Jacobian

Define the Jacobian

Let T:R²→R² be a transformation from the uv-plane to the xy-plane given by T(u,v) = (x(u,v), y(u,v)) for differentiable functions x and y.

The Jacobian of T is defined as … (check notes for answer)

Define a simple polar region

A simple polar region is a region of the plane enclosed between two rays θ = α and θ = β and two curves r = r₁(θ), r = r₂(θ) with 0≤β-α≤2π, r₁(θ)≤r₂(θ)

Double integrals with simple polar regions

Integrate wrt r between the two curves and then integrate wrt θ between the two angles

Triple integrals over cuboids

Integrate wrt z, wrt y and wrt x in any order, treating the others as a constant when you do

Triple integrals over a simple xy-solid

For a sold with xy-cross-section R, bounded above and below by surfaces z = g₁(x,y) and z = g₂(x,y), integrate first wrt z between the two surfaces and then complete the double integral for the xy-cross-section

Change of variables for triple integrals

Works the same way as for double integrals but with a more complicated Jacobian.

The Jacobian is… (see notes to check)

Triple integrals for spherical coordinates

∫∫∫ f(x,y,z) dV = ∫∫∫ g(ρ,θ,∅)ρ²sin∅ dρ d∅ dθ where 0≤∅≤π, 0≤θ≤2π

(Note that the Jacobian is ρ²sin∅)

Triple integrals for cylindrical coordinates

∫∫∫ f(x,y,z) dV = ∫∫∫ g(r,θ,z)r dr dθ dz where 0≤θ≤2π

(Note that the Jacobian is r)

Define an ordinary DE

A D.E. is ordinary if there is only one independent variable

Define the order, n, of a DE

Order, n, is the highest derivative of the DE

Define a linear DE

A DE is linear if it can be written as the sum of products of derivatives of y and functions of x equal to a function of x

Define an explicit solution to a DE

A function y = f(x) that satisfies it

Define an implicit solution to a DE

An equation that relates x and y s.t. they satisfy the DE

Define a general solution to a DE

A solution that includes arbitrary constants.

An nth order DE will have a general solution with n arbitrary constants

Define a particular solution to a DE

A solution that does not contain arbitrary constants

To visualise first order DEs…

you can draw slope fields based off the solutions. Note that the DE solutions (or trajectories) are a family of curves determined by an initial condition.

Define a separable ODE

A first-order ODE is separable if it can be written in the form dy/dx = p(x)q(y)

To solve a separable ODE…

Divide to get the q(y) on one side and p(x) on the other. Then integrate

Define a first-order linear ODE

A first-order ODE is linear if it can be written in the form dy/dx + p(x)y = q(x)

To solve a first-order linear ODE…

Multiply the equation by e^{∫p(x) dx} (the integrating factor) and integrate which will reverse the product rule

A function is (Euler) homogenous if…

∃ n ∈ N s.t. f(tx,ty) = tⁿ f(x,y) for any t ∈ R/{0}

(n is the degree of homogenity)

To solve homogenous ODEs…

Let t = 1/x so f(x,y) = f(1, y/x) = g(u) where u = y/x. Thus use y=ux as a substitution to make the ODE separable and solve as before.

Define a Bernoulli ODE

An ODE in the form dy/dx + p(x)y = q(x)yⁿ, where n ∈ Z

To solve Bernoulli ODEs…

Rearrange to y^-ndy/dx +p(x)y^1-n = q(x) and let z = y^1-n. This makes the ODE linear and you can solve as before but ensure you change back to in terms of y at the end.

Define an exact ODE

An ODE of the form M(x,y) + N(x,y)dy/dx = 0 is exact if there exists some function f(x,y) s.t. the partial derivative of f wrt x is M and wrt y is N.

Then f(x,y) is a potential function for the ODE and f(x,y) = c is the implicit solution of the ODE.

What is the test for exactness?

The eqn M(x,y) + N(x,y)dy/dx = 0 is exact iff the partial derivative of M wrt y equals the partial derivative of N wrt x

To find the potential function of an exact ODE…

Integrate M wrt x which will give you f(x,y) without g(y). Integrate N wrt y which will give you f(x,y) without h(x). Set the results equal to each other to find g(y) and h(x).

This determines f(x,y) completely.

When is a second-order linear ODE forced?

When it is in the form a d²y/dx² + b dy/dx +cy = f(x). f(x) is the forcing function

To determine the arbitrary constants of a unique solution…

you need two conditions

Initial value problems give you the solutions when x = 0.

Boundary value problems give you other combinations

If u(x) and v(x) are solutions to the free second-order linear ODE…

any combination w(x) = Au(x) + Bv(x) is a solution to the ODE

Functions u and v, defined on the same domain are said to be linearly independent…

if there are no non-zero constants A and B s.t. Au(x) + Bv(x) = 0 ∀x

If the auxiliary equation gives b²-4ac > 0…

the roots are real and distinct, λ₁, λ₂

General solution: y = Ae^λ₁x + Be^λ₂x

Graph: exponential graphs depending on sign

If the auxiliary equation gives 0 > b²-4ac…

the roots are complex conjugates, λ₁, λ₂ = α±βi

General solution: y = e^αx(Acosβx + Bsinβx)

Graph: increasing or decreasing oscillations depending on sign of α

If the auxiliary equation gives b²-4ac = 0…

the roots are real and repeated, λ₁ = λ₂

General solution: y = e^λx(A + Bx)

Graph: increasing or decreasing exponential depending on sign

To find the solutions to a forced second-order linear ODE…

Find the general solution of the free ODE, q(x), the complementary function (CF).

Find a particular solution of the forced ODE, p(x), the particular integral (PI).

Construct the general solution: y = CF + PI = q(x) + p(x)

Sinusoidal forcing

If f(x) = αcosγx + βsinγx, then try p(x) = lcosγx + msinγx and find l and m

Exponential forcing

If f(x) = αe^γx, then try p(x) = le^γx and find l

Polynomial forcing

If f(x) = α₀ + α₁x + α₂x², then try p(x) = r₀ + r₁x + r₂x² where r₀, r₁, r₂ are to be found

Finding unknowns

Differentiate p(x) and sub into the ODE before comparing coefficients

If any of the forms of the particular integral are the same as the complementary function…

multiply the p(x) you are trying by x until in a different form

To relate to mechanics…

y(t) is the position or displacement of an object

dy/dt is the velocity

d²y/dt² is the acceleration

If in mechanics in the form d²x/dt²+ax = 0…

it is simple harmonic motion

If in mechanics in the form d²x/dt²+ady/dt + bx = 0…

it is damped harmonic motion

If damped harmonic motion with a forcing function present…

it is forced damped harmonic motion

If damped harmonic motion with complex roots…

it is underdamped

Graph: decreasing oscillations

If damped harmonic motion with real distinct roots…

it is overdamped

Graph: quickly decreasing curve

If damped harmonic motion with a single repeated real root…

it is critically damped

Graph: slightly less quickly decreasing curve

Define pure resonance

A forced second-order ODE exhibit pure resonance if the frequency of its forcing is the same as the natural frequency of the ODE

i.e. a component of the forcing function also appears in the complementary function

Define practical resonance

A forced second-order ODE exhibits practical resonance if the frequency of the forcing function amplifies the oscillations of the solution