Multivariable Calculus Flashcards by Rachel Allan

What is a scalar field?

By a scalar field φ on R³ we shall mean a map φ: R³ → R

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What is a vector field?

By a vector field F on R³ we shall mean a map F: R³ → R³

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A set is said to be open if ?

What is an open ball?

if for every x ∈ U there exists ε > 0 such that
B(x, ε) = {y ∈ Rⁿ : |y − x| < ε} ⊆ U.
and where
|y − x|² = ⁿΣᵢ₌₁|yᵢ − xᵢ|²
We refer to B(x, ε) as the open ball of radius ε,centred at x.

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What is the moment of inertia in terms of a double integral, about an axis vertically through a point (x0, y0)?

∫∫ᵣ ρ(x, y)((x − x₀)² + (y − y₀)²) dA

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Given two co-ordinates u(x, y) and v(x, y) which depend on variables x and y,
we define the Jacobian ∂(u, v)/∂(x, y) to be….

the determinant
| uₓ uᵧ |
| vₓ vᵧ |

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What is the centre of mass of a body occupying a region R and with density ρ(r) at the point with position vector r ?

r(bar) = (xbar, ybar, zbar) = 1/M ∫∫∫ᵣ 𝐫ρ(𝐫) dV

all 𝐫s vectors
r under integrals should be R

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What is the median of a function? Given a function f on a region R ⊆ R³

The median of f is the value of m that satisfies

Vol ({(x, y, z) : f(x, y, z) ≤ m}) = 1/2 Vol(R)

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What does dS refer to?

|∂𝐫/∂u ∧ ∂𝐫/∂v| du dv

𝐫s are vectors

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What does d𝐒 (vector) refer to

or written as 𝐧 dS

∂𝐫/∂u ∧ ∂𝐫/∂v du dv
𝐫s are vectors
𝐧 is the unit normal in the direction of ∂𝐫/∂u ∧ ∂𝐫/∂v

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The surface area is [ ] of the choice of parametrization

independent

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The surface area of r (U) is independent of the choice of parameterization
Prove it

Proof pg 29/30

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If 𝐅 and φ are a vector field and scalar field defined on a parameterized surface
Σ = 𝐫(U), then we may define the following surface integrals:
1) ∫∫Σ 𝐅 · d𝐒

∫∫Σ 𝐅 · d𝐒 = ∫∫ᵤ 𝐅(𝐫(u, v)) · (∂𝐫/∂u ∧ ∂𝐫/∂v) du dv

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If 𝐅 and φ are a vector field and scalar field defined on a parameterized surface
Σ = 𝐫(U), then we may define the following surface integrals:
2) ∫∫Σ 𝐅dS

∫∫Σ 𝐅 · dS = ∫∫ᵤ 𝐅(𝐫(u, v)) |(∂𝐫/∂u ∧ ∂𝐫/∂v)| du dv

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If F and φ are a vector field and scalar field defined on a parameterized surface
Σ = 𝐫(U), then we may define the following surface integrals:
3) ∫∫Σ φ d𝐒

3) ∫∫Σ φ d𝐒 = ∫∫ᵤ φ(𝐫(u, v))(∂𝐫/∂u ∧ ∂𝐫/∂v) du dv

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If F and φ are a vector field and scalar field defined on a parameterized surface
Σ = 𝐫(U), then we may define the following surface integrals:
4) ∫∫Σ φ dS

4) ∫∫Σ φ dS = ∫∫ᵤ φ(𝐫(u, v))|(∂𝐫/∂u ∧ ∂𝐫/∂v)| du dv

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What is a flux integral?

∫∫Σ F · dS

F and S bold

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What is the relationship between dS and dS (one bold)?

dS = ndS
first S and n bold
where n is the outwards!!!! pointing normal from the body

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What is the wordy definition of a solid angle?

The solid angle is the angle an object subtends at a point in three-dimensional
space. More precisely, half-lines from a fixed point (or observer) will either intersect with the object in question or not; those lines of sight that are blocked by the object represent a subset
of the unit sphere centred on the observer. The solid angle is the area of this subset (strictly it is the area of this subset divided by the unit of length squared to ensure the solid angle is dimensionless).

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What is the unit of a solid angle?

The unit of solid angle is the steradian. Given that the surface area of a
sphere is 4π(radius)²
then a whole solid angle is 4π`

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If Σ∗
is a surface and Σ is the subset of Σ∗
facing the unit sphere, then the solid angle Ω
subtended at O by Σ∗
equals, by definition

Ω = ∫∫ (eᵣ · dS)/r² = ∫∫ (r · dS)/r³
Σ Σ

First Integral:
e,S bold
Second Integral:
first r and S bold

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Define a curve

By a curve we shall mean a piecewise smooth function γ : I → R³ defined on
an interval I of R. Notice that order on I also gives the curve γ an orientation.

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What is a simple curve?

We say a curve γ : [a, b] → R³
is simple if γ is 1—1, with the one possible exception that γ(a) = γ(b) may be true; this means that the curve does not cross itself except possibly by its endpoints meeting

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What is a closed curve?

We say a curve γ : [a, b] → R³ is closed if γ(a) = γ(b)

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Let C be a curve in R³, parameterized by γ : [a, b] → R³ and let F be a vector field, whose domain includes C. We define the line integral of F along C as….

∫𝒸 F · dr = ₐ∫ᵇ F(r(t)) · r’(t) dt

Fs and rs bold

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Is the line ∫𝒸 F · dr independent of choice of parametrization?

Yes, if orientated the same

If oriented the same, the line integral ∫𝒸 F · dr is independent of the choice of parameterization Prove it

pg 35/36

If φ is a scalar field defined on a curve C with parameterization γ : [a, b] → R³, then we also define the line integral

∫𝒸 φ ds = ₐ∫ᵇ φ(t)|γ'(t)| dt

If F = (F1, F2, F3) is a vector field defined on the curve C then we define ∫𝒸 F ds

∫𝒸 F ds = (∫𝒸 F1 ds , ∫𝒸 F2 ds, ∫𝒸 F3 ds) | First F is bold

Note that if t is the unit tangent vector field along C, in the same direction as the parameterization, then ∫𝒸 φ ds =

∫𝒸 φ ds = ∫𝒸 (φt) · dr | t and r bold

If C is a curve with parameterization γ : [a, b] → R³ then the arc length of the curve is

∫𝒸 ds = ₐ∫ᵇ |γ'(t)| dt

Define 1) dr (r bold) 2) ds

``` dr = (dx, dy, dz) ds = |dr| ```

If the vector field F represents a force on a particle then ∫𝒸 F · dr ( F and r bold) is ???

The work done by the force in moving the particle along C.

If F = ∇φ and γ : [a, b] → S is any curve such that γ (a) = p, γ (b) = q then ∫γ F · dr = F bold r bold p, q bold

∫γ F · dr = φ (q) − φ (p) F,r,q,p bold In particular, the integral depends only on the endpoints of the curve γ

``` In particular, the integral ∫γ F · dr depends only on the endpoints of the curve γ prove it (not long) ```

pg 39

Let S be an open subset of R³. A vector field F: S → R³ is said to be conservative if F bold

there exists a scalar field φ: S → R such that F = ∇φ | F bold

If F = ∇φ (bold F), what is the potential?

φ is the potential for F

A subset S of R³ is said to be path connected if ....

for any points p, q ∈ S there exists a curve γ : [a, b] → S such that γ(a) = p and γ(b) = q. bold q, p

Prove that If ∇φ = 0 on a path-connected set S then φ is constant. In particular, if φ and ψ are potentials of the conservative field F, defined on S, then φ and ψ differ by a constant

If ∇φ = 0 then for any curve, with endpoints p, q, we have φ (q) − φ (p) = ∫γ ∇φ · dr = 0 (bold r) Given a fixed point p in S, any q in S is connected to p by a curve, and so φ is constant. If F = ∇φ = ∇ψ then ∇ (φ − ψ) = 0 and the result follows

Let S be an open path connected subset of R³ and let F : S → R³ be a vector field. Then the following three statements are equivalent (i) F is conservative (ii) Given any two points p, q ∈ S and curve γ in S, starting at p and ending at q, then the integral ∫γ F · dr is independent of [ ] (iii) For any simple closed curve γ then ∫γ F · dr = [ ]

(i) F is conservative (ii) Given any two points p, q ∈ S and curve γ in S, starting at p and ending at q, then the integral ∫γ F · dr is independent of the choice of curve γ. (iii) For any simple closed curve γ then ∫γ F · dr = 0

Define the gradient

Let φ: R³ → R be a scalar field. Then the gradient of φ written grad φ or ∇φ equals grad φ = ∇φ = (∂φ/∂x, ∂φ/∂y, ∂φ/∂z)

Define divergence

Let F: R³ → R³ be a vector field with F = (F1, F2, F3). Then the divergence of F written div F or ∇ · F equals div F = ∇ · F = ∂F1/∂x + ∂F2/∂y + ∂F3/ ∂z

Define curl

Let F: R³ → R³ be a vector field with F = (F1, F2, F3). Then the curl of F written curl F or ∇ ∧ F equals curl F = ∇ ∧ F = | i j k | | ∂/∂x ∂/∂y ∂/∂z | | F1 F2 F3

For any scalar field φ, what is the Laplacian?

div grad = ∇ · ∇ | ∇²φ = ∂²φ/∂x² + ∂²φ/∂y² + ∂²φ/∂z²

Let F be a vector field on R³ and φ be a scalar field on R³ curl grad φ = div curl F =

curl grad φ = 0 (bold) | div curl F = 0

(Product Rules for div and curl) Let F be a vector field on R³ and φ, ψ be scalar fields on R³. Then div (φF) = curl (φF) = bold Fs

div (φF) = ∇φ · F + φ div F; curl (φF) = φ curl F + ∇φ ∧ F. Bold Fs

For vector fields F, G (All F and G bold) ∇ (F · G) = ∇ · (F ∧ G) =

∇ (F · G) = (F · ∇) G + (G · ∇) F + F ∧ (∇ ∧ G) + G ∧ (∇ ∧ F) ∇ · (F ∧ G) = G · (∇ ∧ F) − F · (∇ ∧ G)

For vector fields F, G (All F and G bold) ∇ ∧ (F ∧ G) = ∇ ∧ (∇ ∧ F) =

∇ ∧ (F ∧ G) = F (∇ · G) − G (∇ · F) + (G · ∇) F − (F · ∇) G ∇ ∧ (∇ ∧ F) = ∇ (∇ · F) − ∇²F

What is Fourier's law?

In an isotropic medium with constant thermal conductivity k, with temperature T (x, t) at position x and at time t, Fourier’s Law states that q = −k∇T where q is the heat flux — that is the amount of energy that flows through a particular surface per unit area per unit time. Bold q and x

What is the divergence theorem?

Let R be a region of R³ with a piecewise smooth boundary ∂R, and let F be a differentiable vector field on R. Then ∫∫∫R div F dV = ∫∫∂R F · dS where dS is oriented in the direction of the outward pointing normal from R F and S bold

Define convex

We say that a region R is convex if for any p, q ∈ R the line segment connecting p and q is contained in R Bold p, q

Let φ be a smooth scalar field defined on a region R ⊆ R³ with a piecewise smooth boundary. Then ∫∫∫R ∇φ dV =??

∫∫∫R ∇φ dV = ∫∫∂R φ dS S bold

Prove that Let φ be a smooth scalar field defined on a region R ⊆ R³ with a piecewise smooth boundary. Then ∫∫∫R ∇φ dV = ∫∫∂R φ dS

``` Let c be a constant vector and F = cφ Then div (cφ) = c · ∇φ + φ div c = c · ∇φ as c is constant. By the Divergence Theorem ..... pg 57 ```

What is Green's Theorem in the plane?

Let D be a closed bounded region in the (x, y) plane, whose boundary C is a piecewise smooth simple closed curve, and that p(x, y), q(x, y) have continuous first-order derivatives in D. Then ∫∫D (∂p/∂x + ∂q/∂y) dx dy = ∫𝒸 (p, q) · n ds = ∫𝒸 p dy - q dx bold n

Prove Green's Thm in the plane

pg58

``` (Uniqueness of the Dirichlet Problem) Let R ⊆ R³ be a path-connected region with piecewise smooth boundary ∂R and let f be a continuous function defined on ∂R. Suppose that φ1 and φ2 are such that ∇²φ1 = 0 = ∇²φ2 in R; φ1 = f = φ2 on ∂R Then .... ```

Then φ1 = φ2 | in R.

Prove the Uniqueness of the Dirichlet Problem

pg 61/62

(Uniqueness, up to a constant, of the Neumann Problem) Let R ⊆ R³ be a path-connected region with piecewise smooth boundary ∂R and let f be a continuous function defined on ∂R. Suppose that φ1 and φ2 are such that ∇²φ1 = 0 = ∇²φ2 in R; ∂φ1/∂n = f = ∂φ2/ ∂n on ∂R Then....

Then φ1 − φ2 | is constant in R.

Prove | Uniqueness, up to a constant, of the Neumann Problem

pg 62

(Robin (or Mixed) Boundary Problem) Let R ⊆ R³ be a path-connected region with piecewise smooth boundary ∂R and let f be a continuous function defined on ∂R. Further let β be a real constant. Suppose that φ1 and φ2 are such that ∇²φ1 = 0 = ∇²φ2 in R; βφ1 + ∂φ1/∂n = f = βφ2 + ∂φ2/ ∂n on ∂R (i) If β > 0 then (ii) If β = 0 then (iii) If β < 0 then

``` (i) If β > 0 then φ1 = φ2 in R. (ii) If β = 0 then φ1 − φ2 is constant in R. (iii) If β < 0 then the solution need not be unique, even up to a constant ```

Prove | Robin (or Mixed) Boundary Problem

pg 63

If φ : R³ → R is a continuous scalar field such that ∫∫∫R φ dV = 0 for all bounded subsets R, then φ ...

φ ≡ 0

If φ : R³ → R is a continuous scalar field such that ∫∫∫R φ dV = 0 for all bounded subsets R, then φ ≡ 0 Prove it

pg 64

Let T (x, t) denote the temperature at position x and at time t in an isotropic medium R with thermal conductivity k, density ρ, specific heat c with heat flow determined by Fourier’s Law which states that q = −k∇T where q is the heat flux The T satisfies the [ ] equation

heat equation ∂T/∂t = k/ρc ∇²T.

pg 65

What is Stoke's theorem?

Let Σ be a smooth oriented surface in R³whose boundary is the curve ∂Σ. Let F be a smooth vector field defined on Σ ∪ ∂Σ. Then ∫∂Σ F · dr = ∫∫Σ curlF · dS

Prove Stoke's thm

pg68/69

If c · v = c · w for all c ∈ R³ then v = all bold

then v = w

If c · v = c · w for all c ∈ R³ then v = w Prove

We have c·(v − w) = 0 for all c and thus for all basis vectors. Thus all the components of v, w are equal, hence so are the vectors

Let Σ be a smooth oriented surface in R³ whose boundary is the curve ∂Σ. Let ψ be a smooth scalar field defined on Σ ∪ ∂Σ. Then ∫∫Σ ∇ψ ∧ dS = ??

∫∫Σ ∇ψ ∧ dS = - ∫∂Σ ψ dr S and r bold

Let Σ be a smooth oriented surface in R³ whose boundary is the curve ∂Σ. Let ψ be a smooth scalar field defined on Σ ∪ ∂Σ. Then ∫∫Σ ∇ψ ∧ dS = - ∫∂Σ ψ dr Prove it

Not long | pg 70

A region R ⊆ R³ is said to be simply connected if ....

every simple, closed curve C can be continuously deformed to a single point. Specifically, if r : [0, 1] → R is a simple closed curve beginning and ending in p then there exists a map H : [0, 1] × [0, 1] → R such that H(t, 1) = r(t) and H(t, 0) = p

Existence of a Potential) Let R be a simply connected region and let F be a smooth vector field on R for which curl F = 0. The F is [ ]

Then F is conservative, i.e. there exists a | potential φ on R such that F = ∇φ.

Firstly note that if F is conservative then F = ∇φ and hence curl F = 0. curl F = 0 is equivalent to (i), (ii) and (iii) (i) F is [ ] (ii) Given any two points p, q ∈ S and curve γ in S, starting at p and ending at q, then the integral ∫𝒸 F (r) · dr is independent of [ ] (iii) For any simple closed curve C then ∫𝒸 F (r) · dr = [ ]

(i) F is [conservative] (ii) Given any two points p, q ∈ S and curve γ in S, starting at p and ending at q, then the integral ∫𝒸 F (r) · dr is independent of [the choice of curve C] (iii) For any simple closed curve C then ∫𝒸 F (r) · dr = [0]

Firstly note that if F is conservative then F = ∇φ and hence curl F = 0. curl F = 0 is equivalent to (i), (ii) and (iii) (i) F is [conservative] (ii) Given any two points p, q ∈ S and curve γ in S, starting at p and ending at q, then the integral ∫𝒸 F (r) · dr is independent of [the choice of curve C] (iii) For any simple closed curve C then ∫𝒸 F (r) · dr = [0] Prove it

pg 75

The gravitational ﬁeld associated with a point mass M at the origin O is...

𝐟 = −(GM/r³)𝐫 = −(GM/r²)𝐞ᵣ = ∇(GM/r) .

The gravitational force on a particle of mass m at the point 𝐫 is...

𝐅 = -(GMm/r²)𝐞ᵣ

The gravitational ﬁeld 𝐟 is [] with gravitational potential

conservative | φ = GM/r

What is curl f, and div f, where f is a gravitational field?

curl 𝐟 = 0 div 𝐟 = ∇²φ = 1/r²(d/dr(r²(dφ/dr))) = 1/r²(d/dr(r²(−GM/r²))) = 0 div 𝐟 = 0

Prove that the gravitational potential φ = GM/r is the amount of work gravity does per unit mass to bring a point object from ∞ to 𝐫.

1/m ∫ʳ∞ 𝐅·d𝐫 = ∫ʳ∞ 𝐟 ·d𝐫 = [φ(r)−φ(∞)] = φ(r)

The gravitational potential energy of a point mass m at 𝐫 equals

−mφ(𝐫).

In the continuous case, we have matter of density ρ(r) occupying a region R; then the potential φ and ﬁeld f are given by...

φ(p) = ∫∫∫ᵣ Gρ(𝐫)/|p−𝐫| dV 𝐟(p) = -∫∫∫ᵣ Gρ(𝐫)(p−𝐫)/|p−𝐫|³ dV

Dirac Delta Function: ∇·(𝐫/r³ ) = [] where δ is the Dirac Delta Function. This function has the important ﬁltering property that... [Says optional in my notes, but not in the lecture notes]

∇·(𝐫/r³ )= 4πδ(r) ∫∫∫ᵣ φ(𝐫)δ(𝐫−a) dV = φ(a).

(Poisson’s Equation) Let φ be the gravitational potential associated with a non-uniform material of density ρ(𝐫) occupying a region R. Then...

∇²φ = −4πGρ(𝐫) Proof non-examinable

(Gauss’ Flux Theorem) For a smooth and bounded region R, which contains matter of total mass M, then... This result is also equivalent to...

∫∫∂ᵣ 𝐟 ·d𝐒 = −4πGM. This result is equivalent to Poisson's equation

Prove Gauss' Flux Theorem

Poisson’s equation gives us that ∇²φ = −4πGρ(𝐫) where ρ(𝐫) is the density of the matter. Applying the Divergence Theorem we have ∫∫∂ᵣ 𝐟 ·d𝐒 = ∫∫∫ᵣ ∇·𝐟 dV = ∫∫∫ᵣ ∇²φ dV = −4πG∫∫∫ᵣ ρ dV = −4πGM. Conversely suppose that we know ∫∫∂ᵣ 𝐟 ·d𝐒 = −4πGM for any bounded region R. Then ∫∫∫ᵣ ∇²φ dV = −4πG∫∫∫ᵣρ dV and so ∫∫∫ᵣ ∇²φ+ 4πGρ dV = 0 for any bounded region R. Hence (at least if ∇²φ and ρ are piecewise continuous) we have ∇²φ + 4πGρ ≡0

Multivariable Calculus Flashcards

(85 cards)