Key concepts of multivariable calculus.

Limits of Functions

For multivariable functions, the limit must exist and be equal along all paths.

The monotone bounded criterion and L'Hôpital's rule cannot be applied to limits of multivariable functions.

Continuity

It suffices that

lim⁡x→x0y→y0f(x,y)=f(x,y)\lim_{x\rightarrow x_0 \atop y\rightarrow y_0}f(x,y) = f(x,y)

Partial Derivatives

lim⁡Δx→0f(x0+Δx,y0)−f(x0,y0)Δx\lim_{\Delta x\to 0 }\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}

Steps to Prove Differentiability

  1. Compute the total increment Δz=f(x0+Δx,y0+Δy)−f(x0,y0)\Delta z = f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0,y_0)

  2. Linear increment: AΔx+BΔyA\Delta x + B \Delta y, where A=fx′(x0,y0),B=fy′(x0,y0)A = f'_x(x_0,y_0),B=f'_y(x_0,y_0)

  3. Take the limit

lim⁡Δx→0Δy→oΔz−A(Δx+BΔy)(Δx)2+(Δy)2\lim_{\Delta x \to 0 \atop \Delta y \to o} \frac{\Delta z- A(\Delta x + B \Delta y)}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}

  1. If the limit equals zero, the function is differentiable; otherwise, it is not.

Extrema of Multivariable Functions

If f(x,y)f(x,y) satisfies certain conditions at the point (x0,y0)(x_0,y_0), denote

fxx′′=A,f′′(xy)=B,f′′(yy)=Cf''_{xx}=A, f''(xy)=B, f''(yy)=C

Then, if AC−B2>0AC-B^2>0, ff attains an extremum at that point. If A<0A<0, it attains a local maximum; if A>0A>0, it attains a local minimum.

Directional Derivatives

x−x0=Δx=tcos⁡αy−y0=Δy=tcos⁡βz−z0=Δz=tcos⁡γ\begin{align} &x-x_0=\Delta x = t\cos{\alpha}\\ &y-y_0=\Delta y = t\cos{\beta}\\ &z-z_0=\Delta z = t\cos{\gamma}\\ \end{align}

Let t=(Δx)2+(Δy)2+(Δz)2t = \sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta{z})^2} denote the distance between PP and P0P_0.

If

lim⁡t→0+u(P)−u(P0)t\lim_{t\to0^+}\frac{u(P)-u(P_0)}{t}

exists, then this limit is the directional derivative at that point.

If all partial derivatives exist, we can also use:

∂u∂l=∇u⋅l∘\frac{\partial u}{\partial l}=\nabla{u}\cdot l^{\circ}

Jacobian Matrix

J=[∂f∂x1⋯∂f∂xn]J=\left[\frac{\partial f}{\partial x_1}\cdots \frac{\partial f}{\partial x_n}\right]

When applied to integration, take the absolute value of its determinant.

Line Integrals of the First Kind

Computation

Express Γ\Gamma using parametric equations, then

∫Γfds=∫βαf(x′)2+(y′)2+(z′)2dt\int_\Gamma f\text{d}{s}=\int^{\alpha}_{\beta}f\sqrt{(x')^2+(y')^2+(z')^2}\text{d}{t}

Line Integrals of the Second Kind

Direct Computation

∫L[PQ]d[xy]=∫βαPtxt′dt+Qtyt′dt\int_L{\left[\begin{matrix} P & Q \end{matrix}\right]\text{d}\left[\begin{matrix} x & y \end{matrix}\right]}=\int^{\alpha}_\beta{P_tx_t'\text{d}t+Q_ty'_t\text{d}t}

Direct computation is a very important method that must not be forgotten.

Green's Theorem

∮LPdx+Qdy=∬D(∂Q∂x−∂P∂y)dσ.\oint_L{P\text{d}x+Q\text{d}y}=\iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\text{d}\sigma.

If

∂Q∂x≡∂P∂y.\frac{\partial Q}{\partial x}\equiv \frac{\partial P}{\partial y}.

then the integral is path-independent. If the path contains singular points, the path can be changed for computation. However, note that the new path must lie in a simply connected region.

If the curve is not closed and

∂Q∂x≠∂P∂y.\frac{\partial Q}{\partial x}\neq\frac{\partial P}{\partial y}.

a simple path can be added to close the curve, then subtract the integral along the added path from the result.

Path Independence of Integrals

∮LPdx+Qdy=0Pdx+Qdy=duPdx+Qdy=0∇u=[PQ]\begin{align} &\oint_L{P\text{d}x+Q\text{d}y}=0\\ &P\text{d}x+Q\text{d}y=\text{d}u\\ &P\text{d}x+Q\text{d}y=0\\ &\nabla{u}=\left[\begin{matrix} P& Q \end{matrix}\right]\\ \end{align}

Stokes' Theorem

∮ΓAds=∬Σ∇×A⋅dσ=∬Σ∇×A⋅n∘ds\oint_{\Gamma}A\text{d}s=\iint_{\Sigma}\nabla\times{A}\cdot\text{d}\sigma=\iint_{\Sigma}{\nabla\times{A}\cdot{n^{\circ}}\text{d}s}

Path Independence

If

∇×F=0,\nabla\times{F} =0,

then the integral is path-independent. Recall the path independence condition in Green's theorem — it is essentially the case in ∇×F\nabla \times F where we set k=0k=0, reducing to the two-dimensional case.

Surface Integrals

∬ΣzdS\iint_{\Sigma}z\text{d}S

General method:

dS=1+z′x2+z′y2dxdy\text{d}S=\sqrt{1+{z'}_x^2+{z'}_y^2}\text{d}x\text{d}y

Alternatively,

dS=1+y′x2+y′z2dxdz\text{d}S=\sqrt{1+{y'}_x^2+{y'}_z^2}\text{d}x\text{d}z

This can be converted to an ordinary double integral.

For oriented surfaces:

∬Σ[PQR]⋅d[xyz]=∬Σ(Pdydz+Qdzdx+Rdxdy)\iint_{\Sigma}\left[\begin{matrix} P & Q & R \end{matrix}\right] \cdot \text{d}\left[\begin{matrix} x & y & z \end{matrix}\right]= \iint_{\Sigma}(P\text{d}y\text{d}z+Q\text{d}z\text{d}x+R\text{d}x\text{d}y)

When the surface is closed and has continuous first-order partial derivatives, Gauss's theorem can be applied:

∯ΣPdydz+Qdzdx+Rdxdy=∭Ω(∂P∂x+∂Q∂y+∂R∂z)dv\displaystyle {\oiint}_{\kern{-12mu}\Sigma}P\text{d}y\text{d}z+Q\text{d}z\text{d}x+R\text{d}x\text{d}y= \iiint_{\Omega}{\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)}\text{d}v

If the surface is closed and contains a singular point in its interior, and away from the singular point

∇⋅F=0\nabla \cdot F = 0

then the surface of integration can be changed (the boundaries need not coincide).

If the curve is not closed, then the boundaries must coincide.