Part 21 - Summary | rezarezvan.com

Limits

Let’s compare limits in one variable to several.

\lim_{x \to a} f(x) = L

\lim_{(x, y) \to (a, b)} f(x, y) = L

They work quite similar, let’s see the limit laws:

\lim_{(x, y) \to (a, b)} f(x, y) \pm h(x, y) = L \pm M

\lim_{(x, y) \to (a, b)} f(x, y) \cdot h(x, y) = L \cdot M

\lim_{(x, y) \to (a, b)} \dfrac{f(x, y)}{h(x, y)} = \dfrac{L}{M}, \quad M \neq 0

Partial derivative

Recall (Partial derivative)

The partial derivative of a function of two variables with respect to $x$ , denoted by $f_x$ , is the function of two variables given by:

f_x(x, y) = \lim_{h \to 0} \dfrac{f(x + h, y) - f(x, y)}{h}

Similarly for $y$ :

f_y(x, y) = \lim_{h \to 0} \dfrac{f(x, y + h) - f(x, y)}{h}

Chain rule

Recall (Chain rule)

Suppose $z = f(x, y)$ is differentiable function of $x, y$ . Where $x = g(t)$ and $y = h(t)$ . Which both are differentiable functions of $t$ . Then:

\boxed{\dfrac{dz}{dt} = \dfrac{\partial z}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial z}{\partial y} \dfrac{dy}{dt}}

General chain rule

Recall (General chain rule)

Let $z$ be a differentiable function of $n$ variables, $x_1, \ldots, x_n$ , where each $x_i$ is differentiable function of $m$ variables, $t_1, \ldots, t_m$ .

Then:

\dfrac{\partial z}{\partial t_i} = \dfrac{\partial z}{\partial x_1} \dfrac{\partial x_1}{\partial t_i} + \dfrac{\partial z}{\partial x_2} \dfrac{\partial x_2}{\partial t_i} + \ldots + \dfrac{\partial z}{\partial x_n} \dfrac{\partial x_n}{\partial t_i}

Implicit differentiation

Recall (Implicit differentiation)

Suppose a function $y = y(x)$ is given implicitly by equation of the form $F(x, y) = 0$ where $F$ is a differentiable function of two variables and $F_y \neq 0$ . Then:

\boxed{\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}}

Implicit differentiation of several variables

Recall (Implicit differentiation of several variables)

Let $z(x, y)$ be given implicitly, by equation of form $F(x, y, z) = 0$ where $F$ is differentiable and $F_z \neq 0$ , then:

\begin{align*} \dfrac{\partial z}{\partial x} & = -\dfrac{F_x}{F_z}, \newline \dfrac{\partial z}{\partial y} & = -\dfrac{F_y}{F_z}. \end{align*}

Definition of directional derivatives

The directional derivative of a function, $f$ , of two variables, at point $(x_0, y_0)$ in the direction of a unit vector, $\vec{u}$ .

\vec{u} = \langle a, b \rangle

We denoted the directional derivative with a $D_{\vec{u}} f$ .

D_{\vec{u}} f(x_0, y_0) = \lim_{h \to 0} \dfrac{f(x_0 + ha, y_0 + hb) - f(x_0, y_0)}{h}

Definition of gradient

The gradient of, $f$ , is the vector-function:

\nabla f = \langle f_x, f_y \rangle

Therefore, we usually write:

\boxed{D_{\vec{u}} f = \nabla f \cdot \vec{u}}

Maximum rate of change

If we ask ourselves the question, in what direction is $D_{\vec{u}} f(x_0, y_0)$ the largest?

If we look at the definition:

D_{\vec{u}} f(x_0, y_0) = \nabla f(x_0, y_0) \cdot \vec{u}

If we use the geometrical definition:

D_{\vec{u}} f(x_0, y_0) = |\nabla f(x_0, y_0)| |\vec{u}| \cos(\alpha)

Gradient and level curves

Recall that level curves are curves with the equation:

f(x, y) = k

Where $k$ is just any constant. For each point $(x_0, y_0)$ , $\nabla f(x_0, y_0)$ is orthogonal to the level curve that contains $(x_0, y_0)$

Critical points

$(a, b)$ is a critical point if $\nabla f(a, b) = 0 \textbf{ or } \nabla f(a, b) \nexists$

In other words, if $(a, b)$ is a critical point of local minimum/maximum then it is a critical point.

However, the other way around this implication is not true. A critical point might not be a local minimum/maximum.

2nd derivative test for functions of two variables

Suppose $(a, b)$ is a critical point of a function, $f$ , of two variables. Suppose 2nd order partial derivatives exists and are continuous.

Let $D = D(a, b) = f_{xx}(a, b) f_{yy}(a,b) - \left(f_{xy} (a,b)\right)^2$

Then:

\begin{align*} D > 0 \text{ and } f_{xx}(a, b) > 0 & \quad \text{local minimum}, \newline D > 0 \text{ and } f_{xx}(a, b) < 0 & \quad \text{local maximum}, \newline D < 0 & \quad \text{neither, this is a saddle point}. \end{align*}

Lagrange multiplier method

Recall (Lagrange multiplier method)

To find minimum/maximum values of function, $f$ , subject to constraint, $g(x, y) = k$ , assuming minimum/maximum points exists and $\nabla g \neq 0$

Find all numbers $x, y$ and $\lambda$ , such that:
$\nabla f(x, y) = \lambda \nabla g(x, y), \quad g(x, y) = k.$
Compute values from step 1, choose the minimum/maximum value(s).

Type I region definition

If $D$ is of type I, then:

\iint_D f(x, y) \, dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx

Type II region definition

If $D$ is of type II, then:

\iint_D f(x, y) \, dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \, dy

General case of variable change

\iint_D f(x, y) \, dA

We want to rewrite this in terms of $u$ and $v$ . Let’s say that:

x = g(u, v), \quad y = h(u, v).

We can say that we have a transformation, $T$ , from the $uv$ -plane to the $xy$ -plane, given by our functions $g$ and $h$ .

We need to make some assumptions of $T$ to make our lives easier.

$T$ is a $C^1$ -transformation, meaning that $g$ and $h$ have continuous partial derivatives.
$T$ is an injective transformation (meaning it is 1-to-1). This means that we can express $u$ and $v$ in terms of $x$ and $y$ .

The Jacobian matrix of a transformation, $T$ , is:

\dfrac{\partial(x, y)}{\partial(u, v)} = \det \begin{bmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \newline \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{bmatrix}

Suppose $T$ is a transformation, from the $uv$ -plane to the $xy$ -plane. Assuming that the Jacobian for $T$ , is non-zero, then:

\iint_D f(x, y) \, dA = \iint_S f(x(u, v), y(u, v)) \left\vert \tfrac{\partial(x, y)}{\partial(u, v)} \right\vert \, dA.

In three variables

Let $x = x(u, v, w), y = y(u, v, w), z = z(u, v, w)$

The Jacobian is:

\dfrac{\partial(x, y, z)}{\partial(u, v, w)} = \det \begin{bmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \newline \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w} \newline \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{bmatrix}

Just as we did before:

\iiint_E f(x, y, z) \, dV = \iiint_S f(x(u, v, w), y(u, v, w), z(u, v, w)) \left\vert \tfrac{\partial(x, y, z)}{\partial(u, v, w)} \right\vert \, dV.

Polar coordinates

\iint_S g(r, \theta) \, dA = \iint_S f(r \cos(\theta), r \sin(\theta)) \cdot r \, dA

It makes sense to change to polar coordinates if we have a circle/circle like shape.

Cylindrical coordinates

\iiint_E f(x, y, z) \, dV = \iiint_S f(r \cos(\theta), r \sin(\theta), z) \cdot r \, dV

It makes sense to change to cylindrical coordinates when we have symmetry around one axis.

Spherical coordinates

\iiint_E f(x, y, z) \, dV = \iiint_S f(\rho \sin(\varphi) \cos(\theta), \rho \sin(\varphi) \sin(\theta), \rho \cos(\varphi)) \cdot \rho^2 \sin(\varphi) \, dV

It makes sense to change to spherical coordinates when our solid is bounded by spheres and/or cones.

Computing mass of solid

Let $E$ be an arbitrary solid, with a density function, $\rho(x, y, z)$

We say that the mass of the solid is:

\iiint_E \rho(x, y, z) \, dV

Center of Mass

One very important application is finding the center of mass of a solid.

Given a physical system with $m_i$ points, the center of mass has the coordinates, $(x_0, y_0, z_0)$ .

Therefore:

x_0 = \dfrac{\sum m_i x_i}{\sum m_i}, \quad y_0 = \dfrac{\sum m_i y_i}{\sum m_i}, \quad z_0 = \dfrac{\sum m_i z_i}{\sum m_i}.

For an arbitrary solid, it’s almost the same:

\begin{align*} x_0 & = \dfrac{\iiint_E \rho(x, y, z) \cdot x \, dV}{\iiint_E \rho(x, y, z) \, dV}, \newline y_0 & = \dfrac{\iiint_E \rho(x, y, z) \cdot y \, dV}{\iiint_E \rho(x, y, z) \, dV}, \newline z_0 & = \dfrac{\iiint_E \rho(x, y, z) \cdot z \, dV}{\iiint_E \rho(x, y, z) \, dV}. \end{align*}

Parameterization over line

Given a line that passes through the point $(x_0, y_0, z_0)$ , with a direction of the vector, $\vec{v} = \langle a, b, c \rangle$ .

If we want to parameterize this line, we can choose another point that this vector passes through as:

\vec{r}(t) = \vec{r}(t_0) + t\vec{v}

(x(t), y(t), z(t)) = (x_0, y_0, z_0) + t(a, b, c)

This means that:

\begin{align*} x(t) & = x_0 + ta, \newline y(t) & = y_0 + tb, \newline z(t) & = z_0 + tc. \end{align*}

Line integral

A curve, $C$ , parameterized as $\vec{r}(t) = \langle x(t), y(t) \rangle$ , for an interval $t \in [a, b]$ and a continuous function, $f: C \rarr \mathbb{R}$ .

Assuming $x(t)$ and $y(t)$ are differentiable.

\int_C f(x, y) \, ds = \int_a^b f(x(t), y(t)) \sqrt{x^\prime(t)^2 + y^\prime(t)^2} \, dt

Non-differentiable lines

There may be some cases where the parameterization isn’t differentiable.

Imagine we have a simple rectangular form. The four corners of the rectangle will not be differentiable.

What we can do is divide $C$ into $n$ subcurves.

\begin{align*} C & = C_1 \cup C_2 \cup \ldots \cup C_n, \newline \int_C f(x, y) \, ds & = \sum_{j = 1}^{n} \int_{C_j} f(x, y) \, ds. \end{align*}

Line integrals with respect to x or y

So far we have integrated with respect to the arc length. But we can integrate with respect to $x$ or $y$ .

\int_C f(x, y) \, dx \quad \text{or} \quad \int_C f(x, y) \, dy

\begin{align*} \int_C f(x, y) \, dx & = \int_a^b f(x(t), y(t)) x^\prime(t) \, dt, \newline \int_C f(x, y) \, dy & = \int_a^b f(x(t), y(t)) y^\prime(t) \, dt. \end{align*}

Line integrals with three variables

There is no difference in two or three variables:

\begin{align*} & C \subset \mathbb{R}^3, \newline & f: C \rarr \mathbb{R}, \newline & \vec{r}(t) = \langle x(t), y(t), z(t) \rangle, \quad a \leq t \leq b. \end{align*}

\int_C f(x, y, z) \, ds = \int_a^b f(x(t), y(t), z(t)) \sqrt{x^\prime(t)^2 + y^\prime(t)^2 + z^\prime(t)^2} \, dt

Vector field

A vector field is a function, $F$ :

\begin{align*} F & : D \subset \mathbb{R}^2 \rarr \mathbb{R}^2 \quad \text{(2D)}, \newline F & : D \subset \mathbb{R}^3 \rarr \mathbb{R}^3 \quad \text{(3D)}. \end{align*}

Gradient field

By this definition, we can say that.

Let $f: D \subset \mathbb{R}^2 \rarr \mathbb{R}$ be a differentiable function. The gradient is $\nabla f(x, y) = \langle f_x(x, y), f_y(x, y) \rangle$ .

The gradient is therefore a vector field!

\nabla f: D \rarr \mathbb{R}^2

Conservative vector field

A vector field, $F$ , is called conservative, if there is a function, $f$ , for which $\nabla f = F$ .

In this case, $f$ , is called a potential for $F$ .

Line integrals with vector fields

If $C$ is parameterized as $\vec{r}(t) = \langle x(t), y(t), z(t) \rangle$ in the interval, $a \leq t \leq b$ . Then:

\int_C F \cdot dr = \int_a^b F(x(t), y(t), z(t)) \cdot \vec{r^\prime}(t) \, dt

Conservative vector field

Let $D \subset \mathbb{R}^2$ be open (every point in $D$ has a small disc around it, which contains $D$ ), connected (consists of one single “piece”).

Let $F: D \rarr \mathbb{R}^2$ be a vector field. Then $F$ is conservative (there exists a $f$ such that $\nabla f = F$ ), if and only if:

\int_C F \cdot dr = 0 \quad \text{for every closed curve } C \subset D.

NB: This only applies for $\mathbb{R}^2$ .

Suppose $F(x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ is a conservative field. Then:

\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}

Green’s Theorem

Let $C$ be a simple, closed curve oriented positively. Let $D$ be the region it surrounds and let $P, Q: D \rarr \mathbb{R}$ .

\int_C P \, dx + Q \, dy = \iint_D \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \, dx \, dy

Equivalently:

\int_{\partial D} P \, dx + Q \, dy = \iint_D \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \, dx \, dy

Parametric surface

A parametric surface is a region, $S \subset \mathbb{R}^3$ , which is the image of a function, $r: D \rarr \mathbb{R}^3$ , defined on a region $D \subset \mathbb{R}^2$ of the plane.

Which means we can parameterize:

\vec{r}(u, v) = \langle x(u, v), y(u, v), z(u, v) \rangle

Tangent planes to surfaces.

Suppose $S$ is a parametric surface with parameterization $r: D \subset \mathbb{R}^2 \rarr \mathbb{R}^3$ .

\vec{r}(u, v) = x(u, v) \vec{i} + y(u, v) \vec{j} + z(u, v) \vec{k}

We define $\vec{r_u}(u, v)$ as:

\vec{r_u} = x_u(u, v) \vec{i} + y_u(u, v) \vec{j} + z_u(u, v) \vec{k}

And for the partial derivative of $v$ :

\vec{r_v} = x_v(u, v) \vec{i} + y_v(u, v) \vec{j} + z_v(u, v) \vec{k}

The tangent plane, $\pi$ , to $S$ , at the point $r(u_0, v_0)$ is the plane that contains $r_u(u_0, v_0)$ and $r_v(u_0, v_0)$ .

Equivalently, it is the plane perpendicular to the cross product of $r_u(u_0, v_0)$ and $r_v(u_0, v_0)$ , which goes through $r(u_0, v_0)$ .

Surface area

Let $r: D \rarr \mathbb{R}^3$ be a parameterization for a surface, $S$ .

\text{Area}(S) = \iint_D | \vec{r_u} \times \vec{r_v} | \, du \, dv.

Surface integrals

Let $r: D \rarr \mathbb{R}^3$ be a parameterization of a surface, $S$ .

Let $f: S \rarr \mathbb{R}$ be a function, then:

\iint_S f(x, y, z) \, dS = \iint_D f(\vec{r}(u, v)) | \vec{r_u} \times \vec{r_v} | \, du \, dv

Surface integrals over vector fields

Let $S$ be a parametric surface, parameterized by, $\vec{r}: D \rarr \mathbb{R}^3$ and $F: \mathbb{R}^3 \rarr \mathbb{R}^3$ , be a vector field. Then:

\iint_S \vec{F} \cdot dS = \iint_D \vec{F}(\vec{r}(u, v)) \cdot (\vec{r_u} \times \vec{r_v}) \, du \, dv.

We call this the flux of $\vec{F}$ through $S$ .

The Divergence Theorem

For a vector field, $\vec{F}: \mathbb{R}^3 \rarr \mathbb{R}^3$ , the divergence is:

\operatorname{div} \vec{F} = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} + \dfrac{\partial R}{\partial z}.

Let $S$ be a hollow surface and let $E$ be the 3-dimensional region it surrounds ( $\partial E = S$ ).

Then:

\iint_S \vec{F} \cdot dS = \iiint_E \operatorname{div} \vec{F} \, dV.

Equivalently:

\iint_{\partial E} \vec{F} \cdot dS = \iiint_E \operatorname{div} \vec{F} \, dV.

Stokes’ Theorem

So far we have seen Green’s Theorem as well as The Divergence theorem. But what is the reason that Green’s Theorem only works in $\mathbb{R}^2$ ?

Well, Stokes’ Theorem is essentially Green’s Theorem but in $\mathbb{R}^3$ .

For $\vec{F}: \mathbb{R}^3 \rarr \mathbb{R}^3$ , we define its curl as:

NB: it is also called rot and sometimes denoted as $\nabla \times \vec{F}$ .

\operatorname{curl} \vec{F} = \left\langle \dfrac{\partial R}{\partial y} - \dfrac{\partial Q}{\partial z}, \dfrac{\partial P}{\partial z} - \dfrac{\partial R}{\partial x}, \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \right\rangle.

If $S \subset \mathbb{R}^3$ is a surface, surrounded by a simple, closed curve in $\mathbb{R}^3$ . Then:

\int_C \vec{F} \cdot dr = \iint_S \operatorname{curl} \vec{F} \cdot dS.