MAT232: Calculus of Several Variables

Notes for MAT232 Summer 2025

This page contains tutorial notes for MAT232H5 Calculus of Several Variables Summer 2025 under the instruction of Dr. Elio Mazzeo.

The notes are completed as of June 13, 2025. If you have any feedback based on your experience over the course of this semester, please share it using this link, which contains 8 short questions. Your input would be greatly appreciated—thank you!

Week 1

Week 1 tutorial notes Here

Week 2

Week 2 tutorial notes Here

Parametric Equations

Suppose there is a third variable \(t\) (called a parameter), we can parameterize \(x\) and \(y\) by the equations \[\begin{align*} x &= f(t) \\ y &= y(t) \end{align*}\] where \(a \le t \le b\) which has initial point \((f(a), g(a))\) and terminal point \((f(b), g(b))\).

Now, we would like to find derivatives and compute arclengths of the parametric curves, much like we did with functions. Suppose that \(f,g\) are differentiable functions and we need to find the tangent line at a point on the parametric curve \(x = f(t)\) and \(y = g(t)\), where y is also a differentiable function of \(x\). If \(dx/dt \ne 0\), then the slopes of the parametric curve is given by \[\begin{align*} \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \end{align*}\]

Furthermore suppose that \(y\) is a twice differentiable function of \(x\), then at any point \(dx/dt \ne 0\) and \[\begin{align*} \frac{d^2y}{dx^2} &= \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{ \dfrac{d}{dt} \left( \dfrac{dy}{dx} \right) }{ \dfrac{dx}{dt} } \end{align*}\]

Suppose that a curve C can be described by the parametric equations \(x = f(t)\) and \(y = g(t)\), where \(\alpha \le t \le \beta\). Additionally, suppose that \(f', g'\) are continuous in \(\alpha \le t \le \beta\). Then the length of C is given by \[\begin{align*} L = \int_{\alpha}^{\beta} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \end{align*}\]

Week 3

Week 3 tutorial notes Here

Polar Coordinates

So far we have been working exclusively in the Cartesian/rectangular coordinate system. As we will see in the near future, this is not always the easiest coordinate system to work in. In this section, we will introduce the polar coordinate system.

\[\begin{align*} x &= r \cos{\theta}, \hspace{0.5cm} y = r \sin{\theta} \\ r^2 &= x^2 + y^2, \hspace{0.5cm} \tan{\theta} = \frac{y}{x} \end{align*}\]

The second line of equation is used to help convert back and forth from the Cartesian coordinate system and the polar coordinate system.

To find a tangent line to the polar curve \(r = f{(\theta)}\), we regard \(\theta\) as a parameter and write its parametric equations as

\[\begin{align*} x = r \cos{\theta} = f(\theta) \cos{\theta}, \hspace{0.5cm} y = r \sin{\theta}\sin{(\theta)} \\ \end{align*}\]

Then following the equation of slope for parametric curves and the product rule,

\[\frac{dy}{dx} = \frac{dy/d \theta}{dx/ d \theta} = \frac{\frac{dr}{d \theta} \sin{\theta} + r \cos{\theta}}{\frac{dr}{d \theta} \cos{\theta} - r \sin{\theta}}\]

We can locate horizontal tangents by finding points where \(dy/d\theta = 0\) given that \(dx/d\theta \ne 0\). Likewise, vertical tangents can be found where \(dx/d\theta = 0\) given that \(dy/d\theta \ne 0\).

Equation of a Parabola, Ellipse, Hyperbola, and Sphere

Parabola: \[\begin{align*} x^2 = 4py, \hspace{0.5cm} \text{or} \hspace{0.5cm} y^2 = 4px \\ \end{align*}\]

Ellipse: \[\begin{align*} \frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1, \hspace{0.5cm} \text{where} \hspace{0.5cm} a \ge b > 0 \hspace{0.5cm} \text{or} \\ \frac{x^2}{b^2} + \frac{y^2}{a^2} &= 1, \hspace{0.5cm} \text{where} \hspace{0.5cm} a \ge b > 0 \end{align*}\]

Hyperbola: \[\begin{align*} \frac{x^2}{a^2} - \frac{y^2}{b^2} &= 1, \hspace{0.5cm} \text{where} \hspace{0.5cm} a \ge b > 0 \hspace{0.5cm} \text{or} \\ \frac{y^2}{b^2} - \frac{x^2}{a^2} &= 1, \hspace{0.5cm} \text{where} \hspace{0.5cm} a \ge b > 0 \end{align*}\]

Sphere: \[\begin{align*} (x-h)^2 + (y-k)^2 + (z-l)^2 = r^2 \hspace{0.5cm} \text{where the center is located at $(h, k, l)$ with radius $r$} \end{align*}\]

Linear Algebra in 3D (Part I)

In 3D space, a point is represented as \((x, y, z)\).

An equation in terms of \(x\) and \(y\) defines a curve in the 2D space \(\mathbb{R}^2\), while an equation in terms of \(x, y\) and \(z\) defines a surface in 3D space \(\mathbb{R}^3\).

Given two points \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\), the distance between them is: \[\begin{align*} |PQ| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \\ \end{align*}\]

A vector is a quantity with both magnitude and direction. Given two points \(P\) and \(Q\), the vector from \(P\) to \(Q\) is defined by the difference of their coordinates: \[\begin{align*} \bar{a} = \langle x_2 - x_1,\, y_2 - y_1,\, z_2 - z_1 \rangle \end{align*}\]

To find the magnitude (or norm) of a vector \(\vec{a} = (a_1, a_2, a_3) \in \mathbb{R}^3\), we compute: \[|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\] The standard basis vectors in \(\mathbb{R}^3\) is given by \[ \vec{i} = \langle 1, 0, 0 \rangle, \quad \vec{j} = \langle 0, 1, 0 \rangle, \quad \vec{k} = \langle 0, 0, 1 \rangle \]

This is particularly useful as any vector in \(\mathbb{R}^3\) can be written as: \[\vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k}\]

A unit vector is any vector with magnitute of \(1\). The standard basis is an example of a unit vector. To convert any vector into a unit vector: \[\vec{u} = \frac{\vec{a}}{|\vec{a}|} = \frac{\vec{a}}{\|\vec{a}\|} = \frac{\vec{a}}{\sqrt{a_1^2 + a_2^2 + a_3^2}}\]

Week 4

Week 4 tutorial notes Here

Linear Algebra in 3D (Part II)

In this section, we continue from last week’s discussion about linear algebra in \(\mathbb{R}^3\) by introducing the dot product, cross product, and the forms of equations of lines and planes.

Let \(\vec{a} = \langle a_1, a_2, a_3 \rangle\) and \(\vec{b} = \langle b_1, b_2, b_3 \rangle\). The dot product is given by \[\vec{a} \cdot \vec{b} = \langle a_1, a_2, a_3 \rangle \cdot \langle b_1, b_2, b_3 \rangle = a_1 b_1 + a_2 b_2 + a_3 b_3\]

Alternatively, if \(\theta\) is an angle between non-zero vectors \(\vec{a}\) and \(\vec{b}\), then \[\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)\]

Note

Theorem: Orthogonality and Dot Product.

Two non-zero vectors are \(\vec{a}\) and \(\vec{b}\) are orthogonal if and only if \(\vec{a} \cdot \vec{b} = 0\)

Let \(\vec{a} = \langle a_1, a_2, a_3 \rangle\) and \(\vec{b} = \langle b_1, b_2, b_3 \rangle\). Then the cross product is given by \[\vec{c} = \vec{a} \times \vec{b} = \langle a_2 b_3 - a_3 b_2,\ a_3 b_1 - a_1 b_3,\ a_1 b_2 - a_2 b_1 \rangle\] Note that this formula comes from the determinant of a matrix

\[\vec{c} = \vec{a} \times \vec{b} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{array} \right| = \vec{i} \left| \begin{array}{cc} a_2 & a_3 \\ b_2 & b_3 \\ \end{array} \right| - \vec{j} \left| \begin{array}{cc} a_1 & a_3 \\ b_1 & b_3 \\ \end{array} \right| + \vec{k} \left| \begin{array}{cc} a_1 & a_2 \\ b_1 & b_2 \\ \end{array} \right|\]

Note

Theorem: Orthogonality and Cross Product.

The vector \(\vec{c}\) is orthogonal to both vectors \(\vec{a}\) and \(\vec{b}\).

Note

Theorem: Geometric Interpretation of the Cross Product.

If \(0 \le \theta \le \pi\) is an angle between \(\vec{a}\) and \(\vec{b}\), then \[|\vec{a} \times \vec{b}| = |\vec{a}|\, |\vec{b}| \sin(\theta)\]

Furthermore, \(\vec{a}\) and \(\vec{b}\) are parallel if and only if \(\vec{a} \times \vec{b} = \vec{0}\).

We now look at the equation of a line in \(\mathbb{R}^3\). Note that the equation \(y = mx+b\) does not describe a line anymore in \(\mathbb{R}^3\), but rather a plane. For now, we introduce two ways to describe a line in \(\mathbb{R}^3\)

The vector form of the equation of a line is given by

\[r = \vec{r}_0 + t \vec{v} = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle\] where

• \(\vec{r}\): a general position vector to any point on the line.
  
• \(\vec{r}_0 = \langle x_0, y_0, z_0 \rangle\): a fixed point (starting point) on the line.
  
• \(\vec{v} = \langle a, b, c \rangle\): a direction vector that gives the direction of the line.
  
• \(t \in \mathbb{R}\): a scalar parameter.

Alternatively, the line can also be represented using its parametric equations, given by:

\[\begin{aligned} x &= x_0 + ta \\ y &= y_0 + tb \\ z &= z_0 + tc \end{aligned}\]

where

• \((x_0, y_0, z_0)\): a known point on the line (starting point).
  
• \((a, b, c)\): direction vector components — the direction in which the line extends.
  
• \(t \in \mathbb{R}\): a scalar parameter.

Note that the parametric equations can be derived by equating each component from the vector form of the line.

Lastly, we look at the equation of a plane in \(\mathbb{R}^3\). We also introduce two ways to describe a plane.

The vector equation of the plane is

\[\vec{n} \cdot \langle \vec{r} - \vec{r}_0 \rangle = \vec{0}\]

where

• \(\vec{n} = \langle a, b, c \rangle\): a normal vector perpendicular to the plane.
  
• \(\vec{r} = \langle x, y, z \rangle\): a general position vector to a point on the plane.
  
• \(\vec{r}_0 = \langle x_0, y_0, z_0 \rangle\): a known point on the plane.

The scalar equation of the plane is \[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]

Note

Example: Vector and Scalar Equations.

Given: \[\vec{r}_0 = \langle 1, 2, 3 \rangle, \quad \vec{n} = \langle 4, -2, 5 \rangle, \quad \vec{r} = \langle x, y, z \rangle\]

The vector equation of the plane is: \[\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0\]

Substitute the vectors: \[\langle 4, -2, 5 \rangle \cdot \left( \langle x, y, z \rangle - \langle 1, 2, 3 \rangle \right) = 0\]

Simplify: \[\langle 4, -2, 5 \rangle \cdot \langle x - 1, y - 2, z - 3 \rangle = 0\]

Dot product gives the scalar equation: \[4(x - 1) - 2(y - 2) + 5(z - 3) = 0 \quad \Rightarrow \quad 4x - 2y + 5z = 21\]

Week 5

Week 5 tutorial notes Here

Vector Functions

A vector function is a function that takes one (or more) variables and returns a vector of functions. We will only discuss the case where the function takes one variable.

In \(\mathbb{R}^3\), a single variable vector function have the form

\[\bar{r}(t) = \langle f(t), g(t), h(t) \rangle = f(t)\bar{i} + g(t)\bar{j} + h(t)\bar{k}\] Let’s now talk about limits, derivatives and integrals with vector functions. For the limit of a vector function,

\[\lim_{t \to a} \bar{r}(t) = \left\langle \lim_{t \to a} f(t), \lim_{t \to a} g(t), \lim_{t \to a} h(t) \right\rangle \quad \newline \text{assuming the limits of all the component functions exist.}\]

Furthermore, we say a vector function \(\bar{r}(t)\) is continuous at \(a\) if \(\lim{t \to a} = \bar{r}(a)\).

Next, let’s talk about the derivatives of vector functions. \[\frac{d\bar{r}}{dt} = \bar{r}'(t) = \lim_{h \to 0} \frac{\bar{r}(t+h) - \bar{r}(t)}{h}\]

Note that the derivative of a vector function is just the vector of the derivatives of its components. That is, \[\bar{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle = f'(t)\bar{i} + g'(t)\bar{j} + h'(t)\bar{k}\]

Note

Theorem: Differential Rules of Vector Functions.

Suppose that \(\vec{u}(t)\) and \(\vec{v}(t)\) are differentiable vector functions, \(c\) is a scalar, and \(f\) is a real-valued function. Then, \[\begin{align} &1.\quad \frac{d}{dt}[\vec{u}(t) + \vec{v}(t)] = \vec{u}'(t) + \vec{v}'(t) \\ &2.\quad \frac{d}{dt}[c \vec{u}(t)] = c \vec{u}'(t) \\ &3.\quad \frac{d}{dt}[f(t)\vec{u}(t)] = f'(t)\vec{u}(t) + f(t)\vec{u}'(t) \\ &4.\quad \frac{d}{dt}[\vec{u}(t) \cdot \vec{v}(t)] = \vec{u}'(t) \cdot \vec{v}(t) + \vec{u}(t) \cdot \vec{v}'(t) \\ &5.\quad \frac{d}{dt}[\vec{u}(t) \times \vec{v}(t)] = \vec{u}'(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v}'(t) \\ &6.\quad \frac{d}{dt}[\vec{u}(f(t))] = f'(t)\vec{u}'(f(t)) \\ \end{align}\]

Given how neatly the limit and derivative rules extend to vector-valued functions, it’s no surprise that similar rules apply to definite (and indefinite) integrals as well. \[\begin{align} \int_a^b \vec{r}(t) , dt &= \left( \int_a^b f(t) , dt \right)\vec{i} + \left( \int_a^b g(t) , dt \right)\vec{j} + \left( \int_a^b h(t) , dt \right)\vec{k} \\ &= \left( F(t) \big|_a^b \right)\vec{i} + \left( G(t) \big|_a^b \right)\vec{j} + \left( H(t) \big|_a^b \right)\vec{k} \\ &= \left( F(b) - F(a) \right)\vec{i} + \left( G(b) - G(a) \right)\vec{j} + \left( H(b) - H(a) \right)\vec{k} \end{align}\]

Finally, the arc length formula extends naturally to vector-valued functions and is given by: \[\begin{align} L &= \int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2 + [h'(t)]^2} \, dt \\ &= \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt \\ &= \int_a^b \left| \vec{r}\,'(t) \right| \, dt \end{align}\]

Week 6

Week 6 tutorial notes Here

Multivariable Functions

A function \(f\) of two variables assigns a unique real number \(f(x,y)\) to each ordered pair \((x, y)\) within a set \(D\). This set \(D\) is called the domain of the function. The range of \(f\) consists of all output values the function takes, that is \[\{f(x,y) | (x,y) \in D\}\]

The graph of \(f\) is the set of all points \((x, y, z) \in R^3\) such that \(z = f(x, y)\) and \((x, y) \in D\).

To visualize \(z = f(x,y)\), one method is to set the equation \(f(x, y) = k\) where \(k\) is a constant. Geometrically, they are like elevation lines on a topographic map where each level marks a specific height.

Derivatives of Multivariable Functions

For a function \(f(x, y)\), the first partial derivatives with respect to \(x\) and \(y\), denoted \(f_x\) and \(f_y\), are defined as follows (provided that the limit(s) exists): \[ f_x(x, y) = \lim_{h \to 0} \frac{f(x + h, y) - f(x, y)}{h} \qquad f_y(x, y) = \lim_{h \to 0} \frac{f(x, y + h) - f(x, y)}{h} \]

These definitions are analogous to the derivative in single-variable calculus, but here we vary only one variable at a time while keeping the other fixed.

There are many alternate notations for partial derivatives. Below are some most common ones used

\[ \begin{aligned} f_x(x, y) = f_x &= \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( f(x, y) \right) = z_x = \frac{\partial z}{\partial x} = D_x f \\ f_y(x, y) = f_y &= \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( f(x, y) \right) = z_y = \frac{\partial z}{\partial y} = D_y f \end{aligned} \]

To compute the derivative with respect to one variable \(x\), we regard the other variable \(y\) as constant and differentiate \(f(x,y)\) with respect to \(x\); the same applies when differentiating with respect to \(y\).

When computing derivatives of second order, we use the following notation which can be extended to three or more variables. These are sometimes called mixed partial derivatives since we are taking derivatives with respect to more than one variable. \[ \begin{aligned} (f_x)_x = f_{xx} = f_{11} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 z}{\partial x^2} \\ (f_y)_x = f_{yx} = f_{21} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 z}{\partial x \partial y} \\ (f_x)_y = f_{xy} = f_{12} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 z}{\partial y \partial x} \\ (f_y)_y = f_{yy} = f_{22} &= \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) = \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 z}{\partial y^2} \end{aligned} \]

Note that the notation might be confusing at first glance, since sometimes it is read from left to right and others right to left.

In general, it is not true that \(f_{xy} = f_{yx}\). However the below theorem states when this is satisfied.

Note

Theorem: Clairaut’s Theorem.

Suppose that \(f\) is defined on a disk \(D\) that contains the point \((a, b)\). If the functions \(f_{x,y}\) and \(f_{y, x}\) are both continuous on \(D\), then \[f_{xy}(a,b) = f_{yx}(a,b)\]

Lastly, if we suppose that \(f\) has continuous partial derivatives, then an equation of the tangent plane to the surface \(z = f(x, y)\) at the point \(P(x_0, y_0, z_0)\) is \[z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\]

Week 7

Week 7 tutorial notes Here

Chain Rule

Let’s look at the chain rule for multivariable functions. There are a couple of cases.

Case 1: \(z = f(x, y)\), \(x = g(t)\), \(y = h(t)\) and compute \(\frac{dz}{dt}\).

This case is analogous to the standard chain rule from MAT135. Since \(z\) is a function of only \(t\) through \(x\) and \(y\), the chain rule for this case is \[\frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}\]

Case 2: \(z = f(x, y)\), \(x = g(s, t)\), \(y = h(s, t)\) and compute \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\).

In this case, \(z\) is a function of both \(t\) and \(s\) through \(x\) and \(y\). So it makes sense that we would have two partial derivatives. The chain rule for these two cases are \[\frac{\partial z}{\partial s} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial s} \qquad \frac{\partial z}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}\]

General case: Suppose that \(u\) is a differentiable function of \(n\) variables \(x_1, x_2, \dots, x_n\), and each \(x_j\), where \(j = 1, 2, \cdots, n\) is a differentiable function of the \(m\) variables \(t_1, t_2, \cdots, t_m\). Then \(u\) is a function of \(t_1, t_2, \cdots, t_m\) and

\[\frac{\partial u}{\partial t_i} = \frac{\partial u}{\partial x_1} \frac{\partial x_1}{\partial t_i} + \frac{\partial u}{\partial x_2} \frac{\partial x_2}{\partial t_i} + \ldots + \frac{\partial u}{\partial x_n} \frac{\partial x_n}{\partial t_i}\]

for each \(i = 1, 2, \cdots, m\).

Directional Derivative

Recall we looked at the two partial derivatives \(f_x(x,y)\) and \(f_y(x,y)\). These two derivatives represents the rate of change of \(f\) as we vary \(x\) while holding \(y\) constant, and vice versa. We now wish to find the rate of change of \(f\) if we allow both \(x\) and \(y\) to vary simultaneously.

The directional derivative of \(f\) at \((x_0, y_0)\) in the direction of a unit vector \(\vec{u} = \langle a, b \rangle\) is \[D_{\vec{u}} f(x_0, y_0) = \lim_{h \to 0} \frac{f(x_0 + ha,\, y_0 + hb) - f(x_0, y_0)}{h}\] provided the limit exist.

Note

Theorem: Gradient Formula for Directional Derivatives.

If \(f\) is differentiable at \((x_0, y_0)\), then \(f\) has a directional derivative at \((x_0, y_0)\) in the direction of any unit vector \(\vec{u} = \langle a, b \rangle\) and \[ D_{\vec{u}} f(x_0, y_0) = f_x(x_0, y_0)a + f_y(x_0, y_0)b. \]

Gradient

The gradient of a two variable function \(f(x,y)\) is the vector function \(\nabla f\) defined by

\[ \nabla f(x, y) = \langle f_x(x, y), f_y(x, y) \rangle = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} \]

Note that \[D_{\vec{u}} f(x_0, y_0) = \nabla f(x_0, y_0) \cdot \vec{u}\] Again, the directional derivative and the gradient can be extended to functions of three or more variables.

Week 8

Week 8 tutorial notes Here

Maximum and Minimum Values

A function of two variables has a local maximum at \((a, b)\) if \(f(x,y) \le f(a, b)\) for all points \((x, y)\) close to \((a, b)\). One can think of this as that \(f(x, y) \le f(a, b)\) for all points \((x, y)\) in some disc centered at \((a, b)\). The value \(f(a, b)\) is called the local/relative maximum value.

Similarly, if \(f(x,y) \ge f(a, b)\) when \((x, y)\) is near \((a, b)\), we call \(f(a, b)\) a local/relative minimum.

Note

Theorem: First Derivative Test for Local Extrema.

If \(f\) has a local maximum or local minimum at \((a, b)\) and the first order partial derivatives of \(f\) exist at \((a, b)\), then \(f_x(a, b) = 0\) and \(f_y(a, b)=0\)

The point \((a, b)\) is a critical point of \(f\) if one of the following is true:

• \(\nabla f(a,b) = \vec{0}\)
  
• \(f_x(a,b)\) and/or \(f_y(a,b)\) does not exist
  

Second Derivative Test

Suppose that the second partial derivatives of \(f\) are continuous on a disk with center \((a, b)\) and \((a, b)\) is a critical point of \(f\). Define \[D = f_{xx}(a,b) f_{yy}(a,b) - [f_{xy}(a,b)]^2\]

Then we can classify the critical points into the following cases:

• If \(D > 0\) and \(f_{xx}(a,b) > 0\), then \((a,b)\) is a relative minimum.
• If \(D > 0\) and \(f_{xx}(a,b) < 0\), then \((a,b)\) is a relative maximum.
• If \(D < 0\), then \((a, b)\) is a saddle point.
• If \(D = 0\), then it is inconclusive (ie. \((a, b)\) may be a relative minimum, relative maximum, or saddle point). Other techniques would need to be used.

Extreme Value Theorem

If \(f\) is continuous on some close and bounded set \(D\) in \(\mathbb{R}^2\), then there are points in \(D\), say \((x_1, y_1)\) and \((x_2, y_2)\), so that \(f(x_1, y_1)\) is the absolute maximum and \(f(x_2, y_2)\) is the absolute minimum of the function in \(D\).

To find the absolute maximum and minimum values of a continuous function \(f\) on \(D\) involves the following steps:

• Find the values of \(f\) at the critical points of \(f\) in \(D\)
• Find the extreme values of \(f\) on the boundaries of \(D\)
• The largest of the values from the previous steps is the absolute maximum; the smallest is the absolute minimum.

Lagrange Multipliers

In this section we take a look at another method for optimizing a function subject to given constraints.

The basic setup is that, we wish to optimize \(f(x, y, z)\) given the constraint \(g(x, y, z) = k\). We then have the following workflow.

Step 1: Find all values of \(x, y, z\) and \(\lambda\) such that \[\nabla f(x,y,z) = \lambda \nabla g(x,y,z), \quad g(x,y,z) = k\] Step 2: Plug in all solutions \((x, y, z)\) from the previous step into \(f(x,y,z)\) to identify the maximum and minimum values.

Note this method can also be applied for a function with two variables.

Week 9

Week 9 tutorial notes Here

Double Integral

The double integral of a function \(f\) over a rectangle \(R = [a, b] \times [c, d]\) is defined as the limit of a Riemann sum: \[ \iint_R f(x, y)\, dA = \lim_{m,n \to \infty} \sum_{i=1}^m \sum_{j=1}^n f(x_{i,j}^*, y_{i,j}^*) \, \Delta A \] where \((x_{i,j}^*, y_{i,j}^*)\) are sample points in \(\mathbb{R}\), and \[\Delta A = \Delta x \Delta y = \left( \frac{b - a}{m} \right)\left( \frac{d - c}{n} \right)\] If \(f(x, y) \ge 0\) for all points in \(\mathbb{R}\), then the volume \(V\) of the solid lying above the rectangle \(R\) and below the surface \(z = f(x, y)\) is: \[ V = \iint_R f(x, y)\, dA \]

Iterated Integrals

Note that the above definition is hard to compute in practice. So let us introduce a more intuitive way to compute double integrals. We will continue to assume that we wish to integrate \(f(x,y)\) over a rectangular region \(R = [a, b] \times [c, d]\).

Then, we can write \[\int_a^b \int_c^d f(x, y)\, dy\, dx = \int_a^b \left[ \int_c^d f(x, y)\, dy \right] dx\] or \[\int_c^d \int_a^b f(x, y)\, dx\, dy = \int_c^d \left[ \int_a^b f(x, y)\, dx \right] dy\]

Note

Theorem: Fubini’s Theorem .

If \(f\) is continuous on \(R = [a, b] \times [c, d]\) then, \[\iint_R f(x, y)\, dA = \int_a^b \int_c^d f(x, y)\, dy\, dx = \int_c^d \int_a^b f(x, y)\, dx\, dy\]

Note that, if \(f(x,y) = f(x) g(y)\) can be written as a product like such, then we can rewrite our integral as \[\iint_R f(x, y)\, dA = \iint_R g(x)\, h(y)\, dA = \left( \int_a^b g(x)\, dx \right) \left( \int_c^d h(y)\, dy \right)\]

Over General Regions

Now we proceed with integrating over regions which might not necessarily be a rectangle. We consider two cases.

In the first case, we are interested in integrating over the domain \[D = \{(x, y) \mid a \le x \le b, \, g_1(x) \le y \le g_2(x)\}\]

The corresponding integral is \[\iint_{D} f(x,y) \, dA = \int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \, dx\] In the second case, the domain is \[D = \{(x, y) \mid h_1(y) \le x \le h_2(y), \, c \le y \le d\}\] The corresponding integral is \[\iint_{D} f(x,y) \, dA = \int_{c}^{d} \int_{h_1(y)}^{h_2(y)} f(x,y) \, dx \, dy\]

Week 10

Week 10 tutorial notes Here

Double Integrals in Polar Coordinates

In some cases, it will be easier to describe the region in terms of polar coordinates. In these cases, it will be harder to describe using rectangular coordinates. A simple example to visualize is when \(D\) is given by a disc with radius of \(2\).

Recall that \[x = r \cos \theta, \space y = r \sin \theta, \space x^2 + y^2 = r^2\] Then, the conversion is given by \[\iint_D f(x, y) \, dA = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} f(r \cos \theta, r \sin \theta) \, r \, dr \, d\theta\] Note it is important not to forget to multiply by the \(r\) term during the conversion. The relation \(dA = r dr d \theta\) can be seen through the change of variables method to be discussed soon.

The Jacobian

Suppose \(T(u,v) = (x,y)\) is a one-to-one transformation from the \(u,v\) plane onto the \(x,y\) plane, then it has an inverse transformation \(T^{-1}\) from the \(x,y\) plane to the \(u,v\) plane represented by \(T^{-1}\). This transformation is given by \[\dfrac{\partial(x,y)}{\partial(u,v)} = \det \begin{pmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{pmatrix}\]

Week 11

Week 11 tutorial notes Here

Change of Variables

Back in MAT135, we discussed the method of u-substitution which allowed us to take an integral in terms of \(x\) and change it in terms of \(u\). Now, we wish to do something similar, but in terms of multivariable functions. The rationale behind using the change of variables technique is similar to that of switching to polar coordinates: it transforms a complex region of integration into a simpler one to work with.

Suppose that \(T\) is a one-to-one transformation, given by \(x = g(u, v)\), \(y = h(u,v)\), with nonzero Jacobian that maps the region \(S\) in the uv-plane onto the region \(R\) in xy-plane. If \(f\) is continuous on \(R\), then \[\iint_R f(x, y) \, dA = \iint_S f\big(g(u, v), h(u, v)\big) \left| \frac{\partial(x, y)}{\partial(u, v)} \right| \, dudv\]

Vector Fields

Let’s begin by introducing a definition that will be vital in the following weeks.

A vector field on three dimensional space is a function \(\vec{F}\) that assigns to each point \((x,y,z)\) a three dimensional vector given by \(\vec{F}(x,y)\). Note this definition also applies in the two dimensional case.

Gradient

Given \(f\) to be a scalar function of three variables, then its gradient \(\nabla f\) is defined by \[\nabla f = f_x(x,y,z) \bar{i} + f_y(x,y,z) \bar{j} + f_z(x,y,z) \bar{k}\], where \(\nabla f\) is a vector in \(\mathbb{R}^3\).

Week 12

Week 12 tutorial notes Here

Line Integrals (in space)

Let \(C\) be a smooth curve parameterized by \(\vec{r}(t) = \langle x(t), y(t), z(t) \rangle\), for \(t \in [a, b]\).
Then the line integral of a scalar function \(f\) along curve \(C\) is defined by:

\[ \int_C f(x, y, z) \, ds = \int_a^b f(\vec{r}(t)) \left\| \frac{d\vec{r}}{dt} \right\| \, dt \]

where

\[ \left\| \frac{d\vec{r}}{dt} \right\| = \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \]

Note that the above can also be applied for functions of two variables.

Line Integrals in Vector Fields

Instead of line integrals of functions, we will now focus on evaluating line integrals of vector fields. Let the vector field be given by \[\vec{F}(x, y, z) = P(x, y, z)\,\vec{i} + Q(x, y, z)\,\vec{j} + R(x, y, z)\,\vec{k}\] and the three-dimensional smooth curve by \[\vec{r}(t) = x(t)\,\vec{i} + y(t)\,\vec{j} + z(t)\,\vec{k} \qquad a \leq t \leq b\]

The line integral of \(\vec{F}\) along curve \(C\) is \[\int_C \vec{F} \cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}\,'(t) \, dt = \int_C \vec{F} \cdot \vec{T} ds\] where \(\vec{T}(t)\) is the unit tangent vector given by \[\vec{T}(t) = \frac{\vec{r}\,'(t)}{\left\| \vec{r}\,'(t) \right\|}\]

The latter part of the line integral in the above definition is equivalent to the middle part, which can be seen simply by subbing in \(\vec{T}(t)\).

The Fundamental Theorem for Line Integrals

In MAT135 we learned about the Fundamental Theorem of Calculus, which gave us a way to evaluate integrals. It turns out a similar version exists for line integrals, when evaluated over certain vector fields.

Given a conservative vector field \(\vec{F}\) along a smooth curve \(C\) given by \(\vec{r}(t) = \left\langle x(t), y(t), z(t) \right\rangle\) for \(a \le t \le b\), then our line integral can be evaluated as

\[ \int_C \nabla f \cdot d\vec{r} = f(\vec{r}(b)) - f(\vec{r}(a)) \]

This means the line integral of \(\vec{F}\) along \(C\) depends only on the endpoints, not on the path taken.

Independent Path

A line integral \[\int_C \vec{F} \cdot d\vec{r}\] is said to be independent of path if the value of the integral only depends on the starting and ending points of the curve \(C\), not on the specific route taken between them.