MAT232: Calculus of Several Variables
This page contains tutorial notes for MAT232H5 Calculus of Several Variables Summer 2025 under the instruction of Dr. Elio Mazzeo.
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 \[\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)\]
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
\[\vec{c} = \vec{a} \times \vec{b} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{array} \right| = \vec{i} \left| \begin{array}{cc} a_2 & a_3 \\ b_2 & b_3 \\ \end{array} \right| - \vec{j} \left| \begin{array}{cc} a_1 & a_3 \\ b_1 & b_3 \\ \end{array} \right| + \vec{k} \left| \begin{array}{cc} a_1 & a_2 \\ b_1 & b_2 \\ \end{array} \right|\]
Theorem: Orthogonality and Cross Product.
The vector \(\vec{c}\) is orthogonal to both vectors \(\vec{a}\) and \(\vec{b}\).
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 \[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]
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, \[\bar{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle = f'(t)\bar{i} + g'(t)\bar{j} + h'(t)\bar{k}\]
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 is a rule that assigns to each ordered pair of real numbers \((x, y)\) in a set \(D\) a unique real number denoted by \(f(x, y)\). The set \(D\) is the domain of \(f\) and its range is the set of values that \(f\) takes on, 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 \(f(x, y)\), we define its first partial derivatives \(f_x\) and \(f_y\) as \[ f_x(x, y) = \lim_{h \to 0} \frac{f(x + h, y) - f(x, y)}{h} \qquad f_y(x, y) = \lim_{h \to 0} \frac{f(x, y + h) - f(x, y)}{h} \] provided that the limit(s) exists. Note its similarity to the definitions of the single variable case.
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.
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
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.
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
If \(f\) is a function of two variables \(x\) and \(y\), then the gradient of \(f\) 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 \(f(x,y) \le f(a, b)\) when \((x, y)\) is near \((a, b)\). 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 number \(f(a, b)\) 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.
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.