Mathematics for Engineering

J. Gordon Leishman

doi:https://doi.org/10.15394/eaglepub.2022.1066.n8

4 Mathematics for Engineering

Introduction

“Omnia de mathematica agitur!“^[1] Indeed, many will say that mathematics is the engineer’s language, so the first thing students need to do before they can study engineering is to learn (or relearn) the essential mathematics. Learning mathematics is a lifelong endeavor for most engineers, but it goes with the territory. While not all types of engineering problem-solving require the use of mathematics, the structure and rigor that mathematics brings to engineering analyses usually shed much light on the eventual solution to any given problem. Mathematics enables engineers to understand and predict the behavior of physical systems, deal with uncertainties, and find optimal solutions to complex problems.

Readers of this ebook will likely be already acquainted with the essential mathematics required to move into various engineering disciplines. However, they will still need to review fundamental mathematical concepts to progress successfully. Geometry, algebra, trigonometry, calculus, and vectors give engineers the essential mathematical tools that keep track of processes and can be used to solve problems. Differential equations, for example, also arise in many aspects of engineering problem-solving. To this end, all engineers must become well-versed in solving various types of differential equations, analytically and numerically. In particular, when doing arithmetical and algebraic problems, engineers must do them cold without making any mistakes. Errores in engineeringo plerumque sunt catastrophes.^[2]

In engineering, quantities with both magnitude and direction are often required, such as forces, velocities, accelerations, and their manipulations involving scalar and vector products. Vector quantities are most conveniently expressed using vector notation, a shorthand version of the corresponding scalar equations. For example, the equations of motion for a fluid are often written as vector equations. Therefore, engineers must become comfortable using vector quantities, including shorthand versions of other vector operators such as the gradient operator, the Laplace operator, and the substantial derivative. Exercitatio perfectos efficit.^[3]

Learning Objectives

Remember the essential mathematics that may have already been forgotten.
Review algebraic manipulations using some simple examples.
Revise the rules of calculus (differentiation and integration) with examples.
Comfortably use the concepts associated with vectors.
Understand the meaning of the primary vector operations such as scalar product, cross-product, gradient, divergence, and curl.
Know how to interpret line integrals, surface integrals, and volume integrals.
Appreciate the meaning of the substantial derivative and the Laplace operator.
Know about the need to use ordinary and partial differential equations in engineering.
Review the numerical rules of accuracy, significant digits, and rounding.

Algebra

There is a lot of algebra in engineering. Algebra is a branch of mathematics that deals with symbols and the arithmetic operations of these symbols. Like most things, doing algebra successfully is an acquired skill that takes practice. A review session for students taking their first classes in engineering might start with them writing down several algebraic equations and going about rearranging or simplifying them. Moving on might require testing their proficiency in homework problems or exams.

Check Your Understanding #1 – Review problems in algebra

Solve for $x$ in the equation $4(x - 2) = 16$ .
Simplify the expression given by $(2x^2 - 3x + 1) - (x^2 + 2x - 5)$ .
Solve the equation given by $2(x + 3) = 8 - x$ .
Simplify the expression $5(2x + 3y) - 2(3x - y)$ .
Solve for $x$ in the expression $\displaystyle{ \frac{{x^2 + 3x - 10}}{{x + 2}}} = 0$ .

Show solution/hide solution.

Distributing the 4, then $4x - 8 = 16$ . Adding 8 to both sides gives $4x = 24$ . Finally, dividing both sides by 4, then $x = 6$ .
Removing parentheses and combining like terms gives $x^2 - 5x + 6$ .
Distributing the 2 gives $2x + 6 = 8 - x$ . Combining like terms gives $3x + 6 = 8$ . Subtracting 6 from both sides gives $3x = 2$ . Finally, dividing both sides by 3 gives $x = 2/3$ .
Distributing the coefficients gives $10x + 15y - 6x + 2y$ . Combining like terms leads to $4x + 17y$ .
First, notice that the expression is undefined when $x = -2$ because it would result in a division by zero. So, $x = -2$ is not a valid solution. First, factor the numerator to get
$\frac{{(x + 5)(x - 2)}}{{x + 2}} = 0$

Then, each factor can be set equal to zero, i.e.,

$x + 5 = 0 \mbox{~or~} x - 2 = 0$

Solving each equation separately gives the solution to the original expression as $x = -5$ and $x = 2$ .

Calculus

Engineering requires calculus, and in the use of calculus, there is a need to know how to do differentiation and integration. Both processes have rules. Success with calculus means first learning the rules, i.e., the mechanical processes usually taught to engineers by the mathematics department. But to succeed with calculus in engineering means much more, in that the process of doing the mathematics must also be physically interpreted. For example, differentiation is akin to finding a rate of change, a slope, or a gradient, and integration is about finding a sum, such as an area, a volume, or a contribution of something that acts or is distributed along a particular path or area. The tangible interpretation of mathematics within the framework of engineering is an acquired skill, but it is a skill needed to succeed as an engineer.

Ordinary Differentiation

Finding the change in one thing with respect to another is the essence of differentiation. The fundamental rules of differentiation are:

1. Power Rule:

(1) $\begin{equation*} \frac{d}{dx}(x^n) = n x^{n-1}~\mbox{~where $n$ is a constant.} \end{equation*}$

2. Constant Rule:

(2) $\begin{equation*} \frac{d}{dx}(c) = 0~\mbox{~where $c$ is a constant.} \end{equation*}$

3. Sum/Difference Rule:

(3) $\begin{equation*} \frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx} (f(x)) \pm \frac{d}{dx} (g(x))~\mbox{~where ${\scriptstyle{f}}$ and $g$ are functions.} \end{equation*}$

4. Product Rule:

(4) $\begin{equation*} \frac{d}{dx}(f(x)g(x)) = f'(x) \, g(x) + f(x) \, g'(x)~\mbox{~where ${\scriptstyle{f}}$ and $g$ are functions.} \end{equation*}$

5. Quotient Rule:

(5) $\begin{equation*} \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x) \, g(x) - f(x) \, g'(x)}{(g(x))^2}~\mbox{~where ${\scriptstyle{f}}$ and $g$ are functions.} \end{equation*}$

6. Chain Rule:

(6) $\begin{equation*} \frac{d}{dx}(f(g(x))) = f'(g(x)) \, g'(x)~\mbox{~where ${\scriptstyle{f}}$ and $g$ are functions.} \end{equation*}$

Having laid down these basic rules, which are the first things to learn, they are necessary but insufficient for engineering. Interpreting and understanding the meaning of these processes is more challenging, but it becomes the only way to be successful in engineering problem-solving. Experienced engineers take these “rules” for granted and do the mathematics valor facialis.^[4]

Check Your Understanding #2 – Review problems using differentiation

Find the derivative of the function $f(x) = 3x^2 + 2x - 1$ .
Find the derivative of the function $g(x) = \sin 2x + \cos x$ .
Find the derivative of the function $h(x) = \dfrac{2}{x^3} \,\mbox{.}$
Find the derivative of the function $y = \ln(3x^2 + 1)$ .
Find the derivative of the function $f(x) = \sqrt{x^3 + 2x}$ .

Show solution/hide solution.

By taking the derivative term by term, then

$f'(x) = \frac{df}{dx} = 6x + 2$

2. Applying the chain rule leads to

$g'(x) = \frac{dg}{dx} = 2\cos 2x - \sin x$

3. Using the power rule gives

$h' (x) = \frac{dh}{dx} = -6x^{-4}$

4. Applying the chain rule gives

$y' (x) = \frac{dy}{dx} = \frac{6x}{3x^2 + 1}$

5. Using the power rule and the chain rule leads to

$f'(x) = \frac{df}{dx} = \frac{3x^2 + 2}{2\sqrt{x^3 + 2x}}$

Partial Differentiation

Partial differentiation is a mathematical operation used to find the rate of change of a multivariable function with respect to one specific variable while keeping all other variables constant. In the context of a function of two variables, $f(x, y)$ , the partial derivative of ${\scriptstyle{f}}$ with respect to $x$ is denoted as $\partial f / \partial x$ , and this represents the rate of change of ${\scriptstyle{f}}$ with respect to $x$ while treating $y$ as a constant. Similarly, the partial derivative of ${\scriptstyle{f}}$ with respect to $y$ is denoted as $\partial f / \partial y$ , which represents the rate of change of ${\scriptstyle{f}}$ with respect to $y$ while treating $x$ as a constant.

In general, to compute a partial derivative, the function is differentiated with respect to the variable of interest, treating all other variables as constants. Partial differentiation is a fundamental tool in calculus. It is extensively used in engineering to analyze how a function changes with respect to specific variables in complex systems with multiple independent variables or degrees of freedom. Indeed, as will become apparent in later chapters of this book, using partial derivatives in engineering becomes everyday practice rather than exceptional.

Check Your Understanding #3 – Determining partial derivatives

Find the partial derivatives of the function $f(x, y) = 2x^2 + 3xy + y^2$ .

Show solution/hide solution.

First, differentiate the function with respect to $x$ while treating $y$ as a constant, giving

$\frac{{\partial f}}{{\partial x}} = 4x + 3y$

Next, differentiate the function with respect to $y$ while treating $x$ as a constant, giving

$\frac{{\partial f}}{{\partial y}} = 3x + 2y$

Integration

Integration is a summation often explained as the opposite of differentiation, i.e., anti-differentiation. Integration is more complex than differentiation, but it comes up in some form in most engineering problem-solving. Integration techniques involve using the power rule, trigonometric and logarithmic integrals, and substitution. Most engineers only remember some of the rules of integration; after all, they want to be something other than pure mathematicians. Still, they can look them up as standard integrals in books, or now they can be solved online using AI (artificial intelligence) such as WolfgramAlpha or ChatGTP. The ability to successfully use calculus should never be guesswork. Labora callide, non fortiter.^[5]

The fundamental rules of integration are:

1. Power Rule:

(7) $\begin{equation*} \int x^n \, dx = \frac{1}{n+1}x^{n+1} + c~\mbox{~where $n$ is $\ne$ $-1$, and $c$ is a constant.} \end{equation*}$

2. Constant Multiple Rule:

(8) $\begin{equation*} \int cf(x) \, dx = c\int f(x) \, dx~\mbox{~where $c$ is a constant.} \end{equation*}$

3. Sum/Difference Rule:

(9) $\begin{equation*} \int (f(x) \pm g(x)) \, dx = \int f(x) \, dx \pm \int g(x) \, dx~\mbox{~where ${\scriptstyle{f}}$ and $g$ are functions.} \end{equation*}$

4. Integration by Parts:

(10) $\begin{equation*} \int u \, dv = uv - \int v \, du~\mbox{~where ${u}$ and ${v}$ are functions.} \end{equation*}$

5. Substitution Rule:

(11) $\begin{equation*} \int f( g(x))\, g'(x) \, dx = \int f(u) \, du~\mbox{~where $u = g(x)$.} \end{equation*}$

Again, the rules are the first things to learn, but the meaning of the integration processes must also be understood in engineering.

Check Your Understanding #4 – Review problems using integration

Evaluate the integral $\displaystyle{ \int (3x^2 + 2x - 1)} \, dx$ .
Evaluate the integral $\displaystyle{\int (2\cos(2x) - \sin x) } \, dx$ .
Evaluate the integral $\displaystyle{\int -6x^{-4}} \, dx$ .
Evaluate the integral $\displaystyle{ \int \frac{6x}{3x^2 + 1} }\, dx$ .
Evaluate the integral $\displaystyle{ \int \frac{3x^2 + 2}{2\sqrt{x^3 + 2x}} } \, dx$ .

Show solution/hide solution.

1. Integrating term by term gives

$\int (3x^2 + 2x - 1) \, dx = \frac{3}{3}x^3 + \frac{2}{2} x^2 - x + c = x^3 + x^2 - x + c$

where $c$ is the constant of integration.

2. Integrating each term separately leads to

$2\sin(2x) + \cos x + c$

3. Applying the power rule of integration gives

$2x^{-3} + c$

4. Using the substitution $u = 3x^2 + 1$ , the integral becomes

$\int \frac{1}{2} \, du$

Integrating then gives

$\frac{1}{2}u + c$

By substituting back $u = 3x^2 + 1$ , the final result is

$\int \frac{6x}{3x^2 + 1} \, dx = \frac{1}{2}(3x^2 + 1) + c$

and simplifying the expression gives

$\int \frac{3x^2}{2\sqrt{x^3 + 2x}} \, dx + \int \frac{2}{2\sqrt{x^3 + 2x}} \, dx$

5. Applying appropriate substitutions and integration techniques gives

$\int \frac{3x^2 + 2}{2\sqrt{x^3 + 2x}} \, dx = \frac{2}{3}(x^3 + 2x)^{3/2} + 2\sqrt{x^3 + 2x} + c$

Working with Vectors

A vector is like a direction, force, or something else with a magnitude and direction. Vectors come up everywhere in engineering. Si vectores nescis, tunc errabis. Vectors are represented by arrows with a length proportional to the magnitude, and the orientation denotes the direction. Vectors are also used to point to something in space, called a position vector. The location of an arbitrary point in space, say $P$ , can be defined by specifying the values of three coordinates, i.e., in terms of $(x, y, z)$ in a standard Cartesian coordinate system. As shown in the figure below, the point $P$ can be located by the position vector, $\vec{r}$ , where

(12) $\begin{equation*} \vec{r} = x \, \vec{i} + y \, \vec{j} + z \, \vec{k} \end{equation*}$

Definition of a position vector $\vec{r}$ in a Cartesian coordinate system.

In general, if $\vec{A}$ is a given vector in a Cartesian coordinate system and $A_x, A_y$ , and $A_z$ are the components of $\vec{A}$ in the $x,y$ and $z$ directions, then

(13) $\begin{equation*} \vec{A} = \vec{A_{x}} + \vec{A_{y}} + \vec{A_{z}} = A_{x}\, \vec{i} + A_{y}\, \vec{j} + A_{z}\, \vec{k} \end{equation*}$

as shown in the figure below.

The position of a point in space is defined in terms of the scalar components in a Cartesian coordinate system.

Vectors can be velocities (a speed in a given direction), a force, an acceleration, or something else. Vectors can be added to find a total effect (the total velocity, force, acceleration, etc.), done component-by-component, nose to tail. If $\vec{A}$ is directed from point $P_1$ to point $P_2$ and a second vector $\vec{B}$ is defined as

(14) $\begin{equation*} \vec{B} = B_{x}\, \vec{i} + B_{y}\, \vec{j} + B_{z}\, \vec{k} \end{equation*}$

that points from $P_2$ to $P_3$ , then the resultant vector $\vec{C}$ that points from $P_1$ to $P_3$ is

(15) $\begin{equation*} \vec{C} = (A_{x} + B_{x}) \, \vec{i} + (A_{y} + B_{y}) \, \vec{j} + (A_{z} + B_{z})\, \vec{k} \end{equation*}$

The process of adding vectors is to connect them “nose-to-tail.”

Alternative coordinate systems may describe any problem, e.g., in a cylindrical or spherical coordinate system instead of a Cartesian coordinate system. Usually, deciding on a suitable coordinate system is a matter of convenience, and it is also important to determine whether or not the mathematics can be appropriately simplified in an alternative coordinate system. For example, in introductory engineering problems, Cartesian coordinate systems are used primarily, but sometimes polar coordinates are used when convenient. It is best to start with Cartesian coordinates to understand the rules of mathematics.

Scalar and Vector Fields

A scalar quantity given as a function of coordinate space $x, y, z$ (and perhaps time $t$ ) is called a scalar field. For example, fluid properties such as pressure $p$ , density $\varrho$ , and temperature $T$ are all scalar quantities, i.e.,

(16) $\begin{eqnarray*} p & = & p\left( x, y,z \right) \\[6pt] \varrho & = & \varrho\left( x, y, z \right) \\[6pt] T & = & T\left( x, y, z \right) \end{eqnarray*}$

Similarly, a vector quantity given as a function of coordinate space and time is called a vector field. e.g., velocity is a vector quantity. A velocity can be written in terms of its scalar components, i.e.,

(17) $\begin{equation*} \vec{V} = V_{x} \, \vec{i} + V_{y} \, \vec{j} + V_{z} \, \vec{k} \end{equation*}$

where the components are

(18) $\begin{eqnarray*} V_{x} & = & V_{x} \left( x, y, z \right) \\[6pt] V_{y} & = & V_{y} \left( x, y, z \right) \\[6pt] V_{z} & = & V_{z} \left( x, y, z \right) \end{eqnarray*}$

More often than not, the velocity vector is written as

(19) $\begin{equation*} \vec{V} = u\, \vec{i} + v \, \vec{j} + w \, \vec{k} \end{equation*}$

where ${u}$ , ${v}$ and ${w}$ are the components in the $x$ , $y$ and $z$ direction, respectively.

Similar expressions can be written for cylindrical and spherical coordinates. In many theoretical aerodynamic and other engineering problems, various scalar and vector fields will be unknowns that must subsequently be obtained in a solution with prescribed initial and other boundary conditions.

Scalar Products

Let the vectors $\vec{A}$ and $\vec{B}$ be given by

(20) $\begin{eqnarray*} \vec{A} & = & A_{x}\, \vec{i} + A_{y}\, \vec{j} + A_{z} \, \vec{k} \\[6pt] \vec{B} & = & B_{x}\, \vec{i} + B_{y}\, \vec{j} + B_{z} \, \vec{k} \end{eqnarray*}$

Then the scalar or “dot” product $\vec{A} \bigcdot\vec{B}$ or “ $\vec{A}$ dot $\vec{B} \,$ ” is given by

(21) $\begin{equation*} \vec{A}\bigcdot\vec{B} = A_x B_x + A_y B_y + A_z B_z \end{equation*}$

Notice that the “centered dot” or “ $\bigcdot$ ” represents the scalar product operator.

As shown in the figure below, physically, the scalar product is the projection of one vector onto another, i.e., the component of one vector in the direction of the other. For example, the scalar projection (or scalar component) of a vector $\vec{A}$ in the direction of a vector $\vec{B}$ is given by

(22) $\begin{equation*} \vec{A}\bigcdot\vec{B} = A_x B_x + A_y B_y + A_z B_z = | \vec{A} | | \vec{B} | \cos \theta \end{equation*}$

where $\theta$ is the angle between the vectors $A$ and $B$ .

The physical interpretation of a scalar product is the projection of one vector onto another.

Notice that because $| \vec{A} | \cos \theta$ is the projection of $\vec{A}$ onto $\vec{B}$ , then $\vec{A} \bigcdot \vec{B}$ is equal to the projection of $\vec{A}$ onto $\vec{B}$ times the magnitude of $\vec{B}$ . Similarly, because $| \vec{B} | \cos \theta$ is the projection of $\vec{B}$ onto $\vec{A}$ , then $\vec{A} \bigcdot \vec{B}$ also equals the projection of $\vec{B}$ onto $\vec{A}$ times the magnitude of $\vec{A}$ . However, the easiest way to remember what the dot product is just $\vec{A} \bigcdot \vec{B} = | \vec{A} | | \vec{B} | \cos \theta$ .

Need some help with scalar products? Here is a short video lesson on how to perform scalar products and an interpretation of the physical meaning of this vector operation from Dr. Leishman’s “Math & Physics Hints and Tips” series.

Vector Products

The vector product $\vec{A} \times \vec{B}$ (cross product) is given by

(23) $\begin{equation*} \vec{A} \times \vec{B} = | \vec{A} | | \vec{B} | \, \vec{n} \end{equation*}$

where the vector product or “cross product” is denoted by the $\bigtimes$ operator, $\theta$ is the angle between the two vectors, and $\vec{n}$ is a unit vector perpendicular to the plane containing the two vectors, as shown in the figure below. Notice that the cross-product is also the area of the shaded parallelogram. Like the dot or scalar product, the concept of a cross-product has many uses and applications in engineering.

The physical interpretation of a vector product is another vector perpendicular to the other two.

In general, the vector product can also be written as

(24) $\begin{equation*} \vec{A} \times \vec{B} = \left| \begin{array}{ccc} \, \vec{i} & \, \vec{j} & \, \vec{k} \\[6pt] A_x & A_y & A_z \\[6pt] B_x & B_y & B_z \end{array} \right| \end{equation*}$

Expanding this operation out in terms of the third-order determinant gives

(25) $\begin{equation*} \vec{A}\times \vec{B} = \left( A_y B_z - B_y A_z \right) \, \vec{i} +\left( B_x A_z - A_x B_z \right) \, \vec{j} + \left( A_x B_y - B_x A_y \right) \, \vec{k} \end{equation*}$

Similar expressions for $\vec{A}\bigcdot\vec{B}$ and $\vec{A}\times\vec{B}$ can be written in cylindrical and spherical coordinate systems.

Need some help with vector products? Here is a short video lesson on how to perform vector products as well as the interpretation of the physical meaning of this vector operation from Dr. Leishman’s “Math & Physics Hints and Tips” series.

Check Your Understanding #5 – Review problems with vector operations

Given two vectors $\vec{A} = -3 \, \vec{i} + 2 \, \vec{j}$ and $\vec{B} = 5 \, \vec{i} - 2 \, \vec{j}$ , then evaluate:

$\mbox{~~~}\vec{A} - 3\vec{B}$
$\mbox{~~~}\vec{A} \bigcdot \vec{B}$
$\mbox{~~~}\vec{A} \times \vec{B}$

Show solution/hide solution.

$\mbox{1.} \quad \vec{A} - 3\vec{B} = \left( -3 \, \vec{i} + 2 \, \vec{j} \right) - 3\left( 5 \, \vec{i} - 2 \, \vec{j} \right) = -18 \, \vec{i} + 8 \, \vec{j}$

$\mbox{2.} \quad \vec{A} \bigcdot \vec{B} = \left(-3 \, \vec{i} + 2 \, \vec{j}\right) \bigcdot \left( 5 \, \vec{i} - 2 \, \vec{j} \right) = \left( -15 - 4 \right) = -19$

$\mbox{3.} \quad \vec{A} \times \vec{B} = \left| \begin{array}{cccc} \vec{i} & \vec{j} & \vec{k} \\[6pt] -3 & 2 & 0 \\[6pt] 5 & -2 & 0 \end{array} \right| = 0 \, \vec{i} + 0 \, \vec{j} + (6 - (-10)) \, \vec{k} = 16 \, \vec{k}$

Differentiating a Vector

To differentiate a vector, the vector must depend on time, i.e., its magnitude changes. Therefore, the vector’s derivative will be its time rate of change. For example, let the vector be $\vec{A} (t)$ , a position in space. The time derivative of $\vec{A}$ will be a velocity, i.e.,

(26) $\begin{equation*} \vec{V} = \frac{ d \vec{A}}{dt} = \left( \frac{ d A_x}{dt} \right) \, \vec{i} + \left( \frac{ d A_y}{dt} \right) \, \vec{j} + \left( \frac{ d A_z}{dt} \right) \, \vec{k} = v_ x \, \vec{i} + v_y \, \vec{j} + v_z \, \vec{k} = \left( v_x, v_y, v_z \right) \end{equation*}$

Therefore, the derivative of a vector is a vector.

In other possible engineering examples, if $\vec{A} (t)$ were the momentum, then the time rate of change of momentum would be a force. If the parameter were to be a velocity, then the time rate of change of velocity is acceleration. If the parameter were to be the quantity of work, then the time derivative would be power. Therefore, differentiating a vector can lead to valuable outcomes for engineering applications.

Check Your Understanding #6 – Differentiating a vector

Differentiate the vector function $\vec{A}(t) = \big( 3t^2, \cos(t), \ln(t) \big)$

Show solution/hide solution.

To differentiate this vector function, it is necessary to take the derivative of each component with respect to $t$ , i.e.,

$\frac{d \vec{A}}{dt} = \frac{d}{dt} \left( 3t^2, \cos(t), \ln(t) \right) = \left( \frac{d}{dt} (3t^2), \, \frac{d}{dt}(\cos(t) ), \, \frac{d}{dt} (\ln(t)) \right)$

By simplifying the derivatives of each component, then

$\frac{d \vec{A}}{dt} = (6t, \, -\sin(t), \, 1/t)$

This example demonstrates how to differentiate a vector function component-wise; the result is another vector.

Line Integrals

Consider a vector field $\vec{A} = \vec{A} \left( x, y, z \right)$ . Also, consider a curve in space connecting two points $a$ and $b$ , as shown in the figure below. Let $ds$ be an elemental length of the curve.

The line integral along the curve from $a$ to $b$ is

(27) $\begin{equation*} \int_{b}^{a}\vec{A}\bigcdot d\vec{s} \end{equation*}$

which is simply a statement that the integral is the sum of the components of the vector field along the direction of the curve $a$ to $b$ .

If the curve is closed, i.e., points $a$ and $b$ are coincident, then the line integral is

(28) $\begin{equation*} \oint_{b}^{a}\vec{A}\bigcdot d\vec{s} = \oint_{C}\vec{A}\bigcdot d\vec{s} \end{equation*}$

where the counterclockwise direction is considered positive according to the convention used in the field of mathematics, as shown in the figure below.

A closed-loop line integral sums the effects around a closed path and frequently occurs in engineering practice.

Check Your Understanding #7 – Line integrals

Evaluate the integral $p (x, y) = x^2 + y$ along the curve $C$ defined by the straight line segment from point (0, 0) to (2, 4).

Show solution/hide solution.

To calculate the line integral, the curve $C$ can be parameterized by

$x(t) = t ~\mbox{~~and~~} y(t) = 2t$

where $t$ varies from 0 to 1, corresponding to the endpoints of the line segment. The line integral is given by

$\int_C p(x, y) \, ds = \int_0^1 (t^2 + 2t) \sqrt{\left(\frac{{dx}}{{dt}}\right)^2 + \left(\frac{{dy}}{{dt}}\right)^2} \, dt$

Substituting the parameterizations and simplifying them gives

$\int_0^1 (t^2 + 2t) \sqrt{1^2 + 2^2} \, dt = \int_0^1 (t^2 + 2t) \sqrt{5} \, dt$

And then evaluating the integral gives

$\int_C p(x, y) \, ds = \left[\frac{{1}}{{3}}t^3 + t^2\right]_0^1 \, \sqrt{5} = \frac{{8}}{{3}}\sqrt{5}$

Surface Integrals

Consider now a surface $S$ bounded by curve $C$ , as shown in the figure below. Let $dS$ be an elemental surface area and let $\vec{n}$ be a unit normal vector, i.e., one perpendicular to the surface. Let $p$ be a scalar field in this space and $\vec{A}$ a vector field. A surface integral over the surface can be defined as

(29) $\begin{equation*} \iint_{S}\,\, p\, d\vec{S} \quad \mbox{or} \quad \iint_{S}\, \vec{A}\bigcdot d\vec{S} \end{equation*}$

where $d\vec{S} = \vec{n} \, dS$ is called the elemental vector area or the unit normal vector area. If the surface is closed, then the integral is written as

(30) $\begin{equation*} \oiint_{S} \, p \, d\vec{S} \quad \mbox{or} \quad \oiint_{S} \, \vec{A}\bigcdot d\vec{S} \end{equation*}$

Line integrals can also be based on scalar functions, e.g., the scalar $p(x, y)$ integrated along a curve $C$ given by

$\int_C \, p (x, y) \, ds$

where $ds$ represents an infinitesimal length along the curve $C$ .

Unit Normal Vector

The meaning of a unit normal vector, also known as a normal unit vector or simply a normal vector, should be understood because it comes up in many engineering problems. The unit normal is a vector perpendicular (or orthogonal) to a surface at a particular point. It has a magnitude of 1, hence the term “unit.” The normal vector points always away from the surface in a perpendicular direction, as shown in the figure below.

The unit normal vector is often denoted by the symbol $\vec{n}$ . For a surface defined by a function $\phi (x, y, z) = 0$ , where $(x, y, z)$ are the coordinates of a point on the surface, then the unit normal vector at that point can be obtained by taking the gradient of the function and normalizing it, as given by

(31) $\begin{equation*} \vec{n} = \cfrac{ \left( \displaystyle{\cfrac{\partial \phi}{\partial x}} , \, \displaystyle{\cfrac{\partial \phi}{ \partial y}} , \, \displaystyle{\cfrac{\partial \phi}{ \partial z}} \right) }{\left| \left( \displaystyle{\cfrac{\partial \phi}{ \partial x}} , \, \displaystyle{\cfrac{\partial \phi}{ \partial y}} , \, \displaystyle{\cfrac{\partial \phi}{ \partial z}} \right) \right| } \end{equation*}$

where $|~~|$ denotes the Euclidean L² norm or magnitude of the vector. By normalizing the gradient vector, the resulting unit normal vector will have a magnitude of 1. This normalization is achieved by dividing the gradient vector by its magnitude.

Different methods exist to find the unit normal from discrete surface points, such as the cross-product approach or the least-squares fitting procedure. The cross-product approach is the easiest. The steps are:

Select three neighboring points on the surface defined by the vectors $\vec{P}_1 = (x_1, y_1, z_1)$ , $\vec{P}_2 = (x_2, y_2, z_2)$ and $\vec{P}_3 = (x_3, y_3, z_3)$ .
Calculate the vectors between these points, i.e., $\vec{A}_1 = \vec{P}_2 - \vec{P_1}$ and $\vec{A}_2 = \vec{P}_3 - \vec{P_1}$ .
Compute the cross product of these vectors giving $\vec{N} = \vec{A}_1 \times \vec{A}_2$ .
Normalize the resulting vector to obtain the unit normal vector, i.e., $\vec{n} = \dfrac{\vec{N}}{|| \vec{N} ||}$ .

Remember that the unit normal vector is a crucial concept in vector calculus and is used in various mathematical and physical applications, such as determining the direction of a force on a surface and solving problems involving surface integrals or differential equations.

Check Your Understanding #8 – Calculating unit normal vectors

In a three-dimensional Cartesian coordinate system, what is the unit normal vector for the direction of the x-axis?
Consider a two-dimensional plane defined by $x + y = 1$ . What is the unit normal to this plane?
Consider a three-dimensional plane defined by $2x - 3y + z = 4$ . What is the unit normal to this plane?

Show solution/hide solution.

This vector is parallel to the $x$ -axis and points in the positive $x$ -direction, so $\vec{n} = (1, 0, 0)$ .
In this case, $\phi(x,y) = x + y - 1 = 0$ . Therefore, at any point on this plane, the unit normal vector will be
$\vec{n} = \cfrac{ \left( \displaystyle{\cfrac{\partial \phi}{\partial x}}, \, \displaystyle{\cfrac{\partial \phi}{ \partial y}}, \, \displaystyle{\cfrac{\partial \phi}{ \partial z}} \right) }{ \left| \left( \displaystyle{\cfrac{\partial \phi}{ \partial x}}, \, \displaystyle{\cfrac{\partial \phi}{ \partial y}}, \, \displaystyle{\cfrac{\partial \phi}{ \partial z}} \right) \right| } = \cfrac{ (1, 1, 0)}{\sqrt{2}} = \left( \cfrac{1}{\sqrt{2}}, \, \cfrac{1}{\sqrt{2}}, \, 0 \right)$
In this final case, $\phi(x,y) = 2x - 3y + z - 4 = 0$ . Therefore, at any point on this plane, the unit normal vector will be
$\vec{n} = \cfrac{(2, -3, 1)}{\sqrt{14}} = \left(\cfrac{2}{\sqrt{14}}, \, \cfrac{-3}{\sqrt{14}}, \, \cfrac{1}{\sqrt{14}} \right)$

Volume Integrals

Consider a volume $\cal{V}$ in space, as shown in the figure below, which contains a small elemental volume $d \cal{V}$ . Let $\varrho$ be a scalar field in this space. The volume integral over the volume $\cal{V}$ of the quantity $\varrho$ is written as

(32) $\begin{equation*} \oiiint_{\cal{V}} \varrho \, d\cal{V} \end{equation*}$

The result of the integration is a scalar.

If $\vec{V}$ is a vector field in space, then the volume integral is written as

(33) $\begin{equation*} \oiiint_{\cal{V}} \vec{V} \, d\cal{V} \end{equation*}$

and the result in this latter case will be a vector. If $\vec{V} = (V_1, V_2, V_3)$ , then the volume integral of $\vec{V}$ over a region $\cal{V}$ is evaluated using

(34) $\begin{equation*} \oiiint_{\cal{V}} \vec{V} \, d{\cal{V}} = \left( \oiiint_{\cal{V}} V_1 \, d{\cal{V}}, \ \oiiint_{\cal{V}} V_2 \, d{\cal{V}}, \ \oiiint_{\cal{V}} V_3 \, d{\cal{V}} \right) \end{equation*}$

which is a vector representing the accumulated effect of the vector field throughout the entire volume.

Need to brush up on your understanding of integration? Here is a short video lesson on the integration processes from Dr. Leishman’s “Math & Physics Hints and Tips” series.

Integral Relations

The need to know some important relations between line, surface, and volume integrals occasionally arises when solving engineering problems. Consider an area $S$ bounded by a closed curve $C$ . Let $\vec{A}$ be a vector field. The line integral of $\vec{A}$ over $C$ is related to the surface integral of $\vec{A}$ over $S$ by using Stokes’ Theorem, i.e.,

(35) $\begin{equation*} \oint_{C}\vec{A}\bigcdot d\vec{S} = \oiint_S \left( \nabla \times \vec{A} \right) \times d \vec{S} \end{equation*}$

Again, consider the volume $\cal{V}$ enclosed by the closed surface $S$ . Then, the surface and volume integrals of the vector field $V$ are related by using Gauss’s divergence theorem, i.e.,

(36) $\begin{equation*} \oiint_{S} \vec{A}\bigcdot d\vec{S} = \oiiint_{\cal{V}} \left( \nabla \bigcdot \vec{A} \right) \, d\cal{V} \end{equation*}$

which connects a volume integral to a surface integral. Finally, if $\varrho$ represents a scalar field, then the gradient theorem states that

(37) $\begin{equation*} \oiint_{S} \varrho \, d\vec{S} = \oiiint _{\cal{V}} \nabla \varrho \, d\cal{V} \end{equation*}$

Gradient of a Scalar Field

If $p = p\left(x, y, z\right)$ , then the gradient $\nabla p$ (or grad $p$ ) is given by

(38) $\begin{equation*} \mbox{grad } p \equiv \nabla p = \frac{\partial p}{\partial x}\, \vec{i}+ \frac{\partial p}{\partial y}\, \vec{j}+\frac{\partial p}{\partial z}\, \vec{k} \end{equation*}$

where the operator $\nabla$ (which is called nabla or del) is defined as

(39) $\begin{equation*} \nabla = \frac{\partial }{\partial x}\, \, \vec{i}+ \frac{\partial }{\partial y}\, \, \vec{j}+\frac{\partial }{\partial z}\, \, \vec{k} \end{equation*}$

Physically, $\nabla p$ is the normal to the surface defined by $p\left(x, y, z\right)$ = constant, as illustrated in the figure below.

A gradient of a scalar field is a vector, i.e., a vector that points in the direction of the slope of the scalar field.

Related to the gradient is the concept of a directional derivative. Consider some scalar quantity $p\left(x,y\right)$ defined at some point over a surface. Choose some arbitrary direction $s$ away from the point and let $\vec{n}$ be a unit vector in that direction. The rate of change of $p$ per unit length in the $s$ direction is

(40) $\begin{equation*} \frac{dp}{ds} = \nabla p \bigcdot \vec{n} \end{equation*}$

The term $\dfrac{d}{ds}$ is called the directional derivative. Therefore, it can be appreciated that the rate of change of $p$ in any arbitrary direction is simply the component of $\nabla p$ in that particular direction.

Check Your Understanding #9 – Gradient of a scalar field

A two-dimensional pressure field is given by $p(x,y) = -2x - y + 10$ . If an isobar is a constant pressure line, plot the isobars for $p$ = 0, 2, 4, 6, and 8.

Show solution/hide solution.

The equations of the isobars are

$\begin{eqnarray*} 0 & = & -2x - y + 10 \implies y = -2x + 10 \\[6pt] 2 & = & -2x - y + 10 \implies y = -2x + 8 \\[6pt] 4 & = & -2x - y + 10 \implies y = -2x + 6 \\[6pt] 6 & = &-2x - y + 10 \implies y = -2x + 4 \\[6pt] 8 & = & -2x - y + 10 \implies y = -2x + 2 \end{eqnarray*}$

which are all straight lines, as shown in the figure below.
The gradient is given by

$\nabla p = \frac{\partial p}{\partial x} \vec{i} + \frac{\partial p}{\partial y} \vec{j}$

so that

$\nabla (-2x - y + 10) = \frac{\partial (-2x - y + 10)}{\partial x} \vec{i} + \frac{\partial (-2x - y + 10)}{\partial y} \vec{j} = -2 \vec{i} - 1 \vec{j}$

and the magnitude of the pressure gradient is

$| \nabla p | = \sqrt{(-2)^2 + (-1)^2} = \sqrt{5 } = 2.34$

The pressure gradient vector represents the magnitude and direction of the pressure change at a specific point in space. It points toward the steepest change in pressure, and its magnitude indicates the rate of pressure change in that direction. In this case, the pressure gradient vector is given by $(-2, -1)$ , which means that the pressure changes most rapidly in the direction $(-2, -1)$ , and the rate of decrease is 2 units in the $x$ direction and 1 unit in the $y$ direction.

To visualize the pressure gradient vectors, they are typically plotted as arrows; the tail of each arrow is placed at a specific point in space, and the direction and length of the arrows indicate the direction and magnitude of the pressure gradient at that point. Some arbitrary points can be selected, and then the corresponding pressure gradient vectors are plotted, e.g.,

$\mbox{\small Point A: } (x, y) = (2, 4) \implies \mbox{\small $\nabla p$ at A: } (-2, -1)$

$\mbox{\small Point B: } (x, y) = (0, 6) \implies \mbox{\small $\nabla p$ at B: } (-2, -1)$

$\mbox{\small Point C: } (x, y) = (3, -2) \implies \mbox{\small $\nabla p$ at C: } (-2, -1)$

Notice that the gradient vector runs perpendicular to the isobars, and at each of the points, A, B, and C, the pressure gradient vector is the same.

Divergence of a Vector Field

Consider a vector field $\vec{V} = \vec{V} \left(x, y, z \right)$ , then

(41) $\begin{eqnarray*} \mbox{div } \vec{V} = \nabla \bigcdot \vec{V} & = & \left(\frac{\partial }{\partial x}\, \vec{i}+\frac{\partial }{\partial y} \, \vec{j}+\frac{\partial }{\partial z}\, \vec{k}\right) \bigcdot \left(V_{x}\, \vec{i}+V_{y}\, \vec{j}+V_{z}\, \vec{k}\right) \nonumber \\[6pt] & = & \frac{\partial V_{x}}{\partial x}+\frac{\partial V_{y}}{\partial y}+\frac{\partial V_{z}}{\partial z} \end{eqnarray*}$

which is a scalar quantity.

In aerodynamics and fluid dynamics, the divergence can be interpreted as a flux density or the amount of flux entering or leaving a point in space. Mass fluxes such as $\varrho u$ , $\varrho v$ , etc. are often used. Divergence is the rate of flux expansion (positive divergence) or flux contraction (negative divergence). By convention, the flux is defined as positive when it leaves a closed surface. So, the divergence is just the net flux per unit volume or flux density of the contents of a given region of space.

Curl of a Vector Field

Again, consider the vector field defined by

(42) $\begin{equation*} \vec{V} = V_{x}\, \vec{i}+V_{y}\, \vec{j}+V_{z}\, \vec{k} \end{equation*}$

Then the curl of $\vec{V}$ is defined as

(43) $\begin{equation*} \mbox{curl } \vec{V} \equiv \nabla \times \vec{V} = \left| \begin{array}{ccc} \, \vec{i} &\, \vec{j} & \, \vec{k} \\[6pt] \displaystyle{\frac{\partial}{\partial x}} & \displaystyle{\frac{\partial}{\partial y}} & \displaystyle{\frac{\partial}{\partial z}} \\[10pt] V_x & V_y & V_z \end{array} \right| \end{equation*}$

and expanding this determinant gives

(44) $\begin{equation*} \nabla \times \vec{V} = \left( \frac{\partial V_{z}}{\partial y} - \frac{\partial V_{y}}{\partial z} \right) \, \vec{i} + \left( \frac{\partial V_{x}}{\partial z} - \frac{\partial V_{z}}{\partial x} \right) \, \vec{j} + \left( \frac{\partial V_{y}}{\partial x} - \frac{\partial V_{x}}{\partial y} \right) \, \vec{k} \end{equation*}$

A physical interpretation of the curl is that it represents the angular velocity of the rotation or “spin” of the contents of a given region of space. In aerodynamics, the curl of a velocity field is called vorticity, a physical concept that can help interpret the behavior of a flow. If the vorticity is zero, then the flow is said to be irrotational, and if the vorticity is non-zero, then the flow is rotational.

Check Your Understanding #10 – Finding the curl of a vector field

In a solution to a specific engineering problem, the solution of the vector field $\vec{A}(x, y, z)$ is

$\begin{equation*} \vec{A} = xz \, \vec{i} - y^2 \, \vec{j} + 2 x^2y \, \vec{k} \end{equation*}$

Find the curl of this field.

Show solution/hide solution.

The curl of a vector requires the evaluation of a determinant, i.e.,

$\nabla \times \vec{A} = \left| \begin{array}{ccc} \, \vec{i} &\, \vec{j} & \, \vec{k} \\[6pt] \displaystyle{\frac{\partial}{\partial x}} & \displaystyle{\frac{\partial}{\partial y}} & \displaystyle{\frac{\partial}{\partial z}} \\[6pt] A_x & A_y & A_z \end{array} \right|$

Evaluating this determinant gives

$\nabla \times \vec{A} = \left( \frac{\partial A_{z}}{\partial y} - \frac{\partial A_{y}}{\partial z} \right) \, \vec{i} + \left( \frac{\partial A_{x}}{\partial z} - \frac{\partial A_{z}}{\partial x} \right) \, \vec{j} + \left( \frac{\partial A_{y}}{\partial x} - \frac{\partial A_{x}}{\partial y} \right) \, \vec{k}$

After differentiation then

$\nabla \times \vec{A} = \left( 2x^2 - 0 \right) \, \vec{i} + \left( x - 4xy \right) \, \vec{j} + \left( 0 - 0 \right) \, \vec{k}$

or

$\nabla \times \vec{A} = 2x^2 \, \, \vec{i} +( x - 4xy) \, \vec{j} - 0 \, \, \vec{k}$

Physical Laws in Engineering

Taking a temporary break from the mathematics review allows one to recall some of the physical laws used in engineering and appreciate their significance in context with the mathematics. Aerospace engineers must be well-versed in understanding the laws of physics and how these are applied in the form of principles and the corresponding equations. In the following chapters of this ebook, the rules and principles of physics will be well used to explain fluid mechanics and other physical behaviors. If you have yet to read or listen to the Feynman Lectures on physics, you should do so.

Conservation Principles

Conservation principles in physics refer to fundamental laws that describe the preservation or constancy of specific quantities in physical systems. These laws state that specific physical quantities remain constant in isolated systems or undergo specific transformations under certain conditions.

1. Conservation of Mass (Conservatio Massae):

“Mass is neither created nor destroyed in a closed system; it remains constant.“

This principle states the total mass of a closed system remains unchanged in any physical or chemical process, i.e.,

$\sum_{\mbox{\tiny Something}} m_{\rm after} = \sum_{\mbox{\tiny Something}} m_{\rm before}$

where $m$ denotes the mass of something.

2. Conservation of Momentum (Conservatio Momentum):

“The total momentum of an isolated system remains constant if no external forces act upon it.“

Momentum, defined as the product of an object’s mass and velocity, is conserved in a closed system where no external forces are present, i.e.,

$\sum_{\mbox{\tiny Something}} p_{\rm after} = \sum_{\mbox{\tiny Something}} p_{\rm before}$

where $p$ denotes linear momentum.

3. Conservation of Angular Momentum (Conservatio Momenti Angulares):

“The total angular momentum of an isolated system remains constant without external torques.“

Angular momentum, which is related to an object’s rotational motion, is conserved when no external torques are applied to the system, i.e.,

$\sum_{\mbox{\tiny Something}} l_{\rm after} = \sum_{\mbox{\tiny Something}} l_{\rm before}$

where l denotes angular momentum.

4. Conservation of Energy (Conservatio Energiae):

“Energy cannot be created or destroyed; it can only be converted from one form to another or transferred between different objects or systems.“

This principle states that the total energy of an isolated system remains constant over time, i.e.,

$\sum_{\mbox{\tiny Something}} e_{\rm after} = \sum_{\mbox{\tiny Something}} e_{\rm before}$

where e denotes energy.

Newton’s Absolutes

Newton’s absolutes refer to the fundamental concepts introduced by Isaac Newton in his formulation of classical mechanics, specifically in his book Philosophiæ Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy). These concepts include absolute space, absolute time, and absolute motion.

1. Absolute time: According to Newton, absolute time flows uniformly and continuously, independent of any events or objects. It is not influenced by the physical processes occurring within it. This concept implies a universal time that is the same everywhere, providing a consistent measure of the duration of events.

2. Absolute space: Newton posited the existence of a fixed, immovable, and infinite space that exists independently of any objects within it. This space provides a reference frame against which the position and motion of objects can be measured. Absolute space remains constant and unchanging, serving as a basis for all physical events.

3. Absolute motion: Newton distinguished between absolute motion and relative motion. Absolute motion refers to the movement of an object through absolute space, while relative motion is the movement of an object relative to other objects. Absolute motion is measured with respect to the fixed framework of absolute space. Newton’s absolutes were foundational to his laws of motion and gravitational theory. However, Einstein’s theory of relativity later challenged and refined these ideas, demonstrating that space and time are not absolute but interconnected and relative to the observer.

Newton’s Laws

Isaac Newton formulated three fundamental laws of motion in his Philosophiæ Naturalis Principia Mathematica. These laws laid the foundation for explaining classical mechanics and have been essential in understanding the science of many fields. While they may seem simple and intuitive today, these laws provided a paradigm shift in thinking and a rigorous formalism that enabled significant advances in science and engineering.

1. Newton’s First Law of Motion (Law of Inertia). Lex prima motus Newtonii (Lex inertiae):

“An object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an external force.“

Inertia is the property of an object that resists changes in its motion. According to the first law, an object will maintain its state of rest or uniform motion unless a net external force acts on it. This law introduced the concept of inertia and the idea that motion does not require a force unless there is a change in motion.

2. Newton’s Second Law of Motion (Law of Acceleration). Lex secunda motus Newtonii (Lex accelerationis):

“The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.“

This law quantifies the relationship between force, mass, and acceleration, providing the basis for analyzing the dynamics of objects. It can be mathematically expressed as

$F = m \, a$

where $F$ is a force, $m$ is a mass, and $a$ is an acceleration, or more specifically

$\vec{F} = \frac{d}{dt} ( m \, \vec{V} ) = \frac{d}{dt} ( \vec{p} )$

and where $\vec{p} \, ( = m \, \vec{V})$ is the momentum of “something.”

3. Newton’s Third Law of Motion (Law of Action-Reaction). Lex tertia motus Newtonii (Lex actionis et reactionis):

“For every action, there is an equal and opposite reaction.“

According to the third law, whenever an object exerts a force on another object, the second object exerts an equal and opposite force on the first object, i.e.,

$\vec{R} = - \vec{F}$

This law explains the interactions between objects and emphasizes that forces always come in pairs.

Newton’s laws of motion have profoundly influenced the development of science and engineering. They provided a systematic framework for understanding the behavior of objects in motion and at rest, leading to advancements in various fields, including mechanics, aeronautics, astronautics, astronomy, all fields of engineering, physics, and chemistry.

Other Mathematical Operators

Many mathematical operators are fundamental in fields such as calculus, differential equations, and physics, and are also essential tools for describing and analyzing engineering problems. In addition to operators such as the gradient operator ${\nabla}$ , divergence ${\nabla \bigcdot \,}$ , and curl ${\nabla \times \,}$ , the Laplace operator and substantial derivative operator are frequently used.

Laplace Operator

The Laplace operator appears in several of the equations of fluid mechanics and is a second-order differential operator defined as the divergence (i.e., $\nabla \bigcdot$ ) of the gradient (i.e., $\nabla$ ). The Laplace operator can be applied to both scalar and vector fields.

In the case of a scalar field, say $\phi$ , the Laplace operator would be written in Cartesian coordinates as

(45) $\begin{equation*} \nabla^2 \,\phi = \nabla \bigcdot \nabla \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} \end{equation*}$

Of course, the Laplace operator appears in the familiar Laplace equation, i.e.,

(46) $\begin{equation*} \nabla^2 \,\phi = 0 \end{equation*}$

In aerodynamics, the Laplace equation is, in fact, the governing equation for an incompressible, irrotational flow. To this end, the linearity of the Laplace equation is helpful in that if $\phi_1$ and $\phi_2$ are both flow solutions to the Laplace equation, then $\phi_1 + \phi_2$ is also a solution, i.e., the idea that incompressible, irrotational flows can be combined using the principle of linear superposition, which is a significant advantage when dealing with complex flows.

The Laplace operator can also be applied to a vector field, i.e., $\vec{V} = u \, \vec{i} + v \, \vec{j} + w \, \vec{k}$ . In this case, then

(47) $\begin{equation*} \nabla^2 \,\vec{V} = \nabla^2 \,u \, \, \vec{i} + \nabla^2 \,v \, \, \vec{j} + \nabla^2 \,w \, \, \vec{k} \end{equation*}$

where

(48) $\begin{eqnarray*} \nabla^2 \,u & = & \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} \\[6pt] \nabla^2 \,v & = & \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} + \frac{\partial^2 v}{\partial z^2} \\[6pt] \nabla^2 \,w & = & \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2} + \frac{\partial^2 w}{\partial z^2} \end{eqnarray*}$

Check Your Understanding #11 – Finding a Laplacian of a scalar field

A two-dimensional function $\psi (x, y) = x^2 + y^2$ , where ${x}$ and ${y}$ are the Cartesian coordinates. Determine the Laplacian of this function.

Show solution/hide solution.

1. To find the Laplacian, the second partial derivatives must be obtained, followed by taking their sum. Calculate the first partial derivatives:

$\frac{\partial \psi}{\partial x} = 2x \mbox{~~and~~}\frac{\partial \psi}{\partial y} = 2y$

2. Calculate the second partial derivatives:

$\frac{\partial^2 \psi}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial \psi}{\partial x}\right) = \frac{\partial}{\partial x}(2x) = 2$

$\frac{\partial^2 \psi}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{\partial \psi}{\partial y}\right) = \frac{\partial}{\partial y}(2y) = 2$

3. Sum the second partial derivatives to obtain the Laplacian:

$\nabla^2 \,\psi = \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} = 2 + 2 = 4$

So, in this example, the Laplacian of the function $\psi (x, y) = x^2 + y^2$ is 4.

Substantial Derivative Operator

The substantial derivative operator in Cartesian coordinates is written as

(49) $\begin{equation*} \frac{D}{Dt} = \frac{\partial}{\partial t} + \left( \frac{\partial}{\partial x} \right) \left( \frac{\partial x}{\partial t} \right) + \left( \frac{\partial}{\partial y} \right) \left( \frac{\partial y}{\partial t} \right) + \left( \frac{\partial}{\partial z} \right) \left( \frac{\partial z}{\partial t} \right) \end{equation*}$

or because the velocity components are given by

(50) $\begin{equation*} u = \left( \frac{\partial x}{\partial t} \right); \quad u = \left( \frac{\partial x}{\partial t} \right); \quad u = \left( \frac{\partial x}{\partial t} \right) \end{equation*}$

then

(51) $\begin{equation*} \frac{D}{Dt} = \frac{\partial}{\partial t} + u\frac{\partial}{\partial x} + v\frac{\partial}{\partial y} + w\frac{\partial}{\partial z} \end{equation*}$

Recall that vector gradient operator $\nabla$ is defined as

(52) $\begin{equation*} \nabla = \frac{\partial}{\partial x} \, \vec{i} + \frac{\partial}{\partial y} \, \vec{j} + \frac{\partial}{\partial z} \, \vec{k} \end{equation*}$

Hence, the substantial derivative operator can be written as

(53) $\begin{equation*} \frac{D }{Dt} = \frac{\partial }{\partial t} + \vec{V} \bigcdot \nabla \end{equation*}$

The substantial derivative can be applied to any flow field variable, scalar, or vector. For example, if ${u}$ is the component of the velocity field $\vec{V} = (u, v, w)$ then $Du/Dt$ can be written as

(54) $\begin{equation*} \frac{Du}{Dt} = \underbrace{ \frac{\partial u}{\partial t}}_{\rm Local~derivative} \hspace*{-3mm} + \underbrace{ u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + w\frac{\partial u}{\partial z}}_{\rm Convective~derivative} \end{equation*}$

The term $\partial/\partial t$ is called the local derivative, which can be interpreted physically as the time rate of change of a given quantity at a fixed point. The term $\vec{V} \bigcdot \nabla$ is called the convective derivative, which can be interpreted as the time rate of change from the movement of the given quantity from one point to another where its properties are spatially different. It will be apparent that terms like $u (\partial /\partial x)$ have units of time $^{-1}$ .

Notice that for a vector $\vec{V}$ then $D\vec{V}/Dt$ is

(55) $\begin{equation*} \frac{D\vec{V}}{Dt} = \frac{\partial \vec{V}}{\partial t} + u\frac{\partial \vec{V}}{\partial x} + v\frac{\partial \vec{V}}{\partial y} + w\frac{\partial \vec{V}}{\partial z} \end{equation*}$

which, in terms of the scalar components, is

(56) $\begin{equation*} \frac{D\vec{V}}{Dt} = \begin{cases} \dfrac{D u}{Dt} = \dfrac{\partial u}{\partial t} + u\dfrac{\partial u}{\partial x} + v\dfrac{\partial u}{\partial y} + w\dfrac{\partial u}{\partial z} \\[16pt] \dfrac{D v}{Dt} = \dfrac{\partial v}{\partial t} + u\dfrac{\partial v}{\partial x} + v\dfrac{\partial v}{\partial y} + w\dfrac{\partial v}{\partial z} \\[16pt] \dfrac{D w}{Dt} = \dfrac{\partial w}{\partial t} + u\dfrac{\partial w}{\partial x} + v\dfrac{\partial w}{\partial y} + w\dfrac{\partial w}{\partial z} \end{cases} \end{equation*}$

Check Your Understanding #12 – Calculating a substantial derivative

A fluid flows through a pipe, and the temperature distribution is given by $T(x, y, z, t)$ = $10x + 5y - 2z + 3t$ . What is the change in temperature at the point where the fluid velocity is $\vec{V} = (2, 3, -1)$ ?

Show solution/hide solution.

The temporal change and the convective terms involving the gradients and velocity components must be obtained to calculate the substantial temperature derivative at that point. The substantial derivative is

$\frac{D T}{Dt} = \frac{\partial T }{\partial t} + u \frac{\partial T }{\partial x} + v \frac{\partial T}{\partial y} + w\frac{\partial T}{\partial z}$

The contributing terms are

(57) $\begin{equation*} \frac{\partial T }{\partial t} = 3 ; \quad \frac{\partial T }{\partial x} = 10; \quad \frac{\partial T }{\partial y} = 5; \quad \frac{\partial T }{\partial z} = -2 \end{equation*}$

Therefore,

$\frac{D T}{Dt} = 3 + 10 (2) + 5 (3) - 2 (-1) = 40~\mbox{units of temperature per unit time}$

So, the substantial temperature derivative at the point in this example is 40.

Notice that the substantial derivative incorporates both the local temporal change in temperature and the convective transport from fluid motion. In this case, the local temporal change in temperature alone contributes a rate of 3 units per unit of time. The remaining terms involving convective transport, which consider the temperature gradients and the fluid velocity, contribute to a total rate of 37 units per unit of time.

Ordinary Differential Equations (ODEs)

An ordinary differential equation (ODE) is an equation that involves ordinary derivatives of a function. A common engineering problem is solving an ODE, i.e., determining what function (or functions) will satisfy a given differential equation or a set of such equations. For example, given the ODE

(58) $\begin{equation*} \frac{dx}{dt} = \sin t \end{equation*}$

then what will be the function $x(t)$ ?

The antiderivative of $\sin t$ is $-\cos t$ , so

(59) $\begin{equation*} x(t) = -\cos t + c \end{equation*}$

where $c$ is some unknown constant. Therefore, solving an ODE is more complicated than just anti-differentiation because the value of the constant $c$ must also be determined. To this end, additional information, such as an initial or boundary condition, is required. Commonly, the value of $x(t)$ will be known at some specific value of $t$ , e.g., $x(t_0) = x_0$ . In this case, it will be apparent that

(60) $\begin{equation*} x(t) = -\cos t + x_0 \end{equation*}$

for all values of t.

Check Your Understanding #13 – Solving an ODE by the method of separation of variables

Use the method of separation of variables to solve the ODE given by

$\begin{equation*} \frac{dx}{dt} = 5x - 3 \end{equation*}$

to determine the function $x(t)$ .

Show solution/hide solution.

The separation of variables is a method frequently encountered in engineering practice. In this case, the separation of variables gives

$\frac{dx}{5x-3} = dt$

Integrating both sides gives

$\begin{eqnarray*} \int \frac{dx}{5x-3} & = & \int dt \\[6pt] \frac{1}{5} \log |5x-3| & = & t + c \\[6pt] 5x-3 & = & \pm \exp(5 t + 5c ) \\[6pt] x & = & \pm \frac{1}{5}\exp(5t+5c) + 3/5 \end{eqnarray*}$

If this ODE has an initial condition, say $x(2) = 1$ , then $c$ must satisfy

$1 = c \exp \left( (5 (2) \right) + \frac{3}{5}$

So, the constant of integration must be

$c = \frac{2}{5} \exp{(-10)}$

Therefore, the particular solution to this ODE is

$x(t) = \frac{2}{5} \exp \left( 5(t - 2) \right) + \frac{3}{5}$

Partial Differential Equations (PDEs)

Partial differential equations (PDEs) deal with equations that have multiple independent variables and their partial derivatives. PDEs model various engineering phenomena, including heat conduction, fluid dynamics, structural dynamics, elasticity, aeroelasticity, and acoustics.

PDEs can be classified into several types based on their order and linearity. The highest order of the partial derivatives involved determines the order of a PDE. For example, a second-order PDE involves second-order partial derivatives. Linearity refers to whether the equation is linear or nonlinear regarding the unknown function and its derivatives.

Examples of PDEs include:

Heat Equation:

$\frac{\partial u}{\partial t} = \alpha \left( \frac{ \partial^2 u}{ \partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{ \partial^2 u }{ \partial z^2} \right)$

Wave Equation:

$\frac{ \partial^2 u }{ \partial t^2} = a^2 \left (\frac{ \partial^2 u}{ \partial x^2} + \frac{ \partial^2 u} {\partial y^2} + \frac{ \partial^2 u }{\partial z^2} \right)$

Navier-Stokes Equations (incompressible, convective form):

$\frac{ \partial \vec{V} }{\partial t} + (\vec{V} \bigcdot \nabla) \vec{V} = -\frac{1}{\varrho} \nabla p + \nu \nabla^2 \,\vec{V}$

Solving PDEs can be challenging, and different techniques are employed depending on the type of equation and its properties. PDEs are used in incompressible flow theory to help find velocity potentials and stream functions. Analytical methods involve finding exact solutions using techniques such as the separation of variables, Fourier series, or Laplace transforms.

However, analytical solutions are not always possible, especially for nonlinear PDEs or complex boundary conditions. To this end, numerical methods are often used to approximate solutions for PDEs. These methods involve discretizing the domain and approximating the derivatives using finite difference, finite element, or finite volume techniques.

Check Your Understanding #14 – PDEs in incompressible flow theory

In a specific incompressible potential flow, the velocity field $(u,v)$ in the $(x,y)$ plane is given by the pair of partial differential equations

$u = \frac{\partial \phi}{\partial x} = 2x \mbox{~~and~~} v = \frac{\partial \phi}{\partial y} = -2y$

Find the velocity potential $\phi$ .

Show solution/hide solution.

Integrating the first equation with respect to $x$ gives

$\phi = x^2 + f_1(y)$

For the second equation, integrating with respect to $y$ gives

$\phi = -y^2 + f_2(x)$

So, there are two equations

$\begin{eqnarray*} \phi & = & x^2 + f_1(y) \\[5pt] \phi & = & -y^2 + f_2(x) \end{eqnarray*}$

By inspection then $f_1(y) = -y^2$ and $f_2(x) = x^2$ , so the velocity potential is

$\phi = x^2 - y^2$

which are the equations of a set of parabolas. The final result can be verified by partial differentiation of $\phi$ .

Other Coordinate Systems

In engineering, several coordinate systems are used for describing and analyzing spatial relationships and physical phenomena. Cartesian coordinates, using orthogonal axes $x, y, z$ , are natural coordinates and are fundamental for geometric modeling and structural analysis in three-dimensional space. Polar coordinates, defined by radial distance $r$ and azimuthal angle $\theta$ , are useful for applications involving radial symmetry. Cylindrical coordinates expand upon polar coordinates with an added vertical dimension $z$ , making them ideal for radial problems with cylindrical and axial symmetry. Spherical coordinates, incorporating radial distance $r$ , polar angle $\theta$ , and azimuthal angle, $\phi$ , arise in some contexts. Lastly, curvilinear coordinates, encompassing systems like elliptic and parabolic coordinates, also find applications in engineering.

Polar Coordinates

Polar coordinates often arise in engineering problem-solving. A polar coordinate system is a two-dimensional coordinate system in which each point in a plane is defined by a distance from a reference point and an angle from a reference direction. This axis system is advantageous for problems involving radial symmetry, such as fluid dynamics and structural analysis.

In the polar system, the radial coordinate, $r$ , is the distance from the origin at point O to another point, P. The corresponding angular coordinate, $\theta$ , is shown in the figure below; $\theta$ is measured counterclockwise from the positive $x$ -axis to the line connecting the origin to the point.

Basis of a polar coordinate system, which is convenient for specific problems.

Polar coordinates are easily converted into Cartesian coordinates by using the transformation

(61) $\begin{equation*} x = r \cos\theta \quad \text{and} \quad y = r \sin\theta \end{equation*}$

and from Cartesian coordinates to polar coordinates using

(62) $\begin{equation*} r = \sqrt{x^2 + y^2} \quad \text{and} \quad \theta = \tan^{-1}\left(\frac{y}{x}\right) \end{equation*}$

Notice that the differential elements in polar coordinate are $ds = r \, d\theta$ for the arc length and $dA = \dfrac{1}{2} r^2 \, d\theta$ for the area element, both of which frequently occur in engineering problem-solving.

The principal value of the inverse tangent function typically returns an angle in the range $-\dfrac{\pi}{2}\,$ to $\dfrac{\pi}{2}\,$ (or -90° to 90°). The tangent function is periodic, and the inverse function needs a unique range for every output value. The inverse tangent function alone does not account for the correct quadrant of the angle because $\tan \theta$ is positive for both the first and third quadrants and negative for both the second and fourth quadrants. To handle this, the atan2 function is used, which is a function available on all scientific calculators, i.e.,

(63) $\begin{equation*} \theta = \text{atan2}(y, x) \end{equation*}$

This function returns the angle in the correct quadrant by considering the signs of both $x$ and $y$ . It returns the angle and its correct sign in the range $-\pi$ to $\pi$ (or -180° to 180°).

In polar coordinates, the commonly used vector operators are the gradient of a scalar field, i.e.,

(64) $\begin{equation*} \nabla p = \frac{\partial p}{\partial r} \, \hat{e}_r + \frac{1}{r} \frac{\partial p}{\partial \theta} \, \hat{e}_\theta \end{equation*}$

where $p$ is a scalar field, and $\hat{e}_r$ and $\hat{e}_\theta$ are the unit vectors in the $r$ and $\theta$ directions, respectively. The divergence of a vector $\vec{V}$ is

(65) $\begin{equation*} \nabla \bigcdot \vec{V} = \frac{1}{r} \frac{\partial}{\partial r}(r V_r) + \frac{1}{r} \frac{\partial V_\theta}{\partial \theta} \end{equation*}$

Finally, the Laplacian operator is

(66) $\begin{equation*} \nabla^2 p = \frac{1}{r} \frac{\partial}{\partial r} \left( r \, \frac{\partial p}{\partial r} \right) + \frac{1}{r^2} \, \frac{\partial^2 p}{\partial \theta^2} \end{equation*}$

Cylindrical & Spherical Coordinates

Cylindrical coordinates are an axisymmetric extension of polar coordinates. They are helpful for modeling problems involving cylindrical objects like pipes, shafts, and columns. In cylindrical coordinates $(r, \theta, z)$ , each component has a specific role in defining the position of a point in three-dimensional space, as shown in the figure below. Spherical coordinates $(r, \theta, \phi)$ the position of a point in space using three values: radial distance, polar angle, and azimuthal angle.

Definitions of cylindrical and spherical coordinate systems, which are convenient for problems involving radial and axial symmetry.

The radial coordinate $r$ represents the radial distance from the origin to the point projected onto the $(x, y)$ -plane. It is similar to the $r$ in polar coordinates but extends vertically into three dimensions. The azimuthal angle, $\theta$ , is the angle measured counterclockwise from the positive $x$ -axis to the projection of the point onto the $(x, y)$ -plane. The vertical coordinate $z$ denotes the vertical height above (or below) the $(x, y)$ -plane.

Converting from Cartesian to cylindrical coordinates is done by using the transformation

(67) $\begin{equation*} r = \sqrt{x^2 + y^2}, \quad \theta = \tan^{-1}\left(\frac{y}{x}\right), \quad \text{and} \quad z = z \end{equation*}$

and from cylindrical to Cartesian coordinates by using

(68) $\begin{equation*} x = r \cos\theta, \quad y = r \sin \theta, \quad \text{and} \quad z = z \end{equation*}$

In spherical coordinates, the radial distance is often denoted by $\rho$ , but also by $r$ in some contexts, mainly because of symbol conflicts with fluid parameters. The polar angle, $\theta$ , is the angle between the positive $z$ -axis and the line from the origin to the point. The azimuthal angle, $\phi$ , is the angle between the positive $x$ -axis and the projection onto the $(x, y)$ -plane.

Converting from Cartesian to spherical coordinates is performed using

(69) $\begin{equation*} r \equiv \rho = \sqrt{x^2 + y^2 + z^2}, \quad \theta = \cos^{-1}\left(\frac{z}{\rho}\right), \quad \phi = \tan^{-1}\left(\frac{y}{x}\right) \end{equation*}$

and spherical to Cartesian coordinates using

(70) $\begin{equation*} x = \rho \sin\theta \cos\phi, \quad y = \rho \sin\theta \sin\phi, \quad z = \rho \cos\theta \end{equation*}$

Transformations

In mathematics, transformations are operations that map a set of points or coordinates defining a body or some equation(s) to another set of points or equations, altering the position, size, shape, or orientation of the initial values. Transformation operations are fundamental to a wide range of engineering applications.

Translation

One type of transformation is a translation, as shown in the figure below for two-dimensional space, which shifts every point of a body defined in space by the same distance in a given direction. For example, translating a point $P(x, y) = P(1, 2)$ by 3 units to the right and 4 units up, the new coordinates $P'$ will be

(71) $\begin{equation*} P'(x', y') = (x + 3, y + 4) = (4, 6) \end{equation*}$

Translation moves bodies or coordinates without altering their orientation or size.

Rotation

Rotation turns a set of points defining a body about a fixed point through a specified angle. For instance, by rotating $P$ 90 $^{\circ}$ counterclockwise (e.g., about the origin), the new coordinates $P'$ are

(72) $\begin{equation*} P'(x', y') = (-y, x) = (-2, 1) \end{equation*}$

Rotation is a transformation used in solving many types of engineering problems.

Reflection

Reflection flips the set of points defining the shape of a body, thereby producing its mirror image. For example, reflecting $P$ over the $y$ -axis, the new coordinates $P'$ are

(73) $\begin{equation*} P'(x', y') = (-x, y) = (-1, 2) \end{equation*}$

This type of transformation is fundamental in understanding and solving problems that use planar symmetry.

Scaling

Scaling, or dilation, enlarges or reduces a body by a scale factor relative to a fixed point. If point $P$ is scaled by a factor of 2 with respect to the origin, the new coordinates $P'$ are

(74) $\begin{equation*} P'(x', y') = (2x, 2y) = (2, 4) \end{equation*}$

Scaling is essential in similarity transformations, which occur frequently in engineering

Shear

Shear distorts the shape of the body by sliding one part of it parallel to a fixed line. For example, a shear transformation where the $x$ -coordinates are unchanged but the $y$ -coordinates are shifted by twice the $x$ -coordinate value, the new coordinates $P'$ are

(75) $\begin{equation*} P'(x', y') = (x, y + 2x) = (1, 4) \end{equation*}$

Shear transformations occur in engineering problems involving stress and deformation.

Affine Transformations

Affine transformations include translation, rotation, scaling, shear, preserving points, straight lines, and planes. An affine transformation is any transformation that preserves collinearity (i.e., all points lying on a line initially still lie on a line after transformation) and ratios of distances (e.g., the midpoint of a line segment remains the midpoint after transformation). Projective transformations extend affine transformations and include perspective projections. An affine transformation can be represented as

(76) $\begin{equation*} T(\vec{x}) = A\, \vec{x} + \vec{b} \end{equation*}$

where $A$ is a matrix representing the linear part of the transformation, $\mathbf{b}$ is a vector representing the translation, and $\mathbf{x}$ is the input vector.

Transformation in the Lifting Line Theory

The lifting line theory is an aerodynamic method used to analyze finite wings. The lifting line theory uses the $\theta$ transformation to relate the spanwise position along the wing to the angle $\theta$ . The spanwise position $y$ along the wing is transformed into an angular coordinate $\theta$ using

$y = \frac{b}{2} \, \cos \theta = s \, \cos \theta$

where $b$ is the wing span, $s$ is the semi-span, $y$ is the spanwise coordinate, ranging from $-\dfrac{b}{2}$ to $\dfrac{b}{2}$ or $-s$ to $s$ , $\theta$ is the angle in the transformed angular coordinate ranging from $0$ to $\pi$ . This transformation maps the spanwise coordinate to the angular coordinate, making it easier to apply Fourier series solution techniques.

Nonlinear Transformations

Nonlinear transformations can take various forms. Examples include polynomial transformations such as

(77) $\begin{equation*} T(x) = a_0 + a_1 \, x + a_2 \, x^2 + \cdots + a_n \, x^n \end{equation*}$

There are also exponential transformations, such as

(78) $\begin{equation*} T(x) = a \, e^{b \, x} = a \, \exp ({b \, x}) \end{equation*}$

as well as logarithmic transformations, i.e.,

(79) $\begin{equation*} T(x) = a \, \log(b \, x) \end{equation*}$

Working with Numbers

In engineering problems, information is often presented as numbers that must be manipulated to find a final answer. Much engineering work is done with numbers, and it is frequently said that anything else can only be an opinion, i.e., the truth can only be expressed in numbers or data. In this process, questions may be raised about whether it is necessary to round the numbers to a certain number of decimal places and/or significant digits. Common sense rules generally prevail, but some guidelines should be followed.

Rounding Decimals

Sometimes, a number must be rounded to a certain number of decimal places or even a whole number. To this end, some numbers must be removed from the decimals at the end of the number. The method is to look at the value of the following number after the one to stop at, then apply the rules for rounding:

Round up if the first digit to be discarded exceeds, e.g., 3.14159 is 3.1416 to 4 decimal places.
Round down if the first digit to be discarded is below 5. e.g., 3.141592 is 3.14159 to 5 decimal places.
If the first discarded digit is 5, round up if a non-zero digit follows it, but round down if it is followed by a zero. For example, 3.14159 is 3.142 to 3 decimal places.

In multi-step calculations, rounding up or down of numbers is generally permitted only at the end of the calculation; the appropriate number of decimal places (or otherwise the accepted accuracy of the known quantities) must be carried forth and then finally round off the number(s), as needed. However, rounding off intermediate numbers may be avoided altogether in many cases. Instead, the known numbers with an acceptable level of accuracy to a certain number of decimal places are used. However, using numbers to 8 decimal places is usually unnecessary.

So, remember that if rounding off numbers is needed, which is often the case, this is usually done only at the end of the calculation. To this end, common-sense rules with rounding numbers generally prevail. Rounding 9.81 to 9.8 may be acceptable (1 decimal place or two significant digits), but rounding to 10.0 is unacceptable. Rounding 0.002378 to 0.00238 may be fine, but rounding to 0.02 is generally unacceptable. Rounding 3.14156 to 3.142 may be acceptable, but rounding to 3.0 is unacceptable.

Significant Digits

Significant digits are the total number used to express a measured or calculated quantity. Like decimal places, significant digits measure how “precise” a number is given. The value with the fewest significant digits sets the standard for the degree of precision.

Generally, a goal is to carry through numerical calculations containing values that have (more or less) the same number of significant digits to arrive at an answer with the needed (or maximum possible) accuracy. There are also some simple rules for significant digits:

All non-zero digits in a number are always considered significant.
Any zeros appearing in a number between two non-zero digits are considered significant.
A final zero or trailing zeros in the decimal portion are considered significant.

Check Your Understanding #15 – Significant digits

How many significant digits are in 14.638?
How many significant digits are in 0.002378?
How many significant digits are in 273.70?

Show solution/hide solution.

Answer = 5; 2. Answer = 4; 3. Answer = 5

Manipulating Numbers

There are a few basic rules that should be followed when manipulating numbers:

Addition and subtraction. Generally, the answer should have the same number of decimal places as the term with the fewest decimal places, e.g., 4.123 + 2.56 = 6.68.
Multiplication and division. Generally, the answer should have the same number of significant digits as the term with the fewest number of significant digits, e.g., 4.121 x 6.62 = 27.3.
Integers are always treated as having an infinite number of significant digits; e.g., 5 is considered for operations as 5.000…

Check Your Understanding #16 – Accuracy

You are being asked to determine the disk area of a propeller used on a drone by using a ruler to measure its diameter and then using the formula for the area of a circle. The nominal diameter of the propeller is 250 mm. How should the area value be reported numerically in terms of accuracy? A ruler can be read to about 0.1 of a millimeter.

Show solution/hide solution.

A ruler can be read to about 0.1 of a millimeter, so if the disk’s diameter were reported as 250.1 mm (4 significant digits), that would be a suitable measurement. Therefore, the calculated area is $\pi (250.1)^2 /4$ = 49,126.7 mm $^2$ . However, if this area value is reported, it implies a higher precision than the original measurement of the diameter. Therefore, by using rounding, the area must only be reported to 4 significant digits as 49,130 mm $^2$ . If the diameter could be measured to 2 decimal places (e.g., 250.13 mm, which has 5 significant digits), the area could be reported as 49,127.0 mm $^2$ .

Finally, common-sense rules should also be applied when working with numbers in engineering calculations. For example, it would be inappropriate to round up “g” (acceleration under gravity) from 9.81 m/s² to 10.0 m/s² when simultaneously using the value of $\pi$ to 4 decimal places in the same calculation. Sometimes, when doing homework, it is necessary to put down the calculator and pencil, apply common sense, and say: “Based on what I just did, does this numerical answer seem right?” Ultimately, getting the “right” numerical answers in problem-solving means understanding the basic concepts of significant digits and rounding and learning how to spot likely numerical errors. Spotting errors requires experience, and experience is only gained by lots of practice.

Summary & Closure

The ability of engineers to bring to bear the necessary mathematical concepts is an essential aspect of problem-solving. Geometry, algebra, trigonometry, calculus, and differential equations give engineers the crucial mathematical tools to help engineers solve complex problems, quantify processes, and make informed decisions. For example, many engineering problems involve vector quantities (e.g., forces, velocities, accelerations, etc.), which may require various vector operations such as scalar and vector products, line and surface integrals, volume integrals, and gradients. In addition, calculus brings other mathematical tools to the table that are often needed for engineering problem-solving. For example, besides pure mathematics, engineers nearly always need numerical results, which must be evaluated accurately and consistently, and a consistent number of significant digits or decimal places based on the available data.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

Explain the fundamentals of a scalar quantity. Make a list of some familiar scalar quantities used in engineering.
Explain the fundamentals of a vector quantity. Make a list of some examples of vector quantities used in engineering.
Explain the physical meaning of the curl of a vector field. Is the curl of a uniform flow positive, negative, or zero?
What happens when the simple rules of significant digits are not used in basic arithmetic calculations?
What happens when numbers are prematurely rounded up or down, and there is still a need to carry forward the needed number of significant digits in complex arithmetic calculations?
If the function A(x,y) = K(x^2 + y^2) where K is a constant, then what is A(r, theta)? Your answer may depend on whether you are biased toward pure mathematics, physics, or engineering.

Other Useful Online Resources

There is more to explore with these mathematical concepts:

This video goes into more depth on vector fields, divergence, and curl.
If you want to learn more about the Laplace equation, this website has a lot of information on that and other mathematical operations.
Here are some more short video lessons on differentiation and partial differentiation from Dr. Leishman’s “Math & Physics Hints and Tips” series:

Here is a short, fun video on the history of bookbag calculators from Dr. Leishman’s “Math & Physics Hints and Tips” series.

"It’s all about mathematics!" Indeed, whether it’s fluid dynamics, aerodynamics, or orbital mechanics, the foundation often comes down to mathematical principles. ↵
"Errors in engineering are often catastrophes." In engineering, even minor miscalculations can lead to severe consequences, which is why double-checking, simulations, and testing are vital to avoiding catastrophic outcomes. ↵
"Practice makes perfect." Consistent practice is essential for mastery. Whether designing a flight control system or finding optimal airfoil shapes, practice refines understanding and long-term skills. ↵
At face value, i.e., on-demand, as needed. ↵
Work smart, not hard." Efficiency and cleverness are critical in accomplishing tasks rather than relying on sheer effort. ↵

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Introduction to Aerospace Flight Vehicles Copyright © 2022–2025 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.n8