4 Mathematics for Engineering

Introduction

Omnia de mathematica agitur!” Indeed, many will say that mathematics is the engineer’s language, so the first thing people need to do before they can study engineering is to learn the essential mathematics. While not all types of engineering problem-solving require mathematics, the structure and rigor that mathematics brings to engineering analyses usually shed much light on the eventual solution to any given situation.

Most people reading this book will already be familiar with most of the basic mathematics needed to progress into most engineering subjects. 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 engineering problem-solving. To this end, all engineers must become well-versed in solving various differential equations. In particular, when doing arithmetical and algebraic problems, engineers must do them cold without making any mistakes. Errores in engineeringo plerumque sunt catastrophes.

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.

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 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. To this end, consider the following examples.

Check Your Understanding #1 – Review problems in algebra

  1. Solve for x in the equation 4(x - 2) = 16.
  2. Simplify the expression given by (2x^2 - 3x + 1) - (x^2 + 2x - 5).
  3. Solve the equation given by 2(x + 3) = 8 - x.
  4. Simplify the expression 5(2x + 3y) - 2(3x - y).
  5. Solve for x in the expression \displaystyle{ \frac{{x^2 + 3x - 10}}{{x + 2}}}  = 0.
Show solution/hide solution.
  1. Distributing the 4, then 4x - 8 = 16. Adding 8 to both sides gives 4x = 24. Finally, dividing both sides by 4, then x = 6.
  2. Removing parentheses and combining like terms gives x^2 - 5x + 6.
  3. 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.
  4. Distributing the coefficients gives 10x + 15y - 6x + 2y. Combining like terms leads to 4x + 17y.
  5. 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. Factor the numerator to get

        \[ \frac{{(x + 5)(x - 2)}}{{x + 2}} = 0 \]

    Now, each factor can be set equal to zero, so

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

    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, and 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. Tangible interpretation of mathematics within the framework of engineering is an acquired skill.

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 just do the mathematics valor facialis.

Check Your Understanding #2 – Review problems using differentiation

  1. Find the derivative of the function f(x) = 3x^2 + 2x - 1.
  2. Find the derivative of the function g(x) = \sin 2x + \cos x.
  3. Find the derivative of the function h(x) = \displaystyle{ \frac{2}{x^3}}.
  4. Find the derivative of the function y = \ln(3x^2 + 1).
  5. Find the derivative of the function f(x) = \sqrt{x^3 + 2x}.
Show solution/hide solution.
  1. 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.

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. 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. The ability to successfully use calculus should never be considered a quiz. Labora callide, non fortiter.

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

  1. Evaluate the integral \displaystyle{ \int (3x^2 + 2x - 1)} \, dx.
  2. Evaluate the integral \displaystyle{\int (2\cos(2x) - \sin(x)) } \, dx.
  3. Evaluate the integral \displaystyle{ \int (-6x^{-4})} \, dx.
  4. Evaluate the integral \displaystyle{ \int \frac{6x}{3x^2 + 1} }\, dx.
  5. 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

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

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.

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:

  1. \mbox{~~~}\vec{A} - 3\vec{B}
  2. \mbox{~~~}\vec{A} \bigcdot \vec{B}
  3. \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 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.

In this case, the concept of a line integral sums up the effects of a vector field along a path.

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}\,\, pd\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} \,\, pd\vec{S} \quad \mbox{or} \quad \oiint_{S} \, \vec{A}\bigcdot d\vec{S} \end{equation*}

Concept of a surface integral.

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 always points away from the surface by convention.

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:

  1. 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).
  2. 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} .
  3. Compute the cross product of these vectors giving \vec{N} = \vec{A}_1 \times \vec{A}_2.
  4. Normalize the resulting vector to obtain the unit normal vector, i.e., \vec{n} = \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

  1. In a three-dimensional Cartesian coordinate system, what is the unit normal vector for the direction of the x-axis?
  2. Consider a two-dimensional plane defined by x + y = 1. What is the unit normal to this plane?
  3. Consider a three-dimensional plane defined by 2x - 3y + z = 4. What is the unit normal to this plane?
Show solution/hide solution.
  1. This vector is parallel to the x-axis and points in the positive x-direction, so \vec{n} = (1, 0, 0).
  2. In this case, \phi(x,y) = x + y - 1 = 0. Therefore, at any point on this plane, the unit normal vector will be

        \[\vec{n} = \cfrac{ \left( \displaystyle{\cfrac{\partial \phi}{\partial x}}, \, \displaystyle{\cfrac{\partial \phi}{ \partial y}}, \, \displaystyle{\cfrac{\partial \phi}{ \partial z}} \right) }{ \left| \left( \displaystyle{\cfrac{\partial \phi}{ \partial x}}, \, \displaystyle{\cfrac{\partial \phi}{ \partial y}}, \, \displaystyle{\cfrac{\partial \phi}{ \partial z}} \right) \right| } = \cfrac{ (1, 1, 0)}{\sqrt{2}} = \left( \cfrac{1}{\sqrt{2}}, \, \cfrac{1}{\sqrt{2}}, \, 0 \right) \]

  3. In this 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}.

Concept of a volume integral.

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 above result 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.

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

Integral Relations

Some important relations between line, surface, and volume integrals occasionally arise. 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.,

(34)   \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.,

(35)   \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

(36)   \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

(37)   \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

(38)   \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

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

The term dp/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

    \begin{equation*} p(x,y) = -2x - y + 10 \end{equation*}

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

(40)   \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

(41)   \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

(42)   \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}} \\[6pt] V_x & V_y & V_z \end{array} \right| \end{equation*}

and expanding this determinant gives

(43)   \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. 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 laws and principles of physics will be well used to explain fluid dynamics and other physical behavior. 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 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.

2. 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 mass.

3. 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.

4. 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.

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 various fields, including 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.

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 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

    \[ \vec{F} = \frac{d}{dt} ( m \vec{V} ) = \frac{d}{dt} ( \vec{p} ) \]

where \vec{p} 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. Forces always occur in pairs, and they act on different objects.

Laplace Operator

The Laplace operator appears in several of the equations of aerodynamics 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

(44)   \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.,

(45)   \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

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

where

(47)   \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

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

(48)   \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

(49)   \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

(50)   \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

(51)   \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

(52)   \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

(53)   \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

(54)   \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

(55)   \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

(56)   \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

(57)   \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

(58)   \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

(59)   \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.

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:

  1. Round up if the first digit to be discarded exceeds 5. e.g., 3.14159 is 3.1416 to 4 decimal places.
  2. Round down if the first digit to be discarded is below 5. e.g., 3.141592 is 3.14159 to 5 decimal places.
  3. 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:

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

Check Your Understanding #15 – Significant digits

  1. How many significant digits are in 14.638?
  2. How many significant digits are in 0.002378?
  3. How many significant digits are in 273.70?
Show solution/hide solution.
  1. Answer = 5;  2. Answer = 4;  3. Answer = 5

Manipulating Numbers

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

  1. 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.
  2. 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.
  3. Integers are always treated as if they have infinite significant digits; e.g., 5 is considered for operations as 5.000…

Check Your Understanding #16 – Rounding & accuracy

You are being asked to determine the area of a disk using a ruler to measure its diameter and then calculate its area. How should this area value be reported numerically?
Show solution/hide solution.

A ruler can be read easily to about 0.1 of a millimeter, so if the disk’s diameter were reported as 50.1 mm, that would be a suitable measurement. However, if the diameter is reported as 50.13 mm, this value implies that you can read the ruler to 0.01 mm, which is perhaps more unlikely. Therefore, the area is \pi (50.1)^2 /4 = 1971.3568, but if this area value is reported, it implies that the diameter was known to 8 significant figures. So, the area using the ruler must only be reported as 1971 mm^2. If the diameter could be measured to 2 decimal places, its area could be reported as 1971.4 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/s to 10.0 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 gained by 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.

License

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Introduction to Aerospace Flight Vehicles Copyright © 2022 – 2024 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