All About the Math

J. Gordon Leishman

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

9 All About the Math

Introduction

“It is all about the math!” a senior engineer once annunciated to his colleagues. Indeed, it is often said by many that mathematics is the language of the engineer. However, while not all types of engineering problem-solving require mathematics, the rigor that mathematics brings to engineering analyses usually sheds much light on the eventual solution to any given situation. Geometry, algebra, trigonometry, and calculus all give engineers the essential mathematical tools that 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 types of differential equations.

In engineering, quantities with both magnitude and direction are often needed, examples being forces, velocities, and accelerations, their manipulations often 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 statics and motion of 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.

Learning Objectives

Comfortably use the concept of vectors.
Understand the meaning of the primary vector operators 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, which frequently occur in engineering theory.
Know about ordinary differential equations and how to solve them.

Position Vectors

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

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

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

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

as shown in the figure below.

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

Remember that vectors are added 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

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

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

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

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

Alternative coordinate systems may be used to describe any given 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 or 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.,

(5) $\begin{eqnarray*} p & = & p\left( x, y,z \right) \\ \varrho & = & \varrho\left( x, y, z \right) \\ T & = & T\left( x, y, z \right) \end{eqnarray*}$

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.,

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

where the components are

(7) $\begin{eqnarray*} V_{x} & = & V_{x} \left( x, y, z \right) \\ V_{y} & = & V_{y} \left( x, y, z \right) \\ V_{z} & = & V_{z} \left( x, y, z \right) \end{eqnarray*}$

More often than not, the velocity vector is written as

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

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 to 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

(9) $\begin{eqnarray*} \vec{A} & = & A_{x}\vec{i} + A_{y}\vec{j} + A_{z} \vec{k} \\ \vec{B} & = & B_{x}\vec{i} + B_{y}\vec{j} + B_{z} \vec{k} \end{eqnarray*}$

Then the scalar (dot) product $\vec{A}\bigcdot\vec{B}$ is given by

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

Notice that the “centered dot” $\dot$ 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

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

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.

Need some help with scalar products? Here is a short video lesson on how to perform scalar products as well as 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

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

where the vector product or “cross product” is denoted by the $\times$ operator.

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

(13) $\begin{eqnarray*} \vec{A}\times \vec{B} & = & \left( A_y B_z - B_y A_z \right) \vec{i} +\left( B_x A_z - A_x B_z \right) \vec{j} +\\ \nonumber & & \quad \quad \left( A_x B_y - B_x A_y \right) \vec{k} \end{eqnarray*}$

Given two vectors $\vec{A}$ and $\vec{B}$ then physically, the cross product is a vector that is perpendicular to both $\vec{A}$ and $\vec{B}$ and so is perpendicular to the plane containing them, as shown in the figure below. 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 that is perpendicular to the other two.

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.

Worked Example #1

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

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

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

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

3. $\mbox{~~~}\vec{A} \times \vec{B} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ -3 & 2 & 0 \\ 5 & -2 & 0 \\ \end{array} \right| = 0 \vec{i} + 0 \vec{j} + (6-10) \vec{k} = -4k$

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 concept of a line integral, which in this case is to sum the effects of a vector field along a path.

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

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

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

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

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 occur frequently in engineering practice. Surface Integrals

Consider a surface $S$ bounded by curve $C$ . Let $dS$ be an elemental area of the surface and let $\vec{n}$ be a unit normal vector, i.e., one that is 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

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

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

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

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

(18) $\begin{equation*} \iiint_{\cal{V}} \varrho \, d\cal{V} \end{equation*}$

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

(19) $\begin{equation*} \iiint_{\cal{V}} \vec{V} \, d\cal{V} \end{equation*}$

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

There are also some important relations between line, surface, and volume integrals that come up from time to time. 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.,

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

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 through the divergence theorem (or Green’s Theorem) using

(21) $\begin{equation*} \oiint_{S} \vec{A}\times d\vec{S} = \iiint_{\cal{V}} \left( \nabla \vec{A} \right) \, d\cal{V} \end{equation*}$

Finally, if $\varrho$ represents a scalar field, then the gradient theorem states that

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

Gradient of a Scalar Field

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

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

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

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

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 $p\left(x,y\right)$ that is 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

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

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.

Worked Example #2

In the solution to a specific engineering problem, the solution to the scalar field $\phi(x,y,z)$ is given by $\phi = x^2 y z^3$ . Find an expression for $\nabla \phi$ .

Recall that the gradient of a scalar field is given by

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

where the operator $\nabla$ is defined as

$\nabla = \frac{\partial }{\partial x}\, \vec{i}+ \frac{\partial }{\partial y}\, \vec{j}+\frac{\partial }{\partial z}\, \vec{k}$

In this case

$\nabla \phi = \frac{\partial (x^2 y z^3) }{\partial x}\, \vec{i}+ \frac{\partial (x^2 y z^3) }{\partial y}\, \vec{j}+\frac{\partial (x^2 y z^3) }{\partial z}\, \vec{k}$

and after partial differentiation then

$\nabla \phi = (2x y z^3) \vec{i} + (x^2 z^3) \vec{j} + (3 x^2 y z^2) \vec{k}$

which is a vector.

Divergence of a Vector Field

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

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

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. In fact, the term mass fluxes is often used, which are terms like $\varrho u$ , $\varrho v$ , etc. Divergence can be thought of as 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

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

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

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

and expanding this determinant gives

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

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, which is 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.

Worked Example #3

In a solution to a specific engineering problem, the solution the vector field $\vec{A}(x,y,z)$ is $\vec{A} = xz \vec{i} - y^2 \vec{j} + 2 x^2y \vec{k}$ . Find the curl of this field.

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

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

and 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

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

Laplace Operator

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

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

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

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

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

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 useful 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 linear superposition.

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

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

where

(33) $\begin{eqnarray*} \nabla^2 u & = & \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} \\ \nabla^2 v & = & \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} + \frac{\partial^2 v}{\partial z^2} \\ \nabla^2 w & = & \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2} + \frac{\partial^2 w}{\partial z^2} \end{eqnarray*}$

Substantial Derivative Operator

The substantial derivative operator in Cartesian coordinates is written as

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

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

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

Hence, if the vector is $\vec{V}$ then the substantial derivative can be written as

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

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

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

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 physically 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

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

Ordinary Differential Equations (ODEs)

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

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

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

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

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

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 is required such as an initial condition or a boundary condition. 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

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

for all values of $t$ .

Consider another example, which is to solve the ODE

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

to determine the function $x(t)$ .

One common approach for solving and ODE is to use separation of variables, a method frequently encountered in practice. In this case, separation of variables gives

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

Integrating both sides gives

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

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

(45) $\begin{equation*} 1 = C \exp \left( (5 (2) \right) + \frac{3}{5} \end{equation*}$

so the constant of integration must be

(46) $\begin{equation*} C = \frac{2}{5} \exp{(-10)} \end{equation*}$

Therefore, the particular solution to this ODE is

(47) $\begin{equation*} x(t) = \frac{2}{5} \exp \left( 5(t - 2) \right) + \frac{3}{5} \end{equation*}$

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 often 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 needs to 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 is greater than 5. 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, then round up if a non-zero digit follows it, but round down if followed by a zero, e.g., 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 rounding off numbers is needed, which is often the case, then this is done (usually) 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 acceptable 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 of digits used to express a measured or calculated quantity. Like decimal places, how “precise” a number is, is measured using significant digits. The value with the least 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 to be 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.

Worked Example #4

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

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 if they have an infinite number of significant digits, e.g., 5 is considered for operations as 5.000…

Worked Example #5

You are being asked to determine the area of a disk using a ruler to measure its diameter and then calculate its area. Numerically, how should this area value be reported?

A ruler can be read easily to about 0.1 of a millimeter, so if the disk’s diameter was reported as 50.1 mm, that would be a reasonable measurement. However, if the diameter is reported as 50.13 mm, then 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 $^2$ . On the other hand, 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.

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 all give engineers the essential mathematical tools to help engineers solve complex problems, quantify the 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 if we do not use the simple rules of significant digits in basic arithmetic calculations?
What happens when we also prematurely round up or down and do not carry forward the needed number of significant digits in complex arithmetic calculations?

Other Useful Online Resources

There is more to explore with these mathematical concepts:

This video goes more in 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 some more short video lessons of differentiation and partial differentiation from Dr. Leishman’s “Math & Physics Hints and Tips” series:

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Introduction to Aerospace Flight Vehicles 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