# Mathematical Background To Solve Physics Problems Solving Equations

## Introduction Mathematical Background

The purpose of this chapter is to provide you with a review of and reference for the mathematical techniques you will need in working the physics problems in this book. Some topics may be familiar to you while others may not. Depending on the mathematical level of your physics course, some topics may not be of interest to you. Each topic is covered in sufficient depth to allow you to perform the mathematical manipulations necessary for a particular problem without getting bogged down in lengthy derivations. It is not our intention to teach mathematics, but to show you how to apply specific mathematical procedures to physics problems.

The most efficient use of this chapter is for you to do a brief review of the chapter, spending time on those sections that are unfamiliar to you and that you know you will need in your course, then refer to specific topics as they are encountered in the solution to problems. With this reference you should be able to perform all the mathematical operations necessary to complete the problems in your physics course. If you need or desire more depth in a particular topic go to an algebra or calculus text.

## Solving Equations

The simplest equations to solve are the linear equations of the form ax + b = 0 which have as solution x = -b / a. You should be very familiar with these.

The next most complicated equations are the quadratics. The simplest quadratic is the type that can be solved by taking square roots directly, without any other manipulations.

An example is 4x2 = 36, which is first divided by 4 to read x2 = 9 and square roots taken to produce x = ±3. Both plus and minus values are legitimate solutions. The reality of the physical problem producing the equation may dictate that one of the solutions be discarded.

The next complication in quadratic equations is the factorable equations such as x2 - x - 6 = 0, which can be factored to (x - 3)(x + 2) = 0. The solutions, the values of x that make each parentheses equal to zero and satisfy the factored equation, are x = 3 and x = -2.

If the quadratic cannot be solved by factoring, the most convenient solution is by quadratic formula, a general formula for solution of any quadratic equation in the form ax2 + bx + c = 0.

The solution according to the quadratic formula is

See any algebra book for a derivation of this formula. The physics problems you are doing should not produce square roots of negative numbers. If your solution to a quadratic produces any square roots of negative numbers, you are probably doing something wrong in the problem. Certain cubic equations such as x3 = 8 can be solved directly producing the single answer x = 2. Cubic equations with quadratic (x2) and linear (x) terms can be solved by factoring (if possible) or approximated using graphical techniques. You most likely will not encounter cubic equations in your early physics courses. Another category of equations you will encounter is simultaneous equations: two independent equations in two unknowns and later three equations in three unknowns. We'll start with two equations in two unknowns. Take two equations

The most direct way of solving these equations is by substitution, solving one equation for one unknown and substituting in the other equation. Looking at these two equations the easiest variable to solve for is x in the second equation

Now substitute equation 2) in equation 1) and solve

Now put this value of y in either original equation and solve for x

These answers can be checked by substituting into both the original equations. Another method, often involving less manipulation, is addition and subtraction where the equations are multiplied in such a way that upon addition or subtraction one of the variables adds away leaving one equation in one unknown. Start with the equations used previously and write equation 1) and  2 times equation 2), and add

This is the same value obtained above and by substitution in either original equation will produce the value for x. The equations could be handled differently by making the y terms add away. Multiply equation 1) by 4 and equation 2) by 3, and add

The use of determinants in solving simultaneous equations is discussed in the next section. Determinants A determinant is a square array of numbers. Determinants are very convenient for solving two equations in two unknowns and three equations in three unknowns. The determinant technique for solving equations, called Cramers Rule, can be derived from the addition and subtraction method of solving simultaneous equations. Use as an example the equations of the previous section.

For the master, or main, determinant the array is the coefficients of the variables.

The numeric equivalent of the determinant is found by multiplying 2 times  4 and subtracting the multiplication of 3 times 1. The numeric equivalence of a 2 by 2 determinant is this first diagonal multiplication minus the second diagonal multiplication. With a little practice this goes very quickly.

Now form the x associated determinant by replacing the x coefficients with the constants.

Perform the same diagonal multiplication minus diagonal multiplication operation: multiply 7 times  4 and subtract the multiplication of 3 times  3.

The y associated determinant is formed by replacing the y associated coefficients with the constants and multiplying and subtracting.

The solutions are

and

If you need practice with determinants write down some sets of equations and solve them by substitution and determinants. After a few manipulations with determinants you will be able to solve simultaneous equations very quickly. Some calculators that solve systems of equations with Cramers Rule ask you to enter the numbers in a determinant format.

Three by three determinants require a little more manipulation. Consider three equations with the master determinant

There are several ways to find the values of this determinant. We'll look at one simple method called expanding the determinant, using the first row and the associated determinants obtained by crossing off the rows and columns associated with each number in the top row. This is easier to see than explain.

Look at the top row of (the 3 by 3) D and write each term times the determinant obtained by blocking off the row and column associated with that term. Also, alternate the signs of the three 2 by 2 determinants so the second number, 1, is changed to  1. The two by two determinants are evaluated as before.

The x associated determinant is (again replacing the x coefficients with the constants)

The y associated determinant is

so

As an exercise find the z associated determinant and calculate z. The value of z = 3 can be verified from any of the original equations.

With a little practice determinants can be a very quick way of solving multiple equations in multiple unknowns.

Determinants as applied to vector products are discussed in the chapter on vectors.

Binomial Expansions.

Squaring (a + b) is done so often that most would immediately write a2 + 2ab + b2.

Cubing (a + b) is not so familiar but easily accomplished by multiplying (a2 + 2ab + b2) by (a + b) to obtain a3 + 3a2b + 3ab2 + b3.

There is a simple procedure for finding the nth power of (a + b). Envision a string of (a + b)'s multiplied together, (a + b)n. Notice that the first term has coefficient 1 with a raised to the nth power, and the last term has coefficient 1 with b raised to the nth power. The terms in between contain a to progressively decreasing powers, n, n   1, n   2, …, and b to progressively increasing powers. The coefficients can be obtained from an array of numbers or more conveniently from the binomial expansion or binomial theorem

The factorial notation may be new to you. The definitions are

As an exercise use the binomial expansion formula to verify (a + b)3.

The real utility of the binomial expansion in physics problems is in finding approximations to expressions where a is equal to 1 and b is less than 1. For this case the expansion looks like

The terms of the series decrease depending on the value of b. Two or three terms is usually a good approximation. Also the “next” term in the expansion is a good measure of the error in using a fixed number of terms of the binomial expansion.

The classic use of this expansion is in special relativity where the expression

regularly occurs.

In special relativity v c is always less than one so this approximation. whether used algebraically or with numbers, is very convenient.

## Coordinate Systems

The standard two dimensional coordinate system works well for most physics problems. In working problems in two dimensions do not hesitate to arrange the coordinate system for your convenience in doing a problem. If a motion is constrained to move up an incline, it may be more convenient to place one axis in the direction of the motion rather than in the traditional horizontal direction. If a projectile is dropped from an airplane, it may be more convenient to place the origin of the coordinate system at the place where the projectile was dropped and have the positive directions down, since the projectile and possibly the distances in the problem are given in reference to the airplane.

Positions in the standard right angle coordinate system are given with two numbers. In a polar coordinate system positions are given by a number and an angle. In the accompanying diagram it is clear that any point (x,y) can also be specified by (r, è). Rather than moving distances in mutually perpendicular directions, the r and è locate points by moving a distance r from the origin along what would be the +x direction, then rotating through an angle è. The relationship between rectangular and polar coordinates is also shown in Fig. I 1.

Three dimensional coordinate systems are usually right handed. In Fig. I 2 imagine your right hand positioned with fingers extended in the +x direction closing naturally so that your fingers rotate into the direction of the +y axis while your thumb points in the direction of the +z axis. It is this rotation of x into y to produce z with the right hand that specifies a right handed coordinate system. Points in the three dimensional system are specified with three numbers (x,y,z).

For certain types of problems, locating a point in space is more convenient with a cylindrical coordinate system. Construct a cylinder with the central axis on the z axis of a right handed coordinate system.

A point is located by specifying a radius measured out from the origin in the +x direction, an angle in the x y plane measured from the x axis, and a height above the x y plane. Thus the coordinates in the cylindrical system are (r,è,z). The relation of these coordinates to x,y,z is given in Fig. I 3.

Spherical coordinates are also convenient in some problems. As the name suggests, points are located on a sphere centered on the origin of an (x,y,z) system. The radius of the sphere is the distance from the origin (to the sphere). The angle between this radius and the z axis is one angle, and the angle between the x axis and the projection of r on the x y plane is the other angle. Thus, the coordinates in the spherical system are (r,è,ö). The relation of these coordinates to x,y,z is given in Fig. I 4.