## Equations and Formulas

Equations and formulas are very similar problem-solving tools. Some texts study them separately, but in this section they will be combined. Equations are a way of showing the relationship between different variables in a problem and are usually derived as needed for each problem. Formulas are equations that are used frequently enough or are some how unique enough that they are remembered and used without a derivation.

### 1.3.3.1. Equations

The solution to some problems requires the derivation of a mathematical equation based on a pattern or another type of relationship between the numbers. These equations will be unique for each problem.

Problem: How much wire is needed to build a single wire fence around a rectangular field measuring 450 ft long and 350 ft wide?

Solution: In this example there are three quantities: length, width, and perimeter. It should be obvious that the perimeter is a function of the other two. Begin by assigning the variables L to represent the length, W to represent the width, and Pr to represent the perimeter. Then, because a rectangle has two lengths and two widths, the perimeter can be found as follows:

Pr = (L + L) + (W + W) Pr = (450 ft + 450 ft) + (350 ft + 350 ft) = 900 ft + 700 ft = 1,600 ft

### 1.3.3.2. Formulas

For some problems the relationships of the variables are fixed and constant, so the equation for that problem is remembered and used. These equations are sometimes called formulas. Another characteristic of formulas is that they usually contain a constant. One example is the area of a circle: A = n r2. The variable n is a constant. There are at least two important considerations in using formulas.

1. You must enter the numbers with the correct units of measure. All formulas are designed with specific units for the numbers, especially if they have a constant. If the units are incorrect, the answer will be incorrect. An example is the equation used to determine the application rate of a boom type sprayer:

gal 5,940 x Flow rate (gal/min)

ac Speed (mi/hr) x nozzle spacing (in)

It should be obvious that the units in the equation do not work (the units when combined do not result in the units for application rate, gal/min). This equation is an example of a situation in which units conversion values, that are always used each time the problem is worked, are combined into a units conversion constant (5,940). If any one of the values is entered in different units, the answer will be incorrect. When we solve for the application rate using units cancellation and conversion values, the source of the constant becomes apparent.

gal gal 60 min 1 hr 1 mi x -x -x ac min 1 hr 1 mi 5,280 ft 43,560 ft2 12 in 1

2. You must be able to rearrange the formula to solve for the unknown value. For example, the application rate equation could be rearranged to solve for nozzle spacing in inches (nsi):

5,940 x Flow rate (gal/min)

Speed (mi/hr) x Application rate (gal/ac)

For the remainder of this text the terms equation and formula will be used as synonyms.