4.6: Working with Multi-Variable Formulas
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Many real-world situations are described by formulas with more than one variable. A formula is just a rule written in mathematical shorthand that connects several quantities. Once you know how to substitute values into a formula and simplify the result, you can use formulas from science, business, statistics, and everyday life without having to derive them yourself.
In this section, we'll work with several common formulas: geometric area and volume formulas, the braking distance of a car, and a margin-of-error formula from statistics. Along the way, we'll introduce subscript notation, which is a small but important piece of mathematical bookkeeping.
Subscripts
A subscript is a small symbol written just below and to the right of a variable. It is used as a label to tell similar variables apart. A subscript is not an operation — you do not multiply, divide, or do anything else with it. It just identifies which version of the variable you are referring to.
For example, in earlier sections we used P₀ (read "P sub-zero") to mean the initial population, and Pₙ ("P sub-n") to mean the population after n time periods. Both are populations, but the subscripts let us tell them apart.
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Subscript Notation A subscript is a label written below and to the right of a variable. Subscripts are used to distinguish between variables that represent similar quantities. Examples: • P₀ means the starting (initial) value of P. • P₁ means the value of P after one time period. • V₀ might mean an initial velocity, while V might mean a current velocity. A subscript is a label, not an operation. |
Example 1
If P₀ = 437 and the population grows by 32 each year, find P₁ and P₂.
P₁ means the population after one year. Since the population grows by 32 each year:
P₁ = P₀ + 32 = 437 + 32 = 469
P₂ means the population after two years:
P₂ = P₁ + 32 = 469 + 32 = 501
Try it Now 1
A car is purchased for $24,000. Each year its value decreases by $1,800. If V₀ represents the original value, find V₁ and V₂.
Area and Volume Formulas
Many real applications require us to calculate the area of a flat region or the volume of a three-dimensional object. The formulas below are the ones we'll use most often in this course.
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Common Area and Volume Formulas Area of a rectangle: A = L · W Area of a triangle: A = ½ · b · h Area of a trapezoid: A = ½ · h · (b₁ + b₂) Area of a circle: A = π · r² Volume of a cylinder: V = π · r² · h In the trapezoid formula, b₁ and b₂ are the two parallel sides (the bases). The subscripts simply label the two different bases. |
Notice the subscripts on the trapezoid formula. There are two parallel sides, and we need names for both, so we use b with two different subscripts. The subscripts do nothing mathematically — they just keep the two bases straight.
Example 2
A courtyard is shaped like a trapezoid. The two parallel sides measure 70 feet and 100 feet, and the distance between them (the height) is 50 feet. Find the area of the courtyard.
Identify each value:
• b₁ = 70 ft
• b₂ = 100 ft
• h = 50 ft
Substitute these into the trapezoid area formula:
A = ½ · 50 · (70 + 100)
A = ½ · 50 · 170 = 4,250 square feet
The courtyard has an area of 4,250 square feet.
Example 3
A cylindrical water cistern has a radius of 10 feet and a height of 35 feet. Find the volume, rounded to the nearest cubic foot.
Identify each value:
• r = 10 ft
• h = 35 ft
Substitute into the volume formula for a cylinder:
V = π · r² · h = π · (10)² · 35
V = π · 100 · 35 = 3,500π
Using a calculator with the π button (do not round π to 3.14 — keep more decimal places to avoid rounding errors):
V ≈ 10,996 cubic feet
Try it Now 2
A garden bed is shaped like a trapezoid with parallel sides of 12 feet and 18 feet. The distance between them is 5 feet. Find the area.
The Grade of a Road
The grade of a road tells you how steep the road is. Truck drivers, bicyclists, and runners all pay attention to grade because it affects braking, fuel consumption, and effort.
Grade measures the rate of vertical change compared to horizontal change. It's written as a fraction with the vertical change on top and the horizontal change on the bottom. After the units divide out, you get a dimensionless number that is usually written as a percent.
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Grade of a Road Grade = (change in height) / (change in horizontal distance) Grade is usually reported as a positive percent, regardless of whether the road is going uphill or downhill. |
Example 4
A road decreases 72 feet in height over a horizontal distance of 600 feet. Find the grade of the road.
Grade = 72 ft / 600 ft = 0.12 = 12%
The road has a 12% grade.
In formulas, the grade is typically used as its decimal equivalent. A 2% grade is written as G = 0.02 when you substitute it into a formula.
Braking Distance of a Car
How far does a car travel after the driver hits the brakes? This depends on three things: how fast the car was going, how slippery the road surface is, and whether the road is going uphill or downhill. Highway engineers and accident investigators use a formula to estimate braking distance:
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Braking Distance Formula d = V² / (2g(f + G)) where: • d = braking distance (in feet) • V = velocity of the car (in feet per second) • g = acceleration due to gravity = 32.2 ft/sec² (a constant) • f = coefficient of friction between tires and road • G = grade of the road (as a decimal) |
The variable g is a constant — it never changes for problems on Earth. The other letters (d, V, f, G) are quantities that depend on the specific situation.
The coefficient of friction f measures how much the tires grip the road. Larger values of f mean more friction, which means a shorter stopping distance. Some typical values:
|
Road condition |
Coefficient of friction (f) |
|
Icy road |
0.10 |
|
Wet road with fair tires |
0.40 |
|
Asphalt road with fair tires |
0.70 |
|
Dry road with great tires |
0.90 |
Example 5
A car is traveling at 60 ft/sec on a road with a 2% grade. The coefficient of friction is f = 0.70 (dry asphalt). Find the braking distance.
First, identify each value:
• V = 60 ft/sec
• g = 32.2 ft/sec² (constant)
• f = 0.70
• G = 0.02 (because 2% = 0.02)
Substitute these into the braking distance formula:
d = (60)² / [2(32.2)(0.70 + 0.02)]
d = 3,600 / [64.4 · 0.72]
d = 3,600 / 46.368 ≈ 77.6 feet
The car will travel about 77.6 feet after the brakes are applied.
Example 6 — Holding Variables Constant
Suppose we want to investigate how the coefficient of friction affects braking distance. To isolate the effect of friction, we hold the other variables constant. Let V = 72 mph and G = 0.02. The grade is fixed at 2% and the velocity is fixed at 72 mph. Only f will change.
Before substituting, we need V in feet per second, not miles per hour. Using dimensional analysis from Section 1.4:
72 mph · (5,280 ft / 1 mi) · (1 hr / 3,600 sec) = 105.6 ft/sec
Now substitute V = 105.6, g = 32.2, and G = 0.02 into the formula, leaving f as the only variable:
d = (105.6)² / [2(32.2)(f + 0.02)]
d = 11,151.36 / [64.4f + 1.288]
This simplified formula now lets us calculate d for any value of f, without having to redo all the arithmetic.
Example 7 — Building a Table of Values
Using the simplified formula from Example 6, fill in the braking distance for each road condition:
|
Road condition |
f |
d (feet) |
|
Icy |
0.10 |
763.2 |
|
Wet, fair tires |
0.40 |
417.5 |
|
Asphalt, fair tires |
0.70 |
239.3 |
|
Dry, great tires |
0.90 |
187.1 |
Sample calculation for f = 0.70:
d = 11,151.36 / [64.4(0.70) + 1.288]
d = 11,151.36 / [45.08 + 1.288] = 11,151.36 / 46.368 ≈ 240.5 ft
(The small differences come from rounding 72 mph to 105.6 ft/sec.)
Notice the pattern: as f increases at a constant rate (going up by 0.20 each row), the braking distance d decreases — and it decreases by a different amount each time. The coefficient of friction does not change the braking distance at a constant rate. This is not a linear relationship.
Try it Now 3
A car is traveling at 88 ft/sec on a road with a 3% grade and a coefficient of friction of f = 0.65. Find the braking distance, rounded to the nearest foot.
Margin of Error in a Survey
When a polling company reports that 71% of voters favor a candidate, they don't really mean exactly 71%. They mean 71% give or take a few percentage points. That "give or take" is called the margin of error.
The margin of error depends on two things: the proportion p̂ ("p-hat") of people who answered a certain way, and the sample size n. The hat above the p tells us that this is an estimate from a sample, not a true value for the whole population.
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Margin of Error Formula (95% Confidence) E = 1.96 · √[ p̂(1 − p̂) / n ] where: • E = margin of error (as a decimal) • p̂ = sample proportion (as a decimal) • n = sample size The number 1.96 corresponds to a 95% confidence level. Different confidence levels use different numbers. |
Example 8
In a survey of 1,300 people, 71% favored stiffer penalties for drunk driving. Find the margin of error, rounded to 3 decimal places.
Identify each value:
• p̂ = 0.71 (because 71% = 0.71)
• n = 1,300
Substitute into the margin of error formula:
E = 1.96 · √[ 0.71(1 − 0.71) / 1,300 ]
E = 1.96 · √[ 0.71(0.29) / 1,300 ]
E = 1.96 · √[ 0.2059 / 1,300 ]
E = 1.96 · √[ 0.0001584... ]
E = 1.96 · 0.01258... ≈ 0.025
The margin of error is about 0.025, or 2.5 percentage points. The survey result could be reported as 71% ± 2.5%.
Try it Now 4
In a survey of 800 people, 42% said they had read a book in the last month. Find the margin of error, rounded to 3 decimal places.
A Strategy for Working with Formulas
When you encounter any formula in this course or in your career, the same basic process applies:
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Steps for Working with a Formula 1. Identify what each variable in the formula represents and what units it uses. 2. Identify which variables are constants (their value never changes) and which depend on the situation. 3. Make sure your given values are in the units the formula expects. Convert if needed. 4. Substitute the values into the formula carefully, using parentheses to keep operations grouped correctly. 5. Simplify using order of operations. Do not round intermediate steps; keep extra decimal places until the end. 6. Check that your final answer has the correct units and is reasonable. |
Knowing what each variable represents is more important than knowing what numbers to substitute. If you understand what the formula is describing, you can spot mistakes and recognize when a number doesn't make sense.
Try it Now Answers
1. V₁ = 24,000 − 1,800 = $22,200. V₂ = 22,200 − 1,800 = $20,400.
2. A = ½ · 5 · (12 + 18) = ½ · 5 · 30 = 75 square feet.
3. d = (88)² / [2(32.2)(0.65 + 0.03)] = 7,744 / 43.792 ≈ 177 feet.
4. E = 1.96 · √[0.42(0.58) / 800] = 1.96 · √[0.0003045] ≈ 0.034, or about 3.4 percentage points.
This section is original content written for Math 1930, drawing on the formula-substitution approach used in the Carnegie Math Pathways Quantway curriculum. Licensed CC-BY-SA 4.0.