The Central Idea
Suppose a fixed trip is
\[ 120\text{ miles}. \]If the average speed is \(v\) miles per hour and the travel time is \(t\) hours, then
\[ vt=120. \]If speed doubles, time must be cut in half to keep the same distance:
\[ (2v)\left(\frac{t}{2}\right)=vt. \]The two quantities move in opposite multiplicative directions while their product stays fixed.
This is the central idea of inverse proportion:
When one quantity is multiplied by a factor, the other is divided by that same factor.
What Is an Inverse Proportion?
Two quantities \(x\) and \(y\) are inversely proportional when their product is a constant:
\[ xy=k, \]where \(k\ne0\).
Solving for \(y\):
\[ y=\frac{k}{x}, \qquad x\ne0. \]This is the standard form of an inverse proportional relationship.
The Constant Product
For a direct proportion, the quotient
\[ \frac{y}{x} \]is constant.
For an inverse proportion, the product
\[ xy \]is constant.
This gives a useful contrast:
| Relationship | Constant Quantity | Equation |
|---|---|---|
| Direct proportion | \(\frac{y}{x}=k\) | \(y=kx\) |
| Inverse proportion | \(xy=k\) | \(y=\frac{k}{x}\) |
Reciprocal Scaling
Suppose
\[ xy=24. \]If
\[ x=3, \]then
\[ y=8. \]If \(x\) doubles from \(3\) to \(6\), then \(y\) must be halved:
\[ y=4. \]Indeed,
\[ 3(8)=24 \]and
\[ 6(4)=24. \]Multiply One, Divide the Other
Suppose \(x\) and \(y\) are inversely proportional.
If \(x\) is multiplied by \(m\), then \(y\) must be multiplied by
\[ \frac{1}{m}. \]That is,
\[ x\longrightarrow mx \]requires
\[ y\longrightarrow\frac{y}{m}. \]The product remains
\[ (mx)\left(\frac{y}{m}\right)=xy. \]Doubling and Halving
In an inverse proportion:
- doubling one quantity halves the other;
- tripling one quantity divides the other by \(3\);
- multiplying one quantity by \(5\) divides the other by \(5\);
- halving one quantity doubles the other.
This reciprocal scaling is the quickest way to recognize many inverse relationships.
Finding the Constant of Inverse Proportionality
If \(x\) and \(y\) are inversely proportional and
\[ x=4,\qquad y=15, \]then
\[ k=xy. \]So
\[ k=4(15)=60. \]The inverse model is
\[ \boxed{y=\frac{60}{x}}. \]Finding a Missing Value
If
\[ y=\frac{60}{x}, \]find \(y\) when
\[ x=12. \]Substitute:
\[ y=\frac{60}{12}=5. \]Therefore,
\[ \boxed{y=5}. \]Finding the Input
If
\[ y=\frac{72}{x} \]and
\[ y=9, \]then
\[ 9=\frac{72}{x}. \]Multiply by \(x\):
\[ 9x=72. \]Thus,
\[ \boxed{x=8}. \]Recognizing Inverse Proportion from a Table
| \(x\) | \(y\) | \(xy\) |
|---|---|---|
| 2 | 12 | 24 |
| 3 | 8 | 24 |
| 4 | 6 | 24 |
| 6 | 4 | 24 |
Because the product is constant,
\[ xy=24. \]Therefore,
\[ \boxed{y=\frac{24}{x}}. \]A Non-Inverse Table
| \(x\) | \(y\) | \(xy\) |
|---|---|---|
| 2 | 10 | 20 |
| 4 | 6 | 24 |
| 5 | 5 | 25 |
The product changes.
So the table does not represent an exact inverse proportion.
Inverse Proportion from an Equation
The equation
\[ y=\frac{18}{x} \]is an inverse proportion with
\[ k=18. \]The equivalent product form is
\[ xy=18. \]Recognizing a Rearranged Inverse Proportion
Suppose
\[ 3xy=48. \]Divide by \(3\):
\[ xy=16. \]Then
\[ y=\frac{16}{x}. \]So the relationship is inversely proportional with
\[ \boxed{k=16}. \]The Graph of an Inverse Proportion
The graph of
\[ y=\frac{k}{x} \]is a rectangular hyperbola.
When \(k>0\), one branch lies in Quadrant I and the other in Quadrant III.
When \(k<0\), the branches lie in Quadrants II and IV.
The Axes Are Not Reached
For
\[ y=\frac{k}{x}, \qquad k\ne0, \]we cannot use
\[ x=0 \]because division by zero is undefined.
Also, \(y\) can never equal zero because
\[ \frac{k}{x}\ne0 \]for finite nonzero \(x\).
The coordinate axes act as asymptotes.
Behavior Near Zero
For positive \(k\), if positive \(x\) becomes very small, then
\[ \frac{k}{x} \]becomes very large.
For example, if
\[ y=\frac{12}{x}, \]then
\[ x=1\Rightarrow y=12, \] \[ x=0.1\Rightarrow y=120, \] \[ x=0.01\Rightarrow y=1200. \]The function grows without bound as positive \(x\) approaches zero.
Behavior for Large Inputs
If
\[ y=\frac{12}{x}, \]then
\[ x=12\Rightarrow y=1, \] \[ x=120\Rightarrow y=0.1, \] \[ x=1200\Rightarrow y=0.01. \]As \(|x|\) becomes very large, \(y\) approaches zero without reaching it.
Direct and Inverse Proportion Behave Differently
For direct proportion
\[ y=kx, \]doubling \(x\) doubles \(y\).
For inverse proportion
\[ y=\frac{k}{x}, \]doubling \(x\) halves \(y\).
This contrast is more important than memorizing the formulas separately.
Direct Versus Inverse Tables
Suppose \(k=12\).
| \(x\) | Direct: \(y=12x\) | Inverse: \(y=12/x\) |
|---|---|---|
| 1 | 12 | 12 |
| 2 | 24 | 6 |
| 3 | 36 | 4 |
| 4 | 48 | 3 |
Direct proportion preserves a quotient. Inverse proportion preserves a product.
Inverse Proportion Is Not Merely “One Goes Up, the Other Goes Down”
Many relationships decrease without being inversely proportional.
For example,
\[ y=10-x \]decreases as \(x\) increases.
But
\[ xy=x(10-x) \]is not constant.
So the relationship is not inverse proportion.
Negative Correlation Is Not the Same Thing
Two observed variables may tend to move in opposite directions.
That is sometimes described as a negative association or negative correlation.
Inverse proportion is much more specific:
\[ xy=k. \]A general downward trend does not guarantee a constant product.
Inverse Proportion Is Not an Inverse Function
The word inverse appears in both ideas, but they are different.
An inverse proportion has form
\[ y=\frac{k}{x}. \]An inverse function reverses the input-output mapping of another function.
Do not confuse the two concepts merely because they share the word “inverse.”
Inverse Proportion and Reciprocals
The formula
\[ y=\frac{k}{x} \]can be written
\[ y=k\left(\frac{1}{x}\right). \]So \(y\) is directly proportional to the reciprocal of \(x\):
\[ y\propto\frac{1}{x}. \]This is why reciprocal behavior lies at the heart of inverse proportion.
Proportionality Notation
We write
\[ y\propto\frac{1}{x} \]to mean that \(y\) is inversely proportional to \(x\).
This implies
\[ y=\frac{k}{x} \]for some nonzero constant \(k\).
Finding \(k\) from Proportionality Notation
Suppose
\[ y\propto\frac1x \]and
\[ y=6 \]when
\[ x=4. \]Write
\[ y=\frac{k}{x}. \]Substitute:
\[ 6=\frac{k}{4}. \]Thus,
\[ k=24. \]The model is
\[ \boxed{y=\frac{24}{x}}. \]Units of the Inverse Constant
Because
\[ k=xy, \]the units of \(k\) are the product of the units of \(x\) and \(y\).
For fixed-distance travel,
\[ vt=d. \]If \(v\) is miles per hour and \(t\) is hours, then
\[ \left(\frac{\text{miles}}{\text{hour}}\right)(\text{hours}) = \text{miles}. \]So \(k\) has units of distance.
Fixed Distance: Speed and Time
For a fixed distance \(d\),
\[ vt=d. \]Solving for time:
\[ t=\frac{d}{v}. \]Thus, travel time is inversely proportional to speed when distance is fixed.
If speed doubles, travel time halves.
Fixed-Distance Example
A trip is 180 miles long.
At 60 mph:
\[ t=\frac{180}{60}=3\text{ hr}. \]At 90 mph:
\[ t=\frac{180}{90}=2\text{ hr}. \]Check the constant product:
\[ 60(3)=180, \] \[ 90(2)=180. \]Workers and Completion Time
Suppose a fixed job requires
\[ 48\text{ worker-hours}. \]If \(n\) equally productive workers divide the work perfectly and \(t\) is the completion time, then
\[ nt=48. \]Thus,
\[ t=\frac{48}{n}. \]Under these assumptions, the number of workers and completion time are inversely proportional.
Work Example
If 6 workers complete a job in 8 hours, then the idealized total work is
\[ 6(8)=48\text{ worker-hours}. \]If 12 equally productive workers perform the same job under the same conditions, then
\[ t=\frac{48}{12}=4\text{ hr}. \]Doubling the workforce halves the time in the idealized model.
Real Workplaces May Not Scale Perfectly
The inverse model for workers and time assumes:
- workers have equal productivity;
- the work can be divided efficiently;
- workers do not interfere with one another;
- setup and coordination costs do not grow;
- the total amount of work remains fixed.
Real systems may violate these assumptions.
Flow Rate and Time
Suppose a tank has fixed volume \(V\) and fills at constant rate \(r\).
Then
\[ rt=V. \]So
\[ t=\frac{V}{r}. \]For a fixed volume, filling time is inversely proportional to flow rate.
Flow Example
A 300-gallon tank is filled at \(25\) gallons per minute.
\[ t=\frac{300}{25}=12\text{ min}. \]If the rate increases to \(50\) gallons per minute,
\[ t=\frac{300}{50}=6\text{ min}. \]Doubling the rate halves the filling time.
Frequency and Period
Frequency \(f\) counts cycles per unit time, while period \(T\) measures time per cycle.
They satisfy
\[ fT=1 \]when the units are reciprocal and consistent.
Thus,
\[ T=\frac1f. \]Frequency and period are inversely proportional.
Pressure and Volume: A Scientific Example
For a fixed amount of ideal gas at constant temperature, Boyle's law gives
\[ PV=k. \]So
\[ P=\frac{k}{V}. \]Under those conditions, pressure and volume are inversely proportional.
The physical conditions matter; changing temperature changes the model.
Area Constraints
Suppose a rectangle has fixed area
\[ A=48. \]If length is \(L\) and width is \(W\), then
\[ LW=48. \]Thus,
\[ W=\frac{48}{L}. \]For a fixed area, length and width are inversely proportional.
Rectangle Example
If a rectangle has area
\[ 60\text{ m}^2, \]then:
\[ L=5\Rightarrow W=12, \] \[ L=10\Rightarrow W=6, \] \[ L=15\Rightarrow W=4. \]Every pair satisfies
\[ LW=60. \]Inverse Scaling from a Known Pair
Suppose \(x\) and \(y\) are inversely proportional and
\[ x=8,\qquad y=15. \]If \(x\) becomes
\[ 20, \]use reciprocal scaling.
The multiplier on \(x\) is
\[ \frac{20}{8}=\frac52. \]So \(y\) is multiplied by
\[ \frac25. \]Therefore,
\[ y=15\left(\frac25\right)=6. \]Solving with the Constant Product
The same example can be solved by finding \(k\):
\[ k=8(15)=120. \]Then
\[ y=\frac{120}{20}=6. \]Reciprocal scaling and the constant-product method are two views of the same structure.
Inverse-Proportion Equation from Two Values
Suppose \(y\) varies inversely with \(x\), and
\[ y=10 \]when
\[ x=6. \]Then
\[ k=xy=60. \]So
\[ \boxed{y=\frac{60}{x}}. \]Use the Model to Predict
Using
\[ y=\frac{60}{x}, \]find \(y\) when
\[ x=15. \]Then
\[ y=\frac{60}{15}=4. \]So
\[ \boxed{y=4}. \]Inverse Proportion with Negative \(k\)
Consider
\[ y=-\frac{12}{x}. \]Then
\[ xy=-12. \]If \(x\) is positive, \(y\) is negative.
If \(x\) is negative, \(y\) is positive.
The relationship remains inverse because the product is constant.
Domain Restrictions
In
\[ y=\frac{k}{x}, \]we must have
\[ x\ne0. \]For \(k\ne0\), we also necessarily have
\[ y\ne0. \]In applied problems, additional restrictions may apply, such as
\[ x>0,\qquad y>0. \]Context Often Restricts the Graph
The algebraic graph of
\[ y=\frac{48}{x} \]includes positive and negative values of \(x\).
But if \(x\) represents number of workers and \(y\) represents hours, negative values are meaningless.
The practical model may therefore use only
\[ x>0. \]If workers are counted individually, we may further require
\[ x\in\mathbb{Z}_{>0}. \]Continuous Versus Discrete Inverse Models
A rectangle's length may vary continuously, so
\[ W=\frac{A}{L} \]can use many positive real values.
A workforce model may allow only positive integers for the number of workers.
The algebraic relationship can be the same while the allowed domain differs.
Not Every Reciprocal-Looking Formula Is a Simple Inverse Proportion
Consider
\[ y=\frac{5}{x+2}. \]This is a reciprocal-type function, but \(y\) is not inversely proportional to \(x\) in the simple sense because
\[ xy \]is not constant.
Instead, \(y\) is inversely proportional to
\[ x+2. \]Shifted Reciprocal Relationships
If
\[ y=\frac{k}{x-a}, \]then
\[ y(x-a)=k. \]The constant-product relationship is between \(y\) and \(x-a\), not between \(y\) and \(x\).
This distinction becomes important when reading more advanced models.
Inverse Powers Are Different Relationships
Suppose
\[ y=\frac{k}{x^2}. \]Then
\[ x^2y=k. \]This is inverse-square variation, not simple inverse proportion.
If \(x\) doubles, \(y\) becomes
\[ \frac14 \]as large, not one-half as large.
General Inverse Variation
The simple inverse proportion
\[ y=\frac{k}{x} \]is one member of a larger family.
More generally, a quantity might vary as
\[ y=\frac{k}{x^n}. \]The exponent determines how strongly the output responds to scaling of the input.
Comparing Direct and Inverse Scaling
Suppose \(x\) is multiplied by \(3\).
In direct proportion:
\[ y\longrightarrow3y. \]In simple inverse proportion:
\[ y\longrightarrow\frac{y}{3}. \]In inverse-square variation:
\[ y\longrightarrow\frac{y}{9}. \]This scaling viewpoint helps distinguish related models.
Checking an Inverse Model
Suppose a model predicts
\[ y=\frac{36}{x}. \]Check the pair
\[ (x,y)=(9,4). \]Substitute:
\[ 4=\frac{36}{9}=4. \]Or check the product:
\[ 9(4)=36. \]Either method verifies the pair.
Check More Than One Data Pair
If a relationship is claimed to be inverse, one pair can determine \(k\), but additional pairs should satisfy the same product.
Suppose the data include
\[ (3,20) \]and
\[ (5,12). \]Check:
\[ 3(20)=60, \] \[ 5(12)=60. \]The matching products support the inverse model
\[ y=\frac{60}{x}. \]Worked Example: Fixed Distance
A driver travels a fixed 240-mile route.
At 60 mph, the time is
\[ t=\frac{240}{60}=4\text{ hr}. \]At 80 mph,
\[ t=\frac{240}{80}=3\text{ hr}. \]The product remains
\[ 60(4)=240 \]and
\[ 80(3)=240. \]Worked Example: Workforce
Eight equally productive workers complete a job in 15 hours.
Assume ideal inverse scaling. How long would 12 workers take?
Find the constant work:
\[ 8(15)=120. \]Then
\[ 12t=120. \]So
\[ \boxed{t=10\text{ hr}}. \]Worked Example: Fixed Area
A rectangle has area
\[ 96\text{ cm}^2. \]If its length is
\[ 12\text{ cm}, \]then its width is
\[ W=\frac{96}{12}=8\text{ cm}. \]If the length doubles to
\[ 24\text{ cm}, \]the width halves to
\[ 4\text{ cm}. \]Worked Example: Determine the Equation
Suppose \(y\) varies inversely with \(x\), and
\[ y=7 \]when
\[ x=9. \]Then
\[ k=xy=63. \]Therefore,
\[ \boxed{y=\frac{63}{x}}. \]Worked Example: Use the Equation
Using
\[ y=\frac{63}{x}, \]find \(y\) when
\[ x=21. \]Then
\[ y=\frac{63}{21}=3. \]Therefore,
\[ \boxed{y=3}. \]A Reliable Recognition Workflow
- Identify the two quantities.
- Ask whether multiplying one by a factor divides the other by that same factor.
- Check whether \(xy\) remains constant.
- Write \(xy=k\) or \(y=k/x\).
- Use known values to find \(k\).
- Use the model to find missing values.
- Check domain and contextual restrictions.
- Verify additional data pairs by testing the product.
A Reliable Problem-Solving Workflow
- Confirm that the relevant total or product is fixed.
- Define the variables and units.
- Write the constant-product equation.
- Find the constant from known information.
- Solve for the requested quantity.
- Interpret the answer in context.
- Check whether the assumptions of inverse scaling are reasonable.
Common Errors
Checking for a Constant Quotient Instead of a Constant Product
Direct proportion keeps
\[ \frac{y}{x} \]constant. Inverse proportion keeps
\[ xy \]constant.
Assuming Any Decreasing Relationship Is Inverse
A decreasing relationship must satisfy the constant-product condition to be a simple inverse proportion.
Using \(x=0\)
The expression
\[ \frac{k}{x} \]is undefined at \(x=0\).
Confusing Inverse Proportion with an Inverse Function
These are different mathematical ideas.
Ignoring Contextual Assumptions
Workers and completion time are only ideally inverse when productivity and divisibility assumptions hold.
Confusing Inverse Proportion with Inverse-Square Variation
Doubling \(x\) halves \(y\) in simple inverse proportion but quarters \(y\) in inverse-square variation.
Forgetting Units of \(k\)
Since
\[ k=xy, \]the units of \(k\) are the product of the variable units.
Practice
1. Find the Constant
If \(y\) varies inversely with \(x\), and
\[ x=5,\qquad y=8, \]find \(k\).
Check Answer
\[ k=xy=5(8)=40. \]Answer:
\[ \boxed{k=40}. \]2. Write the Equation
Using \(k=40\), write the inverse proportional relationship.
Check Answer
\[ \boxed{y=\frac{40}{x}}. \]3. Find a Missing Value
If
\[ y=\frac{40}{x}, \]find \(y\) when
\[ x=10. \]Check Answer
\[ y=\frac{40}{10}=4. \]Answer:
\[ \boxed{4}. \]4. Inverse Table?
Does the following table represent inverse proportion?
| \(x\) | \(y\) |
|---|---|
| 2 | 18 |
| 3 | 12 |
| 6 | 6 |
Check Answer
\[ 2(18)=36, \] \[ 3(12)=36, \] \[ 6(6)=36. \]Yes. The constant product is
\[ \boxed{36}. \]5. Fixed Distance
A 300-mile trip takes 5 hours at one constant average speed. What speed would complete the same trip in 4 hours?
Check Answer
The fixed product is the distance:
\[ vt=300. \]So
\[ v=\frac{300}{4}=75. \]Answer:
\[ \boxed{75\text{ mph}}. \]6. Workforce
Ten workers complete a fixed job in 12 hours. Under ideal inverse scaling, how long would 15 workers take?
Check Answer
\[ 10(12)=120. \] \[ 15t=120. \] \[ t=8. \]Answer:
\[ \boxed{8\text{ hr}}. \]7. Inverse or Merely Decreasing?
Is
\[ y=20-x \]an inverse proportion?
Check Answer
No. Although \(y\) decreases as \(x\) increases, the product
\[ x(20-x) \]is not constant.
Key Ideas
- Inverse proportion has the form \(xy=k\) or \(y=k/x\), with \(k\ne0\).
- The product \(xy\) remains constant.
- If one quantity is multiplied by a factor, the other is divided by the same factor.
- Doubling one quantity halves the other in a simple inverse proportion.
- Tables show inverse proportion when the product of corresponding values remains constant.
- The graph of \(y=k/x\) is a rectangular hyperbola with the coordinate axes as asymptotes.
- A decreasing relationship is not automatically an inverse proportion.
- Inverse proportion is different from an inverse function.
- The constant \(k\) has units equal to the product of the units of \(x\) and \(y\).
- Fixed distance, fixed work, fixed volume, fixed area, frequency-period, and Boyle's-law models can produce inverse relationships under appropriate assumptions.
- Applied domains may restrict the algebraic model to positive or integer values.
- Inverse-square and other reciprocal-power variations are related but distinct from simple inverse proportion.
Where This Leads
Inverse proportion extends proportional reasoning beyond quantities that grow together.
The constant-product viewpoint prepares the way for inverse variation, work and rate problems, fixed-resource models, rational functions, reciprocal graphs, inverse-power laws, and optimization problems in which increasing one quantity forces another to decrease.