Reciprocal Relationships and Inverse Variation
When two quantities are inversely related, one doubles as the other halves, their product is always the same constant $k$. This is inverse variation, modelled by $y = \frac{k}{x}$, and its graph is a hyperbola.
A task takes 12 worker-hours. With $n$ workers, each person works $h$ hours. Complete the pattern: $n = 1 \Rightarrow h = 12$; $n = 2 \Rightarrow h = 6$; $n = 3 \Rightarrow h = 4$; $n = 4 \Rightarrow h = ?$; $n = 6 \Rightarrow h = ?$.
Without a formulafill in the missing values and guess what the equation might be. What do you notice about $n \times h$?
Inverse variation exists when $x \times y$ is constant. That constant is $k$ (the constant of variation), and the relationship is $y = \frac{k}{x}$. As $x$ increases, $y$ decreases proportionally.
The graph of $y = \frac{k}{x}$ is a hyperbolatwo smooth curves, one in each of the quadrants dictated by the sign of $k$. If $k > 0$: curves in quadrants 1 and 3. If $k < 0$: curves in quadrants 2 and 4.
Key facts
- Inverse variation: $y = \frac{k}{x}$ where $k$ is the constant of variation
- Identifying inverse variation: constant product $x \times y = k$
- Graph is a hyperbola with two branches; asymptotes at $x = 0$ and $y = 0$
- $k > 0$: branches in Q1 and Q3; $k < 0$: branches in Q2 and Q4
Concepts
- Why $x \times y = k$ means "as one doubles, the other halves"
- Why the graph can never touch the axes (two asymptotes)
- How to construct a reciprocal model from a practical scenario
- Why the model breaks down near $x = 0$ and has practical limitations
Skills
- Identify inverse variation from a table (constant product test)
- Graph $y = k/x$ using a table of values or graphing technology
- Write the equation $y = k/x$ given any one $(x, y)$ pair
- Solve inverse variation problems graphically or algebraically
- Explain limitations of the reciprocal model in context
Inverse variation is identified by the constant product test: if $x \times y$ gives the same value for every pair in a table, the relationship is $y = k/x$. The graph is a hyperbola, two curved branches, one in Q1 (for $k > 0$) and one in Q3.
Identifying from a table:
For $k = 12$: $x$ = 1, 2, 3, 4, 6, 12 and $y$ = 12, 6, 4, 3, 2, 1. Check: $1 \times 12 = 2 \times 6 = 3 \times 4 = 12$, constant. This confirms $y = 12/x$.
Graphing $y = k/x$:
- Choose several $x$ values (avoid $x = 0$); include both positive and negative values for the full picture.
- Calculate corresponding $y = k/x$ values.
- Plot and draw smooth curves through each branch. Do not connect the two branches.
- Show asymptotes as dashed lines along the $x$- and $y$-axes.
The reciprocal function y = k/x (k ≠ 0) forms a hyperbola with two branches, never touching the axes. Key features: asymptotes at x = 0 and y = 0; as x increases y decreases (if k > 0).
Pause, copy the general form y = k/x, the hyperbola description (two branches, never touches the axes), and both asymptotes: x = 0 (vertical) and y = 0 (horizontal) into your book.
Quick check: A table shows: $x$ = 2, 4, 5, 10 and $y$ = 10, 5, 4, 2. What is the constant of variation $k$, and what is the equation?
The hyperbola y = k/x has asymptotes at x = 0 and y = 0, and two branches, one in the first quadrant and one in the third (for k > 0). To build an inverse variation model from real data, substitute one known (x, y) pair to find k, then use y = k/x to predict any other value, real examples include speed–time for a fixed distance and pressure–volume under Boyle's Law.
Any situation where two quantities have a constant product is an inverse variation problem. Write $y = k/x$, find $k$ from one known pair, and use the equation to find unknown values.
Steps to construct and solve:
- Identify the two varying quantities call them $x$ and $y$ with their units.
- Find $k$ multiply any known pair: $k = x \times y$. Verify with a second pair if possible.
- Write the equation: $y = \frac{k}{x}$.
- Solve by substituting the known value of $x$ (or $y$) to find the unknown.
- Verify graphically: plot the point on the curve to confirm it lies on the hyperbola.
In an inverse variation model y = k/x, find k by substituting a known (x, y) pair. Then use the model to find unknown values. Real examples: speed–time, pressure–volume, density–area.
Pause, copy the k-finding method (substitute a known (x, y) pair into y = k/x and solve for k) and two real-world examples: speed–time for fixed distance and pressure–volume under Boyle's Law into your book.
True or false: In the equation $y = 24/x$, when $x = 8$ the value of $y$ is 3, and you can verify this by checking that $8 \times 3 = 24$.
With k determined from a data point, y = k/x predicts outputs at any x, but the vertical asymptote at x = 0 means the model breaks down for very small x-values, producing unrealistically large y-values. Always state the domain restriction (typically x > 0) and explain why x = 0 is physically excluded from the model.
The reciprocal model is powerful but has specific limitations. Because $y = k/x$ approaches infinity as $x \to 0$ and approaches zero as $x \to \infty$, practical situations will always impose constraints that the model doesn't capture.
Key limitations:
- $x = 0$ is undefined but in practice, some minimum value of $x$ must apply (e.g. you can't have zero workers; a machine can't have zero speed).
- Prediction of extreme values is unrealistic as $x$ gets very small, $y$ theoretically approaches infinity. In practice, this isn't possible (adding 1 million workers to a 12-hour task doesn't make it finish instantaneously).
- Non-integer solutions may be invalid if $x$ must be a whole number (workers, items), fractional answers from the model don't apply in practice.
- The product may not remain exactly constant real situations have inefficiencies, setup costs, diminishing returns. The model assumes a perfectly constant product.
Reciprocal models break down near x = 0 (the asymptote), so they are invalid for x = 0. Always identify the domain restriction and explain why very small x-values produce unrealistically large y-values.
Pause, copy the asymptote limitation (as x → 0, y → ∞; x = 0 is excluded) and the domain restriction rule: state x > 0 (or x ≠ 0) and explain why near-zero x-values produce physically impossible outputs into your book.
Fill the blanks: For $y = k/x$, as $x$ increases, $y$ . The graph has asymptotes at $y = $ and $x = $ . The constant of variation is found by $x$ and $y$.
Worked examples · 3 problems
A car travels a fixed distance. The time $T$ (hours) depends on the average speed $s$ (km/h). The table shows: $s$ = 40, 60, 80, 120 and $T$ = 3, 2, 1.5, 1. (a) Show this is inverse variation. (b) Find $k$ and write the equation. (c) How long does the journey take at 100 km/h?
$40 \times 3 = 120$
$60 \times 2 = 120$
$80 \times 1.5 = 120$
$120 \times 1 = 120$
Constant product → inverse variation ✓
$k = 120$ (from the constant product)
Equation: $T = \dfrac{120}{s}$
$T = 120 \div 100 = 1.2$ hours
For $y = 30/x$: (a) Build a table for $x = 1, 2, 3, 5, 6, 10, 15, 30$. (b) Describe the shape and asymptotes of the graph. (c) Find $x$ when $y = 6$.
$x$: 1, 2, 3, 5, 6, 10, 15, 30
$y$: 30, 15, 10, 6, 5, 3, 2, 1
Note: each $x \times y = 30$ ✓
Graph: two-branch hyperbola (for real $x$); in the practical context ($x > 0$), only the Q1 branch. Asymptotes: $x = 0$ (vertical) and $y = 0$ (horizontal). As $x \to \infty$, $y \to 0$ but never reaches zero.
Method 1 (algebraic): $6 = 30/x \Rightarrow x = 30/6 = 5$
Method 2 (from table): $(5, 6)$ is already in the table ✓
A gas is compressed in a sealed container. Pressure $P$ (kPa) and volume $V$ (L) follow $P = 400/V$ (Boyle's Law). (a) Find $P$ when $V = 5$ L. (b) What happens to $P$ as $V \to 0$? (c) State two limitations of this model.
$P = 400/5 = 80$ kPa
$P = 400/V \to \infty$
The model predicts infinite pressure as the volume approaches zero.
Limitation 2: The model assumes constant temperature (isothermal process). If temperature changes, $P \times V$ is no longer constant, and the equation breaks down.
Match each relationship to its type:
Top 3 list: Name THREE ways an inverse variation model can break down in a real physical situation.
Workers pattern: $n \times h = 12$ always. The equation is $h = 12/n$. At $n = 4$: $h = 3$ hours; at $n = 6$: $h = 2$ hours. The product is constant at 12, that's the constant of variation. The model breaks down at very high $n$ (e.g. 100 workers each doing 0.12 hours = 7.2 minutes, in practice, setup time and coordination overhead mean you can't keep reducing time indefinitely).
SA 1. The table shows values of $x$ and $y$:
| $x$ | 3 | 5 | 6 | 10 |
|---|---|---|---|---|
| $y$ | 20 | 12 | 10 | 6 |
(a) Show that this is an inverse variation relationship. (b) Find the equation. (c) Find $y$ when $x = 15$. (3 marks)
SA 2. A train takes $T$ hours to travel a fixed route at average speed $s$ km/h. When $s = 80$, $T = 3$. (a) Find the length of the route. (b) Write the equation for $T$ in terms of $s$. (c) What speed is required to complete the journey in 2 hours? (d) Why can the train not travel the route in 0 hours, according to this model? (4 marks)
SA 3. The cost $C$ per person for a group booking is modelled by $C = 960/n$ where $n$ is the number of people. (a) Find the cost per person for a group of 8. (b) How many people are needed to reduce the per-person cost to $\$40$? (c) State two limitations of this model and explain why they make the model unrealistic for very large or very small $n$. (4 marks)
📖 Comprehensive answers (click to reveal)
SA 1 (3 marks): (a) Products: $3\times20=60$, $5\times12=60$, $6\times10=60$, $10\times6=60$, constant product 60 confirms inverse variation [1]. (b) $y = 60/x$ [1]. (c) $y=60/15=4$ [1].
SA 2 (4 marks): (a) Distance $= s \times T = 80 \times 3 = 240$ km [1]. (b) $T = 240/s$ [1]. (c) $2 = 240/s \Rightarrow s = 120$ km/h [1]. (d) Setting $T = 0$ requires $s \to \infty$, infinite speed is physically impossible; the model predicts $T = 0$ only when speed is infinite, which cannot occur in practice [1].
SA 3 (4 marks): (a) $C = 960/8 = \$120$ per person [1]. (b) $40 = 960/n \Rightarrow n = 24$ people [1]. (c) Limitation 1: for very small $n$ (e.g. $n=1$), the model gives $C = \$960$, but a single person may not be permitted to book or there may be a minimum booking fee, the model ignores minimum-booking constraints [1]. Limitation 2: for very large $n$, the model predicts near-zero costs per person, but venues have capacity limits and per-person costs include service charges that don't scale to zero, the constant product assumption breaks down at scale [1].
Five timed inverse variation questions. Gold tier requires 90% + speed.
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