Direct and Inverse Variation

Many laws of science have the simple form $y = ax^n$ : the braking distance of a car grows with the square of its speed, the pressure of a gas is inversely proportional to its volume, and light gets dimmer with the square of the distance. These are called variation models. On this page you’ll learn to recognize them, find the constant $a$ from data, read their graphs and asymptotes, and fit cubic models, all while working through the modelling cycle the IB expects you to use.

Key ideas

Direct variation

$y$ varies directly as $x^n$ (written $y \propto x^n$ ) when

y = ax^n, \qquad n \in \mathbb{Z}, \ n \gt 0

for some constant $a \ne 0$ , called the constant of variation (or constant of proportionality).

$y \propto x$ : $y = ax$ , a straight line through the origin.
$y \propto x^2$ : $y = ax^2$ , a parabola with vertex at the origin.
$y \propto x^3$ : $y = ax^3$ , a cubic through the origin.

The key test: the ratio $\dfrac{y}{x^n}$ is the same for every data point, and it equals $a$ .

Inverse variation

$y$ varies inversely as $x^n$ (written $y \propto \dfrac{1}{x^n}$ ) when

y = \frac{a}{x^n} = ax^{-n}, \qquad n \in \mathbb{Z}, \ n \gt 0

So inverse variation is the same family $f(x) = ax^n$ with a negative integer power. The most common cases are $y \propto \dfrac{1}{x}$ (for example Boyle’s law for a gas) and $y \propto \dfrac{1}{x^2}$ (an inverse-square law, like light intensity or gravity).

The key test: the product $x^n y$ is the same for every data point, and it equals $a$ .

Graphs and asymptotes

Direct variation graphs pass through the origin. For a negative power, the

y

-axis is a vertical asymptote and the

x

-axis is a horizontal asymptote.

Model	Passes through the origin?	Asymptotes
$y = ax^n$ , $n \gt 0$	yes	none
$y = ax^n$ , $n \lt 0$	no ( $x = 0$ is not in the domain)	vertical: $x = 0$ (the $y$ -axis); horizontal: $y = 0$ (the $x$ -axis)

For $n \lt 0$ , the domain is $\{x \in \mathbb{R} \mid x \ne 0\}$ . In real contexts the input is usually positive (a volume, a distance), so only the branch with $x \gt 0$ matters.

What happens when x is multiplied

If $y = ax^n$ and $x$ is multiplied by $k$ , then $y$ is multiplied by $k^n$ :

a(kx)^n = k^n \cdot ax^n

So if $y \propto x^2$ , doubling $x$ multiplies $y$ by $4$ . If $y \propto \dfrac{1}{x^2}$ (that is, $n = -2$ ), doubling $x$ multiplies $y$ by $2^{-2} = \dfrac{1}{4}$ .

Cubic models

A cubic model has the form

f(x) = ax^3 + bx^2 + cx + d

It can rise, fall and rise again (or the reverse), so it suits data with up to two turning points: a profit that grows and then falls, the volume of a box as you change one length, or the power produced by a wind turbine. To find the parameters, substitute known points and solve the resulting system of linear equations with your GDC. If the $y$ -intercept is known (for example a starting value at $x = 0$ ), then $d$ is known straight away, and only three equations in $a$ , $b$ and $c$ remain.

The modelling cycle

The IB describes modelling as a cycle, not a one-off calculation:

Develop the model. From the context and the shape of the data, choose a type ( $y = ax^n$ , cubic, exponential, …) and decide on a reasonable domain (what inputs make sense?).
Fit the model. Find the parameters: from one or more data points, from an initial condition, or by solving equations with technology.
Test the model. Compare its predictions with data you didn’t use. Are the differences small?
Reflect. Is the model reasonable in context? Justify your choice from the shape of the data, the properties of the curve, or the situation itself.
Use the model. Read values, interpret parameters and make predictions, remembering that extrapolation (predicting outside the data) is risky.

If the test or reflection shows a problem, go back to step 1 and try again. For more practice choosing between linear, quadratic, exponential and sinusoidal models, see function modelling. For fitting curves to all the data at once (regression), see non-linear regression.

Worked examples

Example 1: Braking distance

The braking distance $d$ metres of a car varies directly as the square of its speed $v$ km/h. At $50$ km/h the braking distance is $12.5$ m.

(a) Find a formula for $d$ in terms of $v$ .
(b) Find the braking distance at $100$ km/h.
(c) Find the speed at which the braking distance is $40$ m.

Solution.

(a) $d \propto v^2$ , so $d = av^2$ . Substitute $v = 50$ , $d = 12.5$ :

12.5 = a(50)^2 \quad\Rightarrow\quad a = \frac{12.5}{2500} = 0.005

So $d = 0.005v^2$ .

(b) $d = 0.005(100)^2 = 50$ m. Notice that doubling the speed multiplied the distance by $2^2 = 4$ : from $12.5$ m to $50$ m.

v^2 = \frac{40}{0.005} = 8000 \quad\Rightarrow\quad v = \sqrt{8000} \approx 89.4 \text{ km/h} \quad (3 \text{ s.f.})

Take the positive root, since a speed is positive here.

Example 2: Boyle’s law from data

A fixed amount of gas is kept at constant temperature. Its pressure $P$ (kPa) is measured at different volumes $V$ (L).

$V$ (L)	$1$	$2$	$4$	$5$	$8$
$P$ (kPa)	$400$	$200$	$100$	$80$	$50$

(a) Show that $P$ varies inversely as $V$ , and write the model.
(b) State the equations of the asymptotes of the graph, and a reasonable domain.
(c) Find the volume when the pressure is $250$ kPa.

Solution.

(a) When $V$ doubles from $1$ to $2$ , $P$ halves, which suggests $P \propto \dfrac{1}{V}$ . Check the products $PV$ :

1 \times 400 = 2 \times 200 = 4 \times 100 = 5 \times 80 = 8 \times 50 = 400

They are all equal, so $P = \dfrac{400}{V}$ (that is, $P = 400V^{-1}$ ).

(b) The vertical asymptote is $V = 0$ and the horizontal asymptote is $P = 0$ . A volume must be positive, so a reasonable domain is $V \gt 0$ (in practice, only the range of volumes the container allows).

Example 3: Which inverse law?

The intensity $I$ (lux) of light from a lamp is measured at distance $x$ metres.

$x$ (m)	$1$	$2$	$3$	$4$
$I$ (lux)	$1440$	$360$	$160$	$90$

(a) Decide whether $I \propto \dfrac{1}{x}$ or $I \propto \dfrac{1}{x^2}$ , and write the model.
(b) Predict the intensity at $6$ m.
(c) At what distance is the intensity $10$ lux?

Solution.

(a) Test both. If $I \propto \dfrac{1}{x}$ , then $xI$ is constant; if $I \propto \dfrac{1}{x^2}$ , then $x^2 I$ is constant.

$x$	$1$	$2$	$3$	$4$
$xI$	$1440$	$720$	$480$	$360$
$x^2 I$	$1440$	$1440$	$1440$	$1440$

Only $x^2 I$ is constant, so this is an inverse-square law:

I = \frac{1440}{x^2}

(b) $I = \dfrac{1440}{36} = 40$ lux. (Check with the scaling rule: from $3$ m to $6$ m the distance doubles, so the intensity is divided by $4$ : $160 \div 4 = 40$ . ✓)

Example 4: A cubic profit model and the modelling cycle

A new café records its monthly profit $P$ , in thousands of dollars, $t$ months after opening. It opened with a loss of $5$ thousand dollars.

$t$ (months)	$0$	$2$	$4$	$6$
$P$ (thousand $)	$-5$	$-4.0$	$1.8$	$7.6$

The owner models the profit with $P(t) = at^3 + bt^2 + ct + d$ .

(a) Find $a$ , $b$ , $c$ and $d$ .
(b) The profit at $t = 8$ was $8.4$ thousand dollars. Test the model with this value.
(c) Find the maximum monthly profit predicted by the model.
(d) Use the model to predict the profit at $t = 12$ , and reflect on the result.

Solution.

(a) At $t = 0$ , $P = d = -5$ . Substitute the other three points into $P(t) = at^3 + bt^2 + ct - 5$ :

\begin{aligned} 8a + 4b + 2c - 5 &= -4.0 &&\Rightarrow\quad 8a + 4b + 2c = 1.0 \\ 64a + 16b + 4c - 5 &= 1.8 &&\Rightarrow\quad 64a + 16b + 4c = 6.8 \\ 216a + 36b + 6c - 5 &= 7.6 &&\Rightarrow\quad 216a + 36b + 6c = 12.6 \end{aligned}

Solve the system with your GDC’s simultaneous-equation solver: $a = -0.1$ , $b = 1.2$ , $c = -1.5$ .

P(t) = -0.1t^3 + 1.2t^2 - 1.5t - 5

(b) $P(8) = -0.1(512) + 1.2(64) - 1.5(8) - 5 = -51.2 + 76.8 - 12 - 5 = 8.6$ . The actual profit was $8.4$ , so the model is off by only $0.2$ thousand dollars. The model passes this test.

(c) Graph $P$ on your GDC and find the maximum: at $t \approx 7.32$ months, $P \approx 9.10$ , so a maximum profit of about $9100 per month (to 3 s.f.).

(d) $P(12) = -0.1(1728) + 1.2(144) - 18 - 5 = -23$ , a loss of $23 000 in a month. In fact the model reaches $P = 0$ at $t = 10$ and falls steeply after that, because every cubic with $a \lt 0$ eventually decreases without limit. Nothing in the data suggests the café will collapse, so this is a warning about extrapolation: the model is reasonable for about $0 \le t \le 8$ (the data), and predictions far beyond that shouldn’t be trusted. To predict further ahead, the owner should collect more data and refit (back to step 1 of the cycle).

Common mistakes

Writing the constant from the wrong ratio. For $y \propto x^2$ , the constant is $a = \dfrac{y}{x^2}$ , not $\dfrac{y}{x}$ . For $y \propto \dfrac{1}{x^2}$ , it’s $a = x^2 y$ . Write the general equation first ( $y = ax^2$ or $y = \dfrac{a}{x^2}$ ), then substitute.

Checking only one pair of points. Any single point gives some value of $a$ . To decide which law fits a table, compute $\dfrac{y}{x^n}$ or $x^n y$ for every point and look for a constant value.

Thinking inverse variation means “subtract”. “As $x$ increases, $y$ decreases” isn’t enough: $y = 10 - x$ also decreases. Inverse variation means $x^n y$ stays constant, so doubling $x$ divides $y$ by $2^n$ .

Forgetting the asymptotes and the domain. For $y = ax^n$ with $n \lt 0$ , $x = 0$ is not in the domain; the $y$ -axis is a vertical asymptote and the $x$ -axis is a horizontal asymptote. In context, restrict the domain further to values that make sense (for example $V \gt 0$ ).

Taking the negative root without thinking. Solving $0.005v^2 = 40$ gives $v = \pm 89.4$ , but a speed or a length must be positive. State which root you keep and why.

Trusting a cubic far outside the data. A cubic always heads to $\pm\infty$ at both ends. A model that fits well on the data can give absurd predictions a short way beyond it. Always state a reasonable domain.

Practice

1. (Warm-up) $y$ varies directly as $x$ , and $y = 18$ when $x = 4$ . Find $y$ when $x = 10$ .

Solution

$y = ax$ with $18 = 4a$ , so $a = 4.5$ . Then $y = 4.5(10) = 45$ .

2. (Warm-up) $y$ is inversely proportional to $x^2$ , and $y = 5$ when $x = 2$ . Find $y$ when $x = 4$ and when $x = 0.5$ .

Solution

$y = \dfrac{a}{x^2}$ with $5 = \dfrac{a}{4}$ , so $a = 20$ and $y = \dfrac{20}{x^2}$ .

$x = 4$ : $y = \dfrac{20}{16} = 1.25$ .

$x = 0.5$ : $y = \dfrac{20}{0.25} = 80$ .

3. (Warm-up) Let $f(x) = 6x^{-1}$ . Write down the equations of the asymptotes of the graph of $f$ , and its domain and range.

Solution

$f(x) = \dfrac{6}{x}$ . Vertical asymptote $x = 0$ ; horizontal asymptote $y = 0$ .

Domain $\{x \in \mathbb{R} \mid x \ne 0\}$ ; range $\{y \in \mathbb{R} \mid y \ne 0\}$ .

4. (Core) Here is a table of values.

$x$	$1$	$2$	$3$	$5$
$y$	$0.6$	$2.4$	$5.4$	$15$

(a) Show that $y \propto x^2$ and find the model.
(b) Find the positive value of $x$ for which $y = 60$ .

Solution

(a) $\dfrac{y}{x^2}$ : $\dfrac{0.6}{1} = 0.6$ , $\dfrac{2.4}{4} = 0.6$ , $\dfrac{5.4}{9} = 0.6$ , $\dfrac{15}{25} = 0.6$ . The ratio is constant, so $y = 0.6x^2$ .

(b) $0.6x^2 = 60$ gives $x^2 = 100$ , so $x = 10$ .

5. (Core) Charles’s law says that, at constant pressure, the volume $V$ of a gas varies directly as its temperature $T$ in kelvin. A balloon has a volume of $2.5$ L at $300$ K.

(a) Find the volume at $360$ K.
(b) Find the temperature, in kelvin and in degrees Celsius, at which the volume is $2.0$ L. (Use $T(\text{K}) = T(^\circ\text{C}) + 273.15$ .)
(c) Explain why the domain of this model must be $T \gt 0$ .

Solution

(a) $V = aT$ with $a = \dfrac{2.5}{300} = \dfrac{1}{120}$ . So $V = \dfrac{360}{120} = 3$ L.

(b) $\dfrac{T}{120} = 2.0$ gives $T = 240$ K, which is $240 - 273.15 = -33.15$ , about $-33.2\,^\circ\text{C}$ (3 s.f.).

(c) Temperatures in kelvin can’t be negative (and a gas would liquefy long before $0$ K), and the model would give a zero or negative volume at $T \le 0$ , which is impossible.

6. (Core) A theatre finds that the number of tickets $q$ it sells for a show varies inversely as the price $p$ dollars. At a price of $8 it sells $450$ tickets.

(a) Find the model, and the number of tickets sold at $12.
(b) Show that the model predicts the same revenue at every price.
(c) Comment on whether the model is reasonable for very low prices.

Solution

(a) $q = \dfrac{a}{p}$ with $a = 8 \times 450 = 3600$ . So $q = \dfrac{3600}{p}$ , and at $12, $q = \dfrac{3600}{12} = 300$ tickets.

(b) Revenue $= pq = p \cdot \dfrac{3600}{p} = 3600$ , so $3600 whatever the price.

(c) Not reasonable. As $p \to 0$ the model predicts $q \to \infty$ , but the theatre has a fixed number of seats. At $5 it already predicts $720$ tickets, which a small theatre may not have. The model should only be used over a limited range of prices.

7. (Core) The temperature $T$ (°C) in a greenhouse $h$ hours after 6 a.m. is modelled by $T(h) = ah^3 + bh^2 + ch + d$ .

$h$	$0$	$2$	$4$	$6$
$T$ (°C)	$12$	$14.5$	$19.4$	$24.3$

(a) Write down the value of $d$ , and set up three equations for $a$ , $b$ and $c$ .
(b) Solve them with your GDC and write down the model.
(c) Find the maximum temperature predicted, and the time it occurs.
(d) Predict the temperature at $h = 16$ and comment.

Solution

(a) $T(0) = d = 12$ . Then

\begin{aligned} 8a + 4b + 2c &= 2.5 \\ 64a + 16b + 4c &= 7.4 \\ 216a + 36b + 6c &= 12.3 \end{aligned}

(b) $a = -0.05$ , $b = 0.6$ , $c = 0.25$ , so $T(h) = -0.05h^3 + 0.6h^2 + 0.25h + 12$ .

(c) Using the GDC’s maximum feature: $h \approx 8.20$ , $T \approx 26.8\,^\circ\text{C}$ . So the maximum is about $26.8\,^\circ\text{C}$ at about 2:12 p.m.

(d) $T(16) = -0.05(4096) + 0.6(256) + 4 + 12 = -204.8 + 153.6 + 16 = -35.2\,^\circ\text{C}$ . That’s absurd for a greenhouse at 10 p.m.: the cubic keeps falling steeply after its maximum. The model is only reasonable for roughly the daytime hours covered by the data (about $0 \le h \le 10$ ).

8. (Challenge) The power $P$ produced by a wind turbine is proportional to the cube of the wind speed $v$ . At $6$ m/s it produces $50$ kW.

(a) Find the power at $12$ m/s.
(b) By what percentage must the wind speed increase to double the power?

Solution

(a) Doubling $v$ multiplies $P$ by $2^3 = 8$ , so $P = 8 \times 50 = 400$ kW. (Or: $a = \dfrac{50}{216}$ , and $P = \dfrac{50}{216}(12)^3 = 400$ .)

(b) If $v$ is multiplied by $k$ , $P$ is multiplied by $k^3$ . We need $k^3 = 2$ , so $k = \sqrt[3]{2} \approx 1.2599$ . The wind speed must increase by about $26.0\%$ (3 s.f.).

9. (Challenge) The gravitational force $F$ between two objects is inversely proportional to the square of the distance $d$ between their centres.

(a) If $d$ increases by $25\%$ , by what percentage does $F$ decrease?
(b) By what factor must $d$ change for $F$ to become $\dfrac{1}{9}$ of its original value?

Solution

(a) $F = \dfrac{a}{d^2}$ . Replacing $d$ by $1.25d$ multiplies $F$ by $\dfrac{1}{1.25^2} = \dfrac{1}{1.5625} = 0.64$ . So $F$ decreases by $36\%$ .

(b) We need $\dfrac{1}{k^2} = \dfrac{1}{9}$ , so $k^2 = 9$ and $k = 3$ (a distance factor is positive). The distance must be tripled.