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Linear and non-linear relations

Algebra · level 6, lesson 6 of 6

A question to start

Drop a ball from a high window. If we round Earth's pull of gravity to 10 metres per second squared and ignore air, the ball has fallen

Time (s) 0 1 2 3 4
Distance fallen (m) 0 5 20 45 80

In the first second it falls 5 m. Does that mean it falls 10 m in two seconds? No: it falls 20 m. In four seconds it falls not 20 m but 80 m. Distance is not proportional to time, and the relation is not even linear: the graph of these points is a curve.

On Mars, where gravity is about 3.71 metres per second squared, the same drop gives about 1.9 m, 7.4 m, 16.7 m and 29.7 m after 1, 2, 3 and 4 seconds: a gentler curve, but still a curve.

How can you tell, from a table, a graph or an equation, whether a relation is a straight line or not? And when you measure something real, how do you decide whether a line or a curve describes it better? That is this lesson.

The idea

What "linear" means, once more

A relation is linear when its graph is a straight line. In level 5 you saw what that means in each view:

  • Table: when \(x\) goes up in equal steps, \(y\) goes up (or down) by the same amount each time.
  • Graph: the points lie on a straight line.
  • Equation: it can be written as \(y = mx + b\).

Everything else is non-linear. This lesson makes each test precise and explains why the three tests agree.

Test 1: first differences in a table

For \(y = 3x + 2\):

\(x\) 0 1 2 3 4
\(y\) 2 5 8 11 14
first difference 3 3 3 3

Why the differences are constant for every line. Take \(y = mx + b\) and increase \(x\) by a step \(h\). The new \(y\) is \(m(x + h) + b = mx + b + mh\). So \(y\) always goes up by exactly \(mh\), no matter where you start. Equal steps in \(x\) give equal steps in \(y\). The reverse is also true: if equal steps always give the same change, the points lie on one straight line (level 5 showed why: identical steps never bend).

For the falling ball, the first differences are \(5, 15, 25, 35\): not constant, so not linear.

Unequal steps: use the rate

The first-difference test only works if the \(x\) values go up in equal steps. If they do not, compare the rate of change between neighbouring points instead:

\[\text{rate} = \frac{\text{change in } y}{\text{change in } x}.\]

The relation is linear exactly when this rate is the same between every pair of neighbours. (This rate is the slope of the segment joining two points; equal slopes mean the segments line up.)

Second differences and ratios: a first look at curves

When the first differences are not constant, take differences again. For the ball: first differences \(5, 15, 25, 35\); second differences \(10, 10, 10\). Constant second differences are the mark of a quadratic relation, one with \(x^2\) in it. Here \(d = 5t^2\). Quadratics are the subject of level 7.

Another common pattern: each \(y\) is the previous one multiplied by the same number, like \(1, 2, 4, 8, 16\). The first differences are \(1, 2, 4, 8\), which never settle down, but the ratios are all 2. That is an exponential relation, \(y = 2^x\), also in level 7.

Pattern in the table Kind of relation Example
constant first differences linear \(y = 3x + 2\)
constant second differences quadratic \(y = 5x^2\)
constant ratios exponential \(y = 2^x\)

Test 2: the graph

On a graph, a linear relation is a straight line; a non-linear one curves, bends or jumps. Be careful with just a few points: three points can look roughly lined up by accident. Check with the table or the equation too.

Test 3: the equation

An equation in \(x\) and \(y\) is linear when it can be rearranged into \(y = mx + b\) (or is a vertical line \(x = c\)). In practice: \(x\) and \(y\) appear only to the first power, are not multiplied together, are not under a square root or in a denominator, and are not in an exponent.

Linear Non-linear
\(y = 3 - 2x\) \(y = x^2 + 1\) (a square)
\(3x - 4y = 12\) (rearranges to \(y = \tfrac34 x - 3\)) \(xy = 12\) (a product)
\(y = \dfrac{x + 1}{2}\) (that is \(\tfrac12 x + \tfrac12\)) \(y = \dfrac{5}{x}\) (\(x\) in the denominator)
\(y = 7\) (a flat line, slope 0) \(y = 2^x\) (\(x\) in the exponent)

Look at the third row: dividing by a number is fine, dividing by \(x\) is not. And sometimes an equation looks non-linear but is not: \(y = (x + 3)^2 - x^2\) expands to \(6x + 9\), which is linear.

Modelling real data

Real measurements never fit a pattern exactly: rulers have limits and hands shake. So with real data the question is not "is it exactly linear?" but "is a straight line a good description?"

Suppose a class hangs masses from a spring and measures its length:

Mass (g) 0 100 200 300 400 500
Length (cm) 10.0 12.1 13.9 16.0 18.1 19.9

The first differences are \(2.1, 1.8, 2.1, 2.1, 1.8\): not identical, but all close to 2. Plotted, the points lie very close to a line. A good model is the line through the start and end, \(L = 10 + 0.02m\) (length in cm, mass in g): the spring is 10 cm long unloaded and stretches 2 cm per 100 g.

0200 400600 1020 mass (g)length (cm) model: L = 10 + 0.02m

Now a curved example. A pendulum's period is the time for one full swing, there and back. For small swings it depends on the length of the string:

Length (m) 0.25 0.5 1 2
Period on Earth (s) 1.00 1.42 2.01 2.84

The \(x\) steps are not equal, so use rates: \(\tfrac{0.42}{0.25} = 1.68\), then \(\tfrac{0.59}{0.5} = 1.18\), then \(\tfrac{0.83}{1} = 0.83\) seconds per metre. The rate keeps falling, so the graph bends over: not linear.

00.5 11.5 2 123 length (m)period (s) dashed: a straight linemisses the middle points

(Physics gives the curve's equation: the period is \(2\pi\sqrt{L/g}\), a square root, which explains the bend. On Mars, with weaker gravity, a 1 m pendulum swings more slowly: about 3.26 seconds per swing instead of 2.01.)

Interpolating and extrapolating

Using a model inside the range of the data (for example, the spring with 250 g) is called interpolation, and is usually reliable. Using it outside the range (the spring with 5 kg) is extrapolation, and is risky: the real spring may stretch differently or snap, and the straight line knows nothing about that. A model is only trusted where it has been tested.

Worked examples

1. A linear table. \(x\): 0, 1, 2, 3, 4 and \(y\): 3, 7, 11, 15, 19.

  • First differences: 4, 4, 4, 4. Constant, so linear. Starting value 3, so \(y = 4x + 3\).

2. A quadratic table. \(x\): 0, 1, 2, 3, 4 and \(y\): 0, 1, 4, 9, 16.

  • First differences: 1, 3, 5, 7: not constant, so non-linear. Second differences: 2, 2, 2: quadratic. Indeed \(y = x^2\).

3. Unequal steps. \(x\): 1, 2, 4, 7 and \(y\): 5, 8, 14, 23.

  • Rates: \(\tfrac{3}{1} = 3\), \(\tfrac{6}{2} = 3\), \(\tfrac{9}{3} = 3\). Constant, so linear with slope 3.
  • Find \(b\) from \((1, 5)\): \(5 = 3 + b\), so \(b = 2\) and \(y = 3x + 2\). Check: \(3 \times 7 + 2 = 23\) ✓.
  • (The plain differences 3, 6, 9 would have suggested "not linear": they are unequal only because the steps are.)

4. Classifying equations. Which are linear? \(y = 3 - 2x\); \(y = x^2 + 1\); \(xy = 12\); \(3x - 4y = 12\); \(y = 2^x\); \(y = \tfrac{x + 1}{2}\).

  • Linear: \(y = 3 - 2x\), \(3x - 4y = 12\), and \(y = \tfrac{x + 1}{2}\). Non-linear: \(y = x^2 + 1\), \(xy = 12\), \(y = 2^x\).

5. Using the spring model. Predict the length with 250 g, and comment on 5 kg.

  • 250 g: \(L = 10 + 0.02 \times 250 = 15\) cm. Interpolation, between 200 g (13.9 cm) and 300 g (16.0 cm): reasonable.
  • 5 kg = 5000 g: \(L = 10 + 100 = 110\) cm. Extrapolation, ten times beyond the data. A spring 10 cm long stretched to over a metre has almost certainly been bent out of shape. Do not trust this.

Try it

  1. Is it linear? If so, find the equation. \(x\): 0, 1, 2, 3, 4 and \(y\): 10, 7, 4, 1, −2.
  2. Is it linear? \(x\): 0, 1, 2, 3, 4 and \(y\): 1, 3, 9, 27, 81. What pattern does it have?
  3. Is it linear? If so, find the equation. \(x\): 1, 3, 4, 8 and \(y\): 2, 8, 11, 23.
  4. Linear or non-linear? (a) \(y = 7x\); (b) \(y = x^3\); (c) \(2x + 5y = 10\); (d) \(y = \dfrac{1}{x + 1}\); (e) \(y = x(x - 2)\); (f) \(y = \dfrac{3x - 6}{3}\).
  5. For squares with side 1, 2, 3, 4 cm, make tables of the perimeter and of the area. Which relation is linear?
  6. Use the spring model \(L = 10 + 0.02m\) to find the mass that makes the spring 17 cm long. Is this interpolation or extrapolation?
  7. Show that the second differences of \(y = x^2\) are always 2. (Hint: work out \((x + 2)^2 - 2(x + 1)^2 + x^2\).)
Answers
  1. First differences \(-3, -3, -3, -3\): linear, going down. \(y = -3x + 10\). Check: \(-12 + 10 = -2\) ✓.
  2. First differences 2, 6, 18, 54: not linear. Each \(y\) is 3 times the one before (constant ratio 3), so it is exponential: \(y = 3^x\).
  3. Rates: \(\tfrac{6}{2} = 3\), \(\tfrac{3}{1} = 3\), \(\tfrac{12}{4} = 3\). Linear with slope 3. From \((1, 2)\): \(2 = 3 + b\), so \(b = -1\) and \(y = 3x - 1\). Check: \(3 \times 8 - 1 = 23\) ✓.
  4. (a) Linear (direct variation). (b) Non-linear (a cube). (c) Linear: \(y = -\tfrac25 x + 2\). (d) Non-linear (\(x\) in the denominator). (e) Non-linear: it expands to \(x^2 - 2x\). (f) Linear: it simplifies to \(y = x - 2\).
  5. Perimeter: 4, 8, 12, 16, differences all 4: linear (\(P = 4s\)). Area: 1, 4, 9, 16, differences 3, 5, 7: non-linear (quadratic, \(A = s^2\); second differences all 2).
  6. \(10 + 0.02m = 17\), so \(0.02m = 7\) and \(m = 350\) g. That is inside the data (0 to 500 g): interpolation.
  7. Expand: \((x^2 + 4x + 4) - 2(x^2 + 2x + 1) + x^2 = x^2 + 4x + 4 - 2x^2 - 4x - 2 + x^2 = 2\). The second difference of three neighbouring values is always 2, wherever you start.

Common mistake

Using first differences when the steps are unequal. In worked example 3 the differences 3, 6, 9 look non-linear, but the \(x\) steps are 1, 2 and 3. Always check that the \(x\) values go up evenly; if not, divide to get rates.

Calling an equation non-linear because it has a fraction or is rearranged. \(3x - 4y = 12\) and \(y = \tfrac{x + 1}{2}\) are both straight lines. What matters is the powers of \(x\) and \(y\) and whether \(x\) is in a denominator or an exponent, not how the equation happens to be written.

Where it leads