Direct variation
A spring stretches further the more weight hangs on it, and a car goes further the more fuel it burns. When is one quantity a fixed number times the other, and what does that number look like on a graph?
Where you are 11 in Linear relationships
Why this exists
Proportional relationships have already been met three ways: a table whose ratios share one constant of proportionality, an equation y = kx with nothing added on the end, and a graph that is a straight line through the origin.
Once Slope is known, the three tests turn out to be one fact: a proportional relationship is a line through the origin, and its constant is the line’s slope. This fact is called direct variation. It settles questions such as: one car’s fuel use is given as a table and another’s as a graph, so which goes further on a liter? Each becomes “which constant is bigger”.
The lesson runs over two days. Part 1 is the idea: y = kx as a line through the origin with slope k, in a table, an equation and a graph. Part 2 uses it: comparing relationships given in different forms, reading k as a unit rate, and finding a missing value.
Definition
y varies directly with x when there is one fixed number k, not zero, such that every pair of values satisfies
The number k is called the constant of variation. It is the same number as the constant of proportionality: “y varies directly with x”, “y is proportional to x” and y \propto x all say the same thing.
Two equivalent ways to say it, each useful in its own place:
- Ratio form. For every pair with x \ne 0, the ratio \displaystyle \frac{y}{x} is the same number k, and when x = 0, y = 0.
- Scaling form. Multiplying x by any number multiplies y by the same number. Double x and y doubles; take a third of x and y becomes a third.
The constant may be negative. The equation y = -2x is a direct variation with k = -2: as x grows, y falls, but each y is still the same fixed multiple of its x.
Intuition
Part 1: the line through the origin with slope k
Take any direct variation y = kx and walk along its graph. Put in x = 0 and out comes y = k \cdot 0 = 0, so the graph passes through the origin. Step from x to x + 1 and y goes from kx to k(x + 1) = kx + k, by the distributive property: one step across always gives k up. Two steps across give 2k up, half a step gives \displaystyle \frac{k}{2}. The rise for each step of 1 across is the same everywhere, so the graph is a straight line, and that rise is its slope. So the graph of y = kx is the line through (0, 0) with slope k.
The constant shows up in each form in its own way.
| Form | What makes it a direct variation | Where k is |
|---|---|---|
| Table | \displaystyle \frac{y}{x} the same for every pair with x \ne 0 | that common ratio |
| Equation | y = kx, with nothing added on the end | the number multiplying x |
| Graph | a straight line through (0, 0) | the slope, or the height above x = 1 |
A line that misses the origin. Now take y = 2x + 3. Each step of 1 across still gives 2 up, so its graph is a straight line with slope 2, parallel to y = 2x. But at x = 0 it gives y = 3, so it misses the origin. Its ratios change: (1, 5) has \displaystyle \frac{5}{1} = 5, while (3, 9) has \displaystyle \frac{9}{3} = 3. The +3 is a fixed amount that does not grow with x, so y is never one fixed multiple of x.
So every direct variation is a line, but not every line is a direct variation. A line y = kx + b is a direct variation exactly when b = 0.
Part 2: the constant as a unit rate
In a real setting the constant carries units, and they tell you what it means. If a car goes d kilometers on f liters of fuel and d = 16f, then k = 16 kilometers per liter: the distance for one liter, which is the unit rate. On the graph of d against f it is the slope: one more liter, sixteen more kilometers.
So two relationships given in different forms (a table and a graph, an equation and a sentence) can each be boiled down to its constant and compared as plain numbers. On shared axes the larger positive constant is the steeper line.
A missing value is just as quick. One known pair fixes \displaystyle k = \frac{y}{x}. Then y = kx gives any y from its x, and \displaystyle x = \frac{y}{k} gives any x from its y. Equivalently, any two pairs have equal ratios, \displaystyle \frac{y_1}{x_1} = \frac{y_2}{x_2}, a proportion to solve.
Results
Graph. y varies directly with x exactly when the graph is a straight line through the origin that is neither axis. The slope of that line is the constant of variation k.
Constant. For any point (x, y) on the graph with x \ne 0, \displaystyle k = \frac{y}{x}. With units, k is the unit rate: the amount of y for one unit of x.
Lines in general. The line y = kx + b is a direct variation exactly when b = 0.
Missing values. If (x_1, y_1) and (x_2, y_2) both fit the same direct variation, with x_1 and x_2 not zero, then \displaystyle \frac{y_1}{x_1} = \frac{y_2}{x_2}.
Why it is true
The slope of y = kx is k. Take any two different points on the graph, (x_1, kx_1) and (x_2, kx_2). The slope between them is
by the distributive property and then canceling the non-zero factor x_2 - x_1. Every pair of points gives the same slope, so the points lie on one straight line of slope k, and (0, 0) is one of them.
Every line through the origin, other than the axes, is a direct variation. Such a line has a slope; call it m. Take any point (x, y) on it other than the origin. The line is not the vertical axis, so x \ne 0, and the slope from (0, 0) to (x, y) is \displaystyle \frac{y - 0}{x - 0} = \frac{y}{x}. A straight line has the same slope between any two of its points (the lesson on slope triangles proves it), so \displaystyle \frac{y}{x} = m, that is y = mx. Every point fits, so y varies directly with x with k = m. The line is not the horizontal axis, so m \ne 0.
Why b \ne 0 breaks it. If y = kx + b were also y = cx for some c, then at x = 0 the first gives y = b and the second gives y = 0, so b = 0.
Equal ratios. Both pairs satisfy y = kx, so \displaystyle \frac{y_1}{x_1} = k and \displaystyle \frac{y_2}{x_2} = k, and two things equal to k are equal to each other.
Worked examples
Examples 1 to 4 belong to Part 1, Examples 5 to 7 to Part 2.
A spring: stretch against length
Radomir hangs masses from a spring that is 12 cm long when nothing hangs on it, and measures it. All the masses are well within the spring’s limit.
| Mass m (g) | 40 | 100 | 160 | 250 |
|---|---|---|---|---|
| Stretch s (cm) | 1.2 | 3 | 4.8 | 7.5 |
| Total length L (cm) | 13.2 | 15 | 16.8 | 19.5 |
(a) Does the stretch vary directly with the mass? If so, write the equation and say what the slope of its graph means.
(b) Does the total length vary directly with the mass?
Solution(a) Divide each stretch by its mass: \displaystyle \frac{1.2}{40} = \frac{3}{100} = \frac{4.8}{160} = \frac{7.5}{250} = 0.03. One ratio for every pair, so the stretch varies directly with the mass, with k = 0.03:
The graph is a line through the origin with slope 0.03: each extra gram stretches the spring a further 0.03 cm, so each extra 100 g stretches it 3 cm.
(b) No. The ratios are \displaystyle \frac{13.2}{40} = 0.33 and \displaystyle \frac{15}{100} = 0.15, already different. Yet the length grows steadily: 1.8 cm from 40 g to 100 g, 1.8 cm from 100 g to 160 g, and 2.7 cm over the 90 g from 160 g to 250 g, each 0.03 cm per gram. So the length’s graph is a line with the same slope, starting from the 12 cm the spring had to begin with:
At m = 0 this gives 12, not 0. Doubling fails too: 40 g gives 13.2 cm, and 80 g gives 0.03 \times 80 + 12 = 14.4 cm, not 26.4 cm. A line, but not a direct variation.
Which equations are direct variations?
Decide which equations show y varying directly with x. For those that do, give k.
(a) \displaystyle y = \frac{2}{5}x (b) y = 4x - 1 (c) 3y = 12x (d) y - 6 = 2(x + 3) (e) y = 5(x + 2) - 10
SolutionAn equation shows direct variation when it can be rewritten, using only steps that keep it true, as y = kx with nothing added on.
(a) Yes, already in the form, with \displaystyle k = \frac{2}{5}.
(b) No. At x = 0 it gives y = -1, so its line crosses the vertical axis at (0, -1).
(c) Yes. Dividing both sides by 3 gives y = 4x, so k = 4. Its line is parallel to that of (b).
(d) No. Distribute and then add 6 to both sides: y - 6 = 2x + 6, so y = 2x + 12. The +12 stays.
(e) Yes. Distribute: y = 5x + 10 - 10 = 5x, so k = 5. The added numbers cancel, which is why an equation has to be simplified before it is judged.
From one point to the whole line
The graph of a direct variation passes through (-6, 9).
(a) Write its equation and give its slope.
(b) Is (10, -15) on the graph? Is (4, -5)?
(c) A line has the same slope and passes through (2, 1). Is it a direct variation?
Solution(a) A direct variation passes through the origin, so the one point fixes the line: \displaystyle k = \frac{9}{-6} = -1.5, and the equation is y = -1.5x. The slope is -1.5: the line falls 1.5 for every 1 across.
(b) -1.5 \times 10 = -15, so (10, -15) is on it. -1.5 \times 4 = -6, not -5, so (4, -5) is not.
(c) No. The only direct variation with slope -1.5 is y = -1.5x, and at x = 2 that gives -3, not 1. The line through (2, 1) is parallel to it and higher: going back 2 steps from x = 2 to x = 0 climbs 2 \times 1.5 = 3, so it crosses the vertical axis at (0, 4). Its equation is y = -1.5x + 4.
A 3D printer that purges first
Before each job, Pradeep’s 3D printer pushes out a 0.6 m purge line of filament to clear its nozzle, then prints using 0.3 m of filament each minute. Let F be the total filament used, in meters, t minutes after printing begins.
(a) Write F in terms of t, and decide whether F varies directly with t.
(b) Which related quantity does vary directly with t?
(c) How much filament does a 90-minute print use in total?
Solution(a) The purge is used once, whatever the time, and then 0.3 m is added for each minute:
This is a line with slope 0.3, but it is not a direct variation, because at t = 0 the printer has already used 0.6 m. The scaling test shows it too: 10 minutes gives 3.6 m and 20 minutes gives 6.6 m. Twice the time, but not twice the filament.
(b) The filament used for printing, P = F - 0.6 = 0.3t, varies directly with t, with k = 0.3 m per minute.
(c) F = 0.3 \times 90 + 0.6 = 27 + 0.6 = 27.6 m, of which 27 m is the print itself.
Two cars: a table against a graph
Nicanor records his car’s fuel use on a long drive at steady speed.
| Fuel used f (liters) | 5 | 12 | 20 | 30 |
|---|---|---|---|---|
| Distance d (km) | 80 | 192 | 320 | 480 |
Quilla’s car, on the same road, is shown by the graph.
(a) Whose car goes further on a liter of fuel?
(b) How much fuel does each car need for 336 km?
(c) Give each car’s fuel use in liters per 100 km.
Solution(a) Find each constant from its own form. Nicanor’s table: \displaystyle \frac{80}{5} = \frac{192}{12} = \frac{320}{20} = \frac{480}{30} = 16, so d = 16f, 16 km per liter. Quilla’s graph is a straight line through the origin and (5, 70), so its slope is \displaystyle \frac{70}{5} = 14, and d = 14f, 14 km per liter. Nicanor’s car goes further, by 2 km on every liter. Drawn on the same axes, his line would be the steeper one.
(b) Solve d = kf for f: \displaystyle f = \frac{d}{k}. Nicanor needs \displaystyle \frac{336}{16} = 21 liters, Quilla \displaystyle \frac{336}{14} = 24 liters.
(c) Turned around, the fuel varies directly with the distance, with the reciprocal constant: \displaystyle f = \frac{1}{16}d for Nicanor, so \displaystyle \frac{100}{16} = 6.25 liters per 100 km; and \displaystyle f = \frac{1}{14}d for Quilla, so \displaystyle \frac{100}{14} \approx 7.14 liters per 100 km.
An equation against a story: two scrap yards
Two scrap yards buy copper wire by weight; both prices are made up for this example. Yard A posts its price as an equation, c = 1.15w, for the dollars c paid for w kilograms. Yard B pays the same amount for every kilogram, with no fixed charge or bonus, and pays Oona 27 dollars for 24 kg.
(a) Which yard pays more for each kilogram?
(b) How many kilograms must be sold at each yard to be paid 34.50 dollars?
Solution(a) Yard A’s constant is in its equation: 1.15 dollars per kilogram. Yard B pays the same for every kilogram with nothing fixed added, so its pay varies directly with the weight, and one sale fixes the constant: \displaystyle \frac{27}{24} = 1.125 dollars per kilogram. Yard A pays more, by 1.15 - 1.125 = 0.025 dollars on every kilogram. On a graph of pay against weight, yard A’s line is the steeper.
(b) Divide by the constant. Yard A: \displaystyle \frac{34.5}{1.15} = 30 kg. Yard B: \displaystyle \frac{34.5}{1.125} \approx 30.67 kg. At the better price, less wire earns the same pay.
A missing mass, and a reading past the limit
Radomir has a second spring, which stretches 4.5 cm when 180 g hangs from it. Its maker states that the stretch varies directly with the mass for loads up to 400 g.
(a) How far does it stretch under 260 g?
(b) An unknown object stretches it 7 cm. What is its mass?
(c) Another object stretches it 12 cm. What can be said about its mass?
Solution(a) \displaystyle k = \frac{4.5}{180} = 0.025 cm per gram, so s = 0.025m, and 260 g gives 0.025 \times 260 = 6.5 cm.
(b) \displaystyle m = \frac{s}{k} = \frac{7}{0.025} = 280 g. Equal ratios agree: \displaystyle \frac{4.5}{180} = \frac{7}{m} gives \displaystyle m = 180 \times \frac{7}{4.5} = 280 g.
(c) The rule s = 0.025m holds only up to 400 g, where the stretch is 0.025 \times 400 = 10 cm. A 12 cm stretch is past that, so the direct variation says nothing about this load, and \displaystyle \frac{12}{0.025} = 480 g is a number with no reason behind it. Saying more needs a fact about the spring that the equation does not give. If we assume that a heavier load always stretches this spring further, then the mass is more than 400 g; that conclusion rests on the assumption, not on s = 0.025m.
Boundaries
Through the origin, or it is not direct variation. A straight line is not enough. A starting amount (a spring’s own length, a purge line, a fee) gives y = kx + b with b \ne 0, which is not a direct variation (Examples 1(b) and 4). Subtracting the starting amount often leaves a quantity that does vary directly.
The constant is not zero. y = 0x makes every output 0, and the graph is the horizontal axis. It is not counted as a direct variation: the output says nothing about the input, and the relationship cannot be turned around to give x from y. The vertical axis is not the graph of y = kx for any k.
Only over the stated range. Real direct variations hold over a range of values: a spring within its limit, a car at steady speed. Outside it the equation still produces numbers, but they no longer describe anything (Example 7(c)).
The quantity must be the right one. Distance driven varies directly with fuel used, but fuel left in the tank does not; stretch varies directly with mass, but total length does not.
Slopes compare by eye only on the same axes. Across two separate graphs, compare the constants as numbers.
Where this goes next
A direct variation is the simplest kind of line, the one through the origin. Every other line is a direct variation moved up or down: slope-intercept form, y = mx + b, adds the starting amount b. The rate of change is the slope’s name in a real setting, and the slope formula measures it for lines through any point. Modeling with linear relationships decides which kind of line a situation needs.
The next lesson, inversely proportional relationships, looks at the opposite pattern, where the product xy stays fixed instead of the ratio. Much later, inverse, joint and combined variation extends both to several quantities.
Exercises
Part 1: tables
Decide whether y varies directly with x. If it does, write the equation; if not, say why.
- x: 2, 5, 8, 10; y: 7, 17.5, 28, 35.
- x: 1, 2, 4, 6; y: 5, 8, 14, 20.
- A spring within its limit. Mass (g): 60, 90, 150, 240; stretch (cm): 2.4, 3.6, 6, 9.6.
- x: -4, -1, 3, 8; y: 10, 2.5, -7.5, -20.
Part 1: equations
Decide whether each equation shows y varying directly with x. If it does, give k.
- \displaystyle y = \frac{x}{6}
- 5y = 8x
- y = 3(x - 4) + 12
- 2x + 3y = 0
Part 1: graphs and points
- The graph of a direct variation passes through (8, 6). Write its equation, give its slope, and name two more points on it.
- The graph of a direct variation passes through (-5, 12). Find k, and decide whether (2.5, -6) is on the graph.
- A line has slope 2.5 and passes through (4, 13). Is it the graph of a direct variation?
- The graph of the filament L meters used by a 3D printer against the printing time t minutes is a straight line through the origin and (20, 7). Write the equation, say what the slope means, and find the filament used in 50 minutes and the time to use 28 m.
Part 2: comparing two relationships
- Car C’s fuel use is given by a table. Fuel (liters): 6, 15, 25; distance (km): 105, 262.5, 437.5. Car D’s distance is d = 19f for f liters. Which car goes further on a liter, and how much further does it go on 40 liters?
- Recipe X for a cookie dough uses 174 g of butter with 150 g of flour, and the butter is proportional to the flour. Recipe Y’s graph of butter against flour is a straight line through the origin and (50, 57). Which recipe uses more butter per gram of flour, and how much more butter does it use with 600 g of flour?
- Spring P stretches 3.6 cm under 80 g and 9 cm under 200 g. Spring Q’s stretch is s = 0.05m. Both are within their limits up to 400 g. How far does each stretch under 300 g? Whose graph is steeper, and which spring is stiffer?
Part 2: missing values
- y varies directly with x, and y = 18 when x = 12. Find y when x = 30, and x when y = 27.
- y varies directly with x, and y = -14 when x = 4. Find y when x = -10, and x when y = 21.
- At one exchange booth, with no fee, Mireille hands over 80 dollars and receives 1360 pesos (a made-up rate). At the same booth, how many pesos does 35 dollars buy, and how many dollars are needed for 2550 pesos?
Reasoning
- Explain, in two different ways, why y = 3x + 5 is the equation of a line but not a direct variation.
- Show that for any two different points on the graph of y = kx, the slope between them is k.
- Show that if y varies directly with x with constant k, then x varies directly with y, and find its constant. What does this say about kilometers per liter and liters per kilometer?
- y varies directly with x with constant a, and z varies directly with y with constant b. Show that z varies directly with x, and give its constant.
Answers
- Yes, y = 3.5x: every ratio is \displaystyle \frac{7}{2} = 3.5.
- No: \displaystyle \frac{5}{1} = 5 but \displaystyle \frac{8}{2} = 4. The points lie on y = 3x + 2, which misses the origin.
- Yes, s = 0.04m: every ratio is \displaystyle \frac{2.4}{60} = 0.04.
- Yes, y = -2.5x: every ratio is \displaystyle \frac{10}{-4} = -2.5.
- Yes, \displaystyle k = \frac{1}{6}, since \displaystyle \frac{x}{6} = \frac{1}{6}x.
- Yes, k = 1.6: dividing by 5 gives \displaystyle y = \frac{8}{5}x.
- Yes, k = 3: y = 3x - 12 + 12 = 3x.
- Yes, \displaystyle k = -\frac{2}{3}: 3y = -2x, so \displaystyle y = -\frac{2}{3}x.
- y = 0.75x, slope 0.75, since \displaystyle \frac{6}{8} = 0.75. For example (4, 3) and (12, 9).
- \displaystyle k = \frac{12}{-5} = -2.4. Yes: -2.4 \times 2.5 = -6.
- No: going back 4 steps to x = 0 drops 4 \times 2.5 = 10, so it is y = 2.5x + 3, crossing at (0, 3).
- L = 0.35t, from \displaystyle \frac{7}{20} = 0.35: 0.35 m of filament each minute. In 50 minutes, 0.35 \times 50 = 17.5 m; 28 m takes \displaystyle \frac{28}{0.35} = 80 minutes.
- Car C: \displaystyle \frac{105}{6} = 17.5 km per liter; car D: 19. Car D, by 19 \times 40 - 17.5 \times 40 = 760 - 700 = 60 km on 40 liters.
- X: \displaystyle \frac{174}{150} = 1.16 g of butter per gram of flour; Y: \displaystyle \frac{57}{50} = 1.14. Recipe X, by 1.16 \times 600 - 1.14 \times 600 = 696 - 684 = 12 g.
- P: \displaystyle \frac{3.6}{80} = 0.045 cm per gram, so 0.045 \times 300 = 13.5 cm. Q: 0.05 \times 300 = 15 cm. Q’s graph is steeper, so P is stiffer: it stretches less per gram.
- k = 1.5, so y = 1.5 \times 30 = 45 and \displaystyle x = \frac{27}{1.5} = 18.
- k = -3.5, so y = -3.5 \times (-10) = 35 and \displaystyle x = \frac{21}{-3.5} = -6.
- \displaystyle k = \frac{1360}{80} = 17 pesos per dollar. 17 \times 35 = 595 pesos; \displaystyle \frac{2550}{17} = 150 dollars.
- First, at x = 0 it gives y = 5, so it misses the origin, which every direct variation passes through. Second, the ratio is not fixed: (1, 8) has \displaystyle \frac{8}{1} = 8 but (5, 20) has \displaystyle \frac{20}{5} = 4. It is still a line, since each step of 1 across adds 3.
- For points (x_1, kx_1) and (x_2, kx_2) with x_1 \ne x_2, the slope is \displaystyle \frac{k(x_2 - x_1)}{x_2 - x_1} = k: factor out k, then cancel x_2 - x_1 \ne 0.
- k \ne 0, so dividing y = kx by k gives \displaystyle x = \frac{1}{k}y: x varies directly with y, with constant \displaystyle \frac{1}{k}. A car going 16 km per liter uses \displaystyle \frac{1}{16} of a liter per kilometer: the two rates are reciprocals.
- y = ax and z = by, so z = b(ax) = (ab)x, by the associative and commutative properties. The number ab is fixed and not zero, since neither a nor b is, so z varies directly with x with constant ab.