Rates of Change in Context

For $y = \dfrac{12}{x}$, work out how much $y$ falls when $x$ goes from $1$ to $2$, and again when $x$ goes from $3$ to $4$. Both are a step of one across. Compare the two falls, and say what that tells you about whether this relationship has a single rate of change.

Every pair of related quantities has a rate of change, and its units are the first quantity's unit per the second's. For direct variation that rate is constant and equals $k$. For inverse variation it is different at every point.

Most rates met in school involve time, but nothing in the definition requires it. A rate of change relates any two quantities.

The units say what it measures, and reading them is the fastest way to know what you have:

dollars per kilogram is a price;
grams per cubic centimetre is a density;
kilometres per litre is a fuel efficiency;
people per square kilometre is a population density;
degrees per kilometre is a temperature gradient;
dollars per hour is a wage.

In each case the rate is calculated the same way, as the change in the first quantity divided by the change in the second, which on a graph is the gradient.

The order matters and the units record it. Kilometres per litre and litres per kilometre are reciprocals of each other and describe the same car, but a bigger number means a better car in the first and a worse one in the second. Naming the units prevents that confusion entirely.

For $y = kx$, take any two points $(x_1, kx_1)$ and $(x_2, kx_2)$. The average rate of change between them is

$$\frac{kx_2 - kx_1}{x_2 - x_1} = \frac{k(x_2 - x_1)}{x_2 - x_1} = k$$

The answer does not depend on which two points were chosen. So a direct variation has a single rate of change, valid everywhere, and it is the constant of variation.

This is the same fact as Lesson 3 of the previous focus area, where the gradient of $y = kx$ turned out to be $k$. Three descriptions coincide:

the constant of variation;
the gradient of the graph;
the rate of change.

In the left panel of the diagram, each step of one across produces the same step up, which is that fact drawn. Equal causes, equal effects, everywhere.

In context this is a strong statement. If the cost of fabric varies directly with length, then the extra cost of one more metre is the same whether you already have one metre or a hundred. That is exactly what a fixed price per metre means.

For $y = \dfrac{k}{x}$, the same calculation behaves quite differently. Between $x_1$ and $x_2$:

$$\frac{\dfrac{k}{x_2} - \dfrac{k}{x_1}}{x_2 - x_1} = \frac{k\left(\dfrac{x_1 - x_2}{x_1x_2}\right)}{x_2 - x_1} = -\frac{k}{x_1x_2}$$

The result depends on $x_1$ and $x_2$ themselves, not only on the gap between them. So the rate of change is different in different places.

Check it against the diagram, with $k = 12$. From $x = 1$ to $x = 2$: the rate is $-\dfrac{12}{1 \times 2} = -6$, and indeed $y$ falls from $12$ to $6$. From $x = 3$ to $x = 4$: the rate is $-\dfrac{12}{3 \times 4} = -1$, and $y$ falls only from $4$ to $3$.

The formula also explains the shape. As $x$ grows, the product $x_1x_2$ grows, so the size of the rate shrinks: the curve flattens. And the rate is always negative for positive $k$, so the curve always falls.

In the language of Lesson 3, an inverse relationship is decreasing at a decreasing rate. Both halves of that phrase are now justified rather than asserted.

The contrast is not a technicality. It changes what advice you would give.

Consider fuel use. If fuel consumed varies directly with distance, then saving $10$ km saves the same fuel whether the trip was $50$ km or $500$ km, because the rate is constant.

Now consider journey time against speed, which is inverse. Increasing from $40$ to $50$ km/h on a $120$ km trip saves $3 - 2.4 = 0.6$ hours, which is $36$ minutes. Increasing from $100$ to $110$ km/h on the same trip saves $1.2 - 1.09 \approx 0.11$ hours, under $7$ minutes. The same $10$ km/h increase saves five times as much at the low end.

The general result is the formula above: the saving depends on $\dfrac{1}{x_1x_2}$, so it collapses as the speeds get larger.

The same structure appears wherever a fixed total is being shared. Going from two workers to three cuts the time by a third; going from twenty to twenty-one barely changes it. Diminishing returns is the everyday name for an inverse relationship's shrinking rate of change.

Five items on rates in context. Work each one, then reveal the answer.

1. 1 What rate has units of grams per cubic centimetre?

   Show answer
   Mass divided by volume.Density

2. 2 For $y = 7x$, what is the rate of change of $y$ with respect to $x$?

   Show answer
   In direct variation the rate is the constant.$7$

3. 3 For $y = \dfrac{20}{x}$, find the average rate between $x = 2$ and $x = 5$.

   Show answer
   Use $-\tfrac{k}{x_1x_2} = -\tfrac{20}{10}$.$-2$

4. 4 Does inverse variation have a single rate of change?

   Show answer
   Check the rate at two different places.No

5. 5 Speed rises by $25\%$ on a fixed route. By what percentage does the time fall?

   Show answer
   Time is multiplied by $\tfrac{1}{1.25} = 0.8$.$20\%$