3.4 Derivatives as Rates of Change

Big idea

The rules of the previous sections make derivatives cheap to compute. This section is about what the number means once you have it. A derivative is a rate: the output changes so many of its units for every one unit of change in the input, measured at a single value of the input rather than averaged over a stretch.

Units carry most of the interpretation. If $C(x)$ is the cost in dollars of producing $x$ items, then $C'(x)$ is in dollars per item, and that alone tells you it answers “what does one more cost?” rather than “what do they all cost?”. If $T(h)$ is temperature in degrees at altitude $h$ metres, then $T'(h)$ is degrees per metre. Getting the units right first makes a wrong interpretation hard to write down.

Motion is the setting where the language is oldest and the layering is clearest. Position is a function of time; its derivative is velocity, which carries a sign telling you which way the object is heading; the absolute value of velocity is speed; and the derivative of velocity is acceleration, the rate at which the rate is changing. Three questions that sound like variations on one another - where is it, which way is it going, is it gaining ground - are a function and its first two derivatives.

Economics borrows the same idea under the name marginal. Marginal cost is defined in textbooks as the cost of producing one additional unit, and the derivative approximates it. The approximation is good because one unit is a small step compared with a production run of hundreds, and the derivative is what a small step, scaled up to a full unit, predicts.

Decoder

If $y = f(x)$, then $f'(a)$ is the instantaneous rate of change of $y$ with respect to $x$ at $x = a$, measured in units of $y$ per unit of $x$.

“With respect to $x$” is the phrase students skip and then miss. It names the denominator of the rate, and changing it changes the number entirely: the rate of change of volume with respect to time is not the rate of change of volume with respect to height. Whenever a rate appears in a sentence, find its “per what”.

“Instantaneous” distinguishes $f'(a)$ from the average rate over an interval. The derivative is what the average rate approaches as the interval shrinks to nothing, so it describes the behaviour at one moment. This is why $f'(a)$ can be large while the total change over an hour is small, or negative at one instant on a trip whose net displacement is positive.

The classic mistake is confusing velocity with speed. Velocity is signed and tells direction; speed is its absolute value. An object moving backwards faster and faster has decreasing velocity and increasing speed at the same time. The test for speeding up is not the sign of the acceleration but whether velocity and acceleration share a sign.

Definitions and results

Rate of change. For $y = f(x)$, the quantity $f'(a)$ is the instantaneous rate of change of $y$ with respect to $x$ at $a$, in units of $y$ per unit of $x$.

Linear estimate of a change. For a small step $\Delta x$ away from $a$,

$$ f(a + \Delta x) \approx f(a) + f'(a)\,\Delta x $$

This is the sentence “the derivative predicts the change” written as arithmetic, and it is what every interpretation below rests on.

Position, velocity, acceleration. If $s(t)$ is position at time $t$, then $v(t) = s'(t)$ is velocity, $|v(t)|$ is speed, and $a(t) = v'(t) = s''(t)$ is acceleration.

Direction and rest. The object moves in the positive direction where $v(t) > 0$ and in the negative direction where $v(t) < 0$. It is momentarily at rest where $v(t) = 0$, and a sign change of $v$ there means it turns around.

Speeding up and slowing down. The object speeds up when $v(t)$ and $a(t)$ have the same sign and slows down when their signs differ. Acceleration alone does not decide it.

Displacement against distance. Displacement over an interval is $s(b) - s(a)$, and average velocity is that divided by $b - a$. Total distance travelled adds the lengths of each leg separately and is larger whenever the object reverses.

Marginal cost, revenue, profit. With $C(x)$, $R(x)$ and $P(x) = R(x) - C(x)$ as functions of the number of units, the marginal cost is $C'(x)$, the marginal revenue is $R'(x)$, and the marginal profit is $P'(x) = R'(x) - C'(x)$. Each approximates the effect of producing one more unit.

Population and other growth rates. If $P(t)$ counts a population at time $t$, then $P'(t)$ is the growth rate in individuals per unit time, positive while the population grows and negative while it shrinks.

Worked examples

A particle that turns around

A particle moves along a line with position $s(t) = t^3 - 6t^2 + 9t$ feet after $t$ seconds, for $t \ge 0$. Differentiating twice:

$$ v(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3), \qquad a(t) = 6t - 12 $$

The velocity is zero at $t = 1$ and $t = 3$, so those are the moments the particle can reverse. The factored form gives the signs directly: $v > 0$ on $[0,1)$, $v < 0$ on $(1,3)$, and $v > 0$ after $t = 3$. So the particle moves forward to $s(1) = 4$ feet, backs up to $s(3) = 0$ feet, and then moves forward again.

Acceleration is negative before $t = 2$ and positive after. Comparing signs, the particle is speeding up on $(1,2)$, where velocity and acceleration are both negative, and on $(3, \infty)$, where both are positive. It is slowing down on $(0,1)$ and on $(2,3)$. The interval $(1,2)$ is the one worth staring at: the acceleration is negative there and yet the particle is gaining speed, because it is moving backwards.

Total distance in the first three seconds is $4$ feet out and $4$ feet back, so $8$ feet, while the displacement is $s(3) - s(0) = 0$.

Marginal cost against the actual cost

A workshop’s cost for $x$ units is $C(x) = 0.02x^2 + 5x + 400$ dollars. The marginal cost is

$$ C'(x) = 0.04x + 5 $$

so $C'(100) = 9$ dollars per unit. The claim is that the $101$st unit costs about nine dollars. Check it exactly: $C(101) = 204.02 + 505 + 400 = 1109.02$ and $C(100) = 200 + 500 + 400 = 1100$, a difference of $9.02$ dollars. The derivative misses by two cents, which is the corner that the linear estimate cuts.

A draining tank

A tank holds $V(t) = 200 - 10t + \tfrac{1}{8}t^2$ gallons of water $t$ minutes after a valve opens, valid for $0 \le t \le 40$. At $t = 0$ it holds $200$ gallons and at $t = 40$ it holds $200 - 400 + 200 = 0$, so the tank empties exactly at the end of the interval. The rate of change is

$$ V'(t) = -10 + \tfrac{1}{4}t $$

in gallons per minute, negative for $t < 40$ because the tank is losing water. At $t = 0$ water leaves at $10$ gallons per minute; at $t = 24$ the rate is $-10 + 6 = -4$ gallons per minute. The draining slows steadily, which is what falling pressure in a tank does, and $V'(40) = 0$ says the last of the water leaves at a trickle.

Average against instantaneous growth

A pond’s fish population is modelled by $P(t) = 3t^2 + 20t + 1000$ individuals $t$ years after stocking. Over the first five years the population rises from $1000$ to $3(25) + 100 + 1000 = 1175$, so the average growth rate is $175/5 = 35$ fish per year. The instantaneous rate is $P'(t) = 6t + 20$, which gives $20$ fish per year at stocking and $50$ fish per year at the end of year five. The average sits between the two endpoint rates, and no single year matched it except the one in the middle, where $P'(2.5) = 35$.

Practice

First, interpretation. These give you a function with units attached and ask what its derivative means, what sign to expect, and what a stated value predicts about a small change.

Then motion. You are given a position function and asked for velocity, speed, acceleration, when the object is at rest, which way it is moving, and when it is gaining speed.

Quiz

Five items on interpreting a derivative with units, and on position, velocity and acceleration.