Rates of change and related rates

Summary

This lesson reads the derivative as what it measures in applications: how fast one quantity changes when another changes. It distinguishes the average, the instantaneous and the relative rate of change, with their units, and characterizes growth at a constant relative rate. It studies motion along a line, with position, velocity, speed and acceleration, and proves what the sign of the derivative at an instant says, and what it does not say. Finally, when several quantities that depend on time are tied together by a relation, the chain rule also ties together their rates of change: these are related rates, which are solved with an explicit method.

Learning objectives

Interpret the derivative as an instantaneous rate of change, with its units, and distinguish it from the average rate of change and from the relative one.
Describe a rectilinear motion with position, velocity, speed and acceleration, and prove what the sign of the derivative at an instant says about nearby values.
Solve related rates problems with an explicit method, justifying that the quantities being differentiated are differentiable.
Relate the relative rates of change of products and powers to the logarithmic derivative, and accelerations to the second derivative of a relation.

The derivative as a rate of change

The lesson on the derivative as a limit arrived at the same definition from two questions: the slope of the tangent and the instantaneous velocity. Both are cases of a more general question. Whenever one quantity depends on another, the area of a circle on its radius, the position of a moving point on time, or the mass of a piece of rod on its length, it is worth knowing how fast the first changes when the second changes.

DefinitionRate of change

Let $f$ be a function defined on an open interval $I$ , which expresses a quantity $y = f(x)$ as a function of another quantity $x$ , and let $a$ and $a + h$ be points of $I$ , with $h \neq 0$ . The average rate of change of $y$ with respect to $x$ between $a$ and $a + h$ is the quotient

\frac{\Delta y}{\Delta x} = \frac{f(a + h) - f(a)}{h}

If $f$ is differentiable at $a$ , the instantaneous rate of change (or simply rate of change) of $y$ with respect to $x$ at $a$ is $f'(a)$ ; if moreover $f(a) \neq 0$ , the relative rate of change at $a$ is the quotient $\frac{f'(a)}{f(a)}$ .

The average rate is the slope of a secant, and the instantaneous rate, its limit: the slope of the tangent. Both are measured in units of $y$ per unit of $x$ : meters per second, if $y$ is a length and $x$ a time; kilograms per meter, if $y$ is the mass of a piece of rod and $x$ its length. In this last case, if $M(x)$ is the mass of the piece between one end and the point at distance $x$ , the rate of change $M'(x)$ is the linear density of the rod at that point: the mass per unit length that a very short piece around it would have. The relative rate, by contrast, is measured only in reciprocal units of $x$ (per second, per day): it does not depend on the unit in which $y$ is measured, and for that reason it allows one to compare the change of quantities of different nature or of different size. It is the logarithmic derivative of the lesson on the variable in the exponent, and it is often expressed as a percentage: a relative rate of $0.03$ per day is 3 percent per day.

A geometric example shows how a rate of change is read. The area of the circle of radius $r$ is $A = \pi r^2$ , and its rate of change with respect to the radius is $\frac{\mathrm{d}A}{\mathrm{d}r} = 2\pi r$ , the circumference, as proposed problem 1 of the lesson on the derivative as a limit asks one to check. The reason is clear: when the radius increases by $h$ , the area increases by a ring of width $h$ , whose area is almost that of the circumference times $h$ .

PropositionGrowth at a constant relative rate

Let $y_0 > 0$ and $k \neq 0$ be real numbers, and let $y(t) = y_0 e^{kt}$ for every real $t$ . Then the relative rate of change of $y$ equals $k$ at every instant, and for every $t$ and every $T$ :

y(t + T) = e^{kT} \, y(t)

In particular, if $k > 0$ , the quantity doubles in every time interval of length $\frac{\ln 2}{k}$ , whatever the instant at which it begins; and if $k < 0$ , it is halved in every interval of length $\frac{\ln 2}{\abs{k}}$ .

Proof

$y'(t) = y_0 e^{kt} \cdot k = k \, y(t), \qquad \frac{y'(t)}{y(t)} = \resaltar{k}$
By the corollary on compositions with the exponential and the logarithm, from the lesson on the variable in the exponent, with $u(t) = kt$ , the function $y$ is differentiable on all of $\mathbb{R}$ and its derivative is $k$ times the function itself, which does not vanish.
$\frac{y(t + T)}{y(t)} = \frac{y_0 e^{kt} e^{kT}}{y_0 e^{kt}} = \resaltar{e^{kT}}$
By the law of exponents, the quotient between the values separated by an interval of length $T$ does not depend on $t$ .
$e^{kT} = 2 \Leftrightarrow T = \frac{\ln 2}{k}, \qquad e^{kT} = \frac{1}{2} \Leftrightarrow T = \resaltar{\frac{\ln 2}{-k}}$
Since the exponential is injective, the quotient equals $2$ if and only if $kT = \ln 2$ , and equals $\frac{1}{2}$ if and only if $kT = -\ln 2$ .

A population of bacteria in an unrestricted medium or a mass of radioactive material changes, to a good approximation, at a constant relative rate, and the proposition explains why they have a fixed doubling time or half-life. That the functions $y_0 e^{kt}$ are the only ones with constant relative rate $k$ is true, but its proof requires knowing that a function with zero derivative on an interval is constant, a consequence of the mean value theorem of the following unit.