Rates of Change in the Natural and Social Sciences, calculus lesson

A physicist asks how fast a particle moves, a chemist how fast a product forms, a biologist how fast a colony grows, and an economist what one more unit of output will cost. Each is asking for a derivative. The mathematics is the same every time. What changes is the meaning of the variables, the units, and the assumptions that make a smooth function a fair model.

Key terms

Rates in physics

Definition If $s(t)$ is the position of an object moving along a line at time $t$ and $s$ is differentiable, its velocity is $v(t) = s'(t)$ . If $v$ is differentiable, its acceleration is $a(t) = v'(t) = s''(t)$ . The speed is $\lvert v(t) \rvert$ . The object moves in the positive direction when $v(t) > 0$ , in the negative direction when $v(t) < 0$ , and is momentarily at rest when $v(t) = 0$ .

In plain words Velocity is how fast position changes, with a sign for direction. Acceleration is how fast velocity changes.

Example A particle has position $s(t) = t^3 - 6t^2 + 9t$ meters at time $t$ seconds. Then

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

It is at rest at $t = 1$ and $t = 3$ and moves backward between them. At $t = 2$ its velocity is $-3$ meters per second and its acceleration is $0$ .

Rates in chemistry

Definition For a reaction $\mathrm{A} + \mathrm{B} \to \mathrm{C}$ , write $[\mathrm{C}](t)$ for the concentration of the product in moles per liter at time $t$ . The rate of reaction is

\frac{d[\mathrm{C}]}{dt} = -\frac{d[\mathrm{A}]}{dt} = -\frac{d[\mathrm{B}]}{dt},

because each molecule of $\mathrm{C}$ formed uses up one molecule of $\mathrm{A}$ and one of $\mathrm{B}$ .

In plain words The reaction rate says how quickly the product builds up, which is the same as how quickly each reactant is used up.

Example Suppose the product concentration after $t$ seconds is

[\mathrm{C}](t) = \frac{t}{t + 10}

moles per liter. Then

\frac{d[\mathrm{C}]}{dt} = \frac{(t + 10) - t}{(t + 10)^2} = \frac{10}{(t + 10)^2}.

At $t = 10$ the rate is $10/400 = 0.025$ moles per liter per second, a quarter of the starting rate $0.1$ .

Rates in biology

Definition If $n(t)$ is the size of a population at time $t$ , modeled by a differentiable function, the growth rate is $n'(t)$ , measured in individuals per unit time. When $n(t) > 0$ the relative growth rate is

\frac{n'(t)}{n(t)},

measured per unit time.

In plain words The growth rate says how many individuals are being added per hour or per year at a given moment.

Example A bacteria culture starts with $500$ cells and doubles every hour, so $n(t) = 500 \cdot 2^t$ with $t$ in hours. Then

n'(t) = 500 \cdot 2^t \ln 2.

At $t = 3$ there are $4000$ cells, and they are increasing at $4000 \ln 2 \approx 2773$ cells per hour.

Marginal cost

Definition In economics a marginal quantity is a derivative with respect to the amount produced. If $C(x)$ is the total cost of producing $x$ units, the marginal cost is $C'(x)$ . Since $C(x + 1) - C(x) \approx C'(x)$ when $C$ is differentiable and its slope changes little over one unit, the marginal cost is about the cost of producing one more unit.

In plain words Marginal cost is the extra cost of the next item, at the current level of production.

Example Let $C(x) = 10000 + 5x + 0.01x^2$ dollars. Then $C'(x) = 5 + 0.02x$ , and $C'(500) = 15$ dollars per unit. The exact cost of the 501st unit is

C(501) - C(500) = 5 + 0.01 \cdot 1001 = 15.01

dollars, very close.

Marginal revenue and profit

Definition If $R(x)$ is the revenue from selling $x$ units, the marginal revenue is $R'(x)$ . With profit $P(x) = R(x) - C(x)$ , the marginal profit is $P'(x) = R'(x) - C'(x)$ . If the profit has a maximum at an interior point $x$ of its domain where $P$ is differentiable, then $P'(x) = 0$ , that is, marginal revenue equals marginal cost.

In plain words Keep producing while the next unit brings in more than it costs. The best output is where the two balance.

Example Suppose the units sell at price $p = 50 - 0.01x$ dollars when $x$ units are offered, so $R(x) = 50x - 0.01x^2$ and $R'(x) = 50 - 0.02x$ . With the cost above, $R'(x) = C'(x)$ means $50 - 0.02x = 5 + 0.02x$ , so $x = 1125$ . Since $P''(x) = -0.04 < 0$ , this output gives the largest profit.

Elasticity of demand

Definition Let the quantity demanded be a differentiable function $q = f(p)$ of the price $p$ , with $f(p) > 0$ . The price elasticity of demand is

E(p) = \frac{p}{f(p)} f'(p) = \frac{dq/q}{dp/p}.

Demand is called elastic at $p$ if $\lvert E(p) \rvert > 1$ and inelastic if $\lvert E(p) \rvert < 1$ . Revenue $R = pq$ satisfies

\frac{dR}{dp} = f(p) + p f'(p) = q\bigl(1 + E(p)\bigr),

so when $E(p) < -1$ raising the price lowers revenue, and when $-1 < E(p) < 0$ it raises revenue.

With this convention $E(p)$ is usually negative, because demand falls as the price rises. Some books define elasticity with the opposite sign,

-\frac{p}{q}\frac{dq}{dp},

so that it comes out positive. The meaning of elastic and inelastic is the same either way, since both compare the size $\lvert E \rvert$ with $1$ .

In plain words Elasticity compares percentage changes. It says roughly by what percent demand changes when the price goes up by one percent, and it has no units. A negative value means demand drops.

Example If $q = 1000 - 20p$ and $p = 30$ , then $q = 400$ and $f'(30) = -20$ , so

E(30) = \frac{30}{400}(-20) = -1.5.

A one percent price rise cuts demand by about $1.5$ percent. Demand is elastic, and a price rise would lower revenue.

Units and modeling assumptions

Definition If $y = f(x)$ , the derivative $f'(a)$ is measured in units of $y$ per unit of $x$ , and for a small change $\Delta x$ the change in $y$ is approximately $f'(a)\,\Delta x$ . Every such model assumes that the quantity can be described by a differentiable function, which is an idealization when the quantity is really a whole number (people, items) or is measured only at separate times.

In plain words Always say the units of a rate. And remember that a smooth formula for a count of cells or cars is an approximation that works when the numbers are large.

Example Let $T(h)$ be the air temperature in degrees Celsius at height $h$ kilometers. The statement $T'(2) = -6.5$ means that near $2$ kilometers the air gets colder by about $6.5$ degrees Celsius per kilometer of climb. In the bacteria example, $n'(3) \approx 2773$ cells per hour does not mean exactly that many cells appear in the next hour, both because the rate itself keeps growing and because real cells divide one at a time.

Check yourself

A particle moves with $s(t) = t^2 - 4t + 3$ meters after $t$ seconds. When is it at rest, and what is its acceleration then?

Show the answer

The velocity $v(t) = 2t - 4$ is zero at $t = 2$ seconds. The acceleration is $a(t) = 2$ meters per second squared at every time, including $t = 2$ .

With $C(x) = 2000 + 3x + 0.02x^2$ dollars, find the marginal cost at $x = 100$ and say what it means.

Show the answer

The marginal cost is $C'(x) = 3 + 0.04x$ , so $C'(100) = 7$ dollars per unit. Producing the 101st unit costs about $7$ dollars.

Find the elasticity of demand for $q = 200 - 4p$ at $p = 10$ . Is demand elastic there?

Show the answer

At $p = 10$ the demand is $q = 160$ and $f'(10) = -4$ , so

E(10) = \frac{10}{160}(-4) = -0.25.

Since $\lvert E \rvert < 1$ , demand is inelastic at that price.

Rates of Change in the Natural and Social Sciences

Theory, part 81

Practice, part 82

Key terms

Rates in physics

Rates in chemistry

Rates in biology

Marginal cost

Marginal revenue and profit

Elasticity of demand

Units and modeling assumptions

Check yourself

How this lesson connects

Builds on