A key application of the derivative

In this section we'll look at a one of the most important applications of the derivative, using it to represent the rate of change of some interesting function.

Any function has an average rate of change over some part of its domain. In this function, for example,

The derivative allows us to find the instantaneous rate of change of a function at any point in its domain. Any well-behaved function has a slope at some point, and that slope represents how fast the dependent variable (or variables if we're doing multidimensional calculus) is changing as the independent variable changes.

The function below shows calculated derivatives evaluated two points, x = a and x = b. These are just the slopes of the lines tangent to those points at x = a and x = b.

In most of the examples that follow, we'll make time ( t ) the independent variable. Here are some examples:

The slope of a distance (or displacement, we'll call it x) vs. time graph is the speed, so speed is the derivative of x(t), where x(t) is the displacement.

The slope of a velocity vs. time graph is the acceleration, so acceleration is (1) the derivative of v(t), and (2) the second derivative of x(t).

The slope of a graph of the gross national product (GNP) of the United States vs. time is a representation of the rate of economic growth.

The derivative has many other uses in economics, including determining the marginal cost, minimizing cost and maximizing profit, and analyzing supply and demand.

There are many other examples, and we'll cover a good many by example in this section.

Example 1

The position along the y-axis of an object changes over time according to the function f(t) = 3t2 - 4t + 7. Calculate the speed of the object at time t = 5.

Solution: The function that gives us distance (from f(t) = 0) with respect to time is:

$$f(t) = 3 t^2 - 4t + 7$$

f(t) is plotted in the top panel of the graph. Speed is the first derivative of the position with respect to time (df/dt). That derivative is

$$f'(t) = 6t - 4$$

So the speed at $t = 5s$ is

$$f'(5) = 6(5) - 4 = \bf 26 \: units/s$$

The derivative is plotted in the bottom panel. The green dashed line passes through the vertex of the parabolic graph of f(t), and it's also the zero of the linear derivative graph. To the left of the vertex the value of the derivative, and thus the slope, is less than zero (negative slope), while to the right of the vertex it's positive. This is consistent with a glance at the graphs.

Example 2

Economics: Marginal cost and average cost of production – simple example

Consider a factory that produces widgets*. The table below outlines some key numbers for the production of up to 500 widgets by such a factory. The manufacturer has fixed costs of operation that amount to $500.00, say, per day. Those could include rents, salaries, &c. The variable costs are those that are generated by the production of each widget. In this case that's $6.30 per widget (3rd column of the table). I'll explain the rest of the table below.

If we let n be the number of widgets produced and C(n) be the total cost of n widgets, then the cost function for this example would be

$$C(n) = 500 + 6.3 n$$

The marginal cost is the cost to add an (n+1)th widget when n have already been manufactured. It's equal to the change in total cost divided by the change in number of widgets (see box below).

Well, this is just the derivative of C(n) with respect to n:

$$C'(n) = 6.3 $$

Notice that in this case, the marginal cost is constant. That isn't always so, but in this example it is: it costs an additional $6.30 to produce any new widget.

Here's a plot of the total cost function for this manufacturer, a linear graph.

Another important measure in economics is the average cost of production, C(n)/n. For this scenario, that is

$$\frac{C(n)}{n} = \frac{500}{n} + 6.3$$

where the n associated with 6.3 has been divided away. Now our average cost function is a rational function. It has a horizontal asymptote at $C(n) = 6.3$

The plot below shows that the average cost of widgets approaches 6.3, the marginal cost, as the number of widgets increases. Another way to think about it is that as we increase n, we spread the $500.00 of fixed costs over more and more widgets, reducing the per-unit cost, and perhaps allowing us to sell them for less.

* "widget" is a generic term used by economists to mean whatever item or bit of service a business produces. It could be cars or hours of legal advice ...

Marginal cost

The marginal cost in economics is the cost generated by producing one more unit of output. For example, if a company manufactures mountain bikes, then the marginal cost of production is the cost of producing one more bike. The marginal cost can be fixed or it can change with the number of units produced.

The marginal cost is the cost of producing an (n+1)th unit of output after n units have already been produced.

Example 3

Marginal cost – a more complicated example

Now consider a business that produces widgets with some sample costs given in the table below. In this example the fixed cost (say, per day) is $120 and the variable costs actually increase as the number of units of output grow. This is sometimes a valid model for certain firms.

The total cost function for this business is

$$C(n) = 10 n^2 + 120$$

where n is the number of units of output (number of widgets) and 120 (dollars) is the fixed cost. You can see that the cost grows with the number of units made, just like in the first example.

The marginal cost is the first derivative of the cost function:

$$\frac{d}{dx} C(n) = 20 n$$

So we see that the cost of adding additional widgets, C'(n) increases with n:

$$C'(n) = 20 n$$

The table above shows that the change in marginal cost from unit-to-unit is constant ($20). This is the second derivative of the cost function. Check it for yourself.

The average cost is the total cost divided by the number of units:

$$\frac{C(n)}{n} = \frac{120}{n} + 10 n$$

The graph of this marginal cost function is not constant. It actually has a minimum, in this case at about 3 or 4 units, so that number would presumably provide the optimum profit for the business.

Example 4

Freefall: The acceleration of freely-falling objects

A freely-falling object, dropped from an initial velocity of zero (from rest), falls along the y-axis according to the equation below, where g = 9.81 m·s-2 is the acceleration of gravity near the surface of Earth (g would be different on another planet).

The so-called freefall formula is:

$$y = \frac{1}{2} gt^2$$

The graph shows a plot of falling distance (y) vs. time (t). The curve is quadratic in time.

The first derivative of the freefall equation is the linear function y' = gt:

$$\frac{dy}{dt} = gt$$

The graph is shown below (dy/dt vs. t), and its slope is g.

The second derivative of the falling distance, with respect to time, is the constant g, the acceleration of gravity.

$$\frac{d^2 y}{d t^2} = g$$

In general, the first derivative with respect to time of any position function is the speed (velocity if it's a vector) and the second derivative is the acceleration.

Example 5

A question on an old AP calculus exam

This question is from an old (2013) AP calculus exam, but a similar question usually appears on every AP exam.*

The rate, in tons per hour, at which unprocessed gravel arrives at a gravel processing plant is modeled by

$$G(t) = 90 + 45\cdot cos \left( \frac{t^2}{18} \right)$$

where t is in hours and 0 ≤ t ≤ 8. Find G'(5) and, using correct units, interpret your answer in the context of the problem.

*You can access many years worth of AP calculus open response items at AP Central

Solution: The quantity G'(5) is a rate, in this case the rate of gravel accumulation, evaluated at a particular time, 5 hours after hour zero.

We first need the derivative of G(x):

$$G'(t) = -45 \cdot sin \left( \frac{t^2}{18} \right) \left( \frac{2t}{18} \right)$$

Note that we needed the chain rule for that derivative. It reduces to

$$G'(t) = -5 t\cdot sin \left( \frac{t^2}{18} \right)$$

Now we need to evaluate that derivative at t = 5 hours by plugging in 5 for t:

$$ \begin{align} G'(5) &= -25 \cdot sin \left( \frac{25}{18} \right) \\ \\ &= \bf -24.59 \: tons/h \end{align}$$

The interpretation is that at t = 5 hours, 24.59 tons of gravel is being lost per hour. The lost part is due to the negative sign that comes out of finding the derivative.

Here is a graph of our derivative vs. time over the domain of the function, [0, 8]. Notice that we lose gravel for the first 5.5 hours, then gain it back.

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