# 3.4 Derivatives as rates of change  (Page 3/15)

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## Population change

In addition to analyzing velocity, speed, acceleration, and position, we can use derivatives to analyze various types of populations, including those as diverse as bacteria colonies and cities. We can use a current population, together with a growth rate, to estimate the size of a population in the future. The population growth rate is the rate of change of a population and consequently can be represented by the derivative of the size of the population.

## Definition

If $P\left(t\right)$ is the number of entities present in a population, then the population growth rate of $P\left(t\right)$ is defined to be ${P}^{\prime }\left(t\right).$

## Estimating a population

The population of a city is tripling every 5 years. If its current population is 10,000, what will be its approximate population 2 years from now?

Let $P\left(t\right)$ be the population (in thousands) $t$ years from now. Thus, we know that $P\left(0\right)=10$ and based on the information, we anticipate $P\left(5\right)=30.$ Now estimate ${P}^{\prime }\left(0\right),$ the current growth rate, using

${P}^{\prime }\left(0\right)\approx \frac{P\left(5\right)-P\left(0\right)}{5-0}=\frac{30-10}{5}=4.$

By applying [link] to $P\left(t\right),$ we can estimate the population 2 years from now by writing

$P\left(2\right)\approx P\left(0\right)+\left(2\right){P}^{\prime }\left(0\right)\approx 10+2\left(4\right)=18;$

thus, in 2 years the population will be 18,000.

The current population of a mosquito colony is known to be 3,000; that is, $P\left(0\right)=3,000.$ If ${P}^{\prime }\left(0\right)=100,$ estimate the size of the population in 3 days, where $t$ is measured in days.

3,300

## Changes in cost and revenue

In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost, revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the use of derivatives. The marginal cost is the derivative of the cost function. The marginal revenue is the derivative of the revenue function. The marginal profit is the derivative of the profit function, which is based on the cost function and the revenue function.

## Definition

If $C\left(x\right)$ is the cost of producing x items, then the marginal cost     $MC\left(x\right)$ is $MC\left(x\right)={C}^{\prime }\left(x\right).$

If $R\left(x\right)$ is the revenue obtained from selling $x$ items, then the marginal revenue $MR\left(x\right)$ is $MR\left(x\right)={R}^{\prime }\left(x\right).$

If $P\left(x\right)=R\left(x\right)-C\left(x\right)$ is the profit obtained from selling x items, then the marginal profit     $MP\left(x\right)$ is defined to be $MP\left(x\right)={P}^{\prime }\left(x\right)=MR\left(x\right)-MC\left(x\right)={R}^{\prime }\left(x\right)-{C}^{\prime }\left(x\right).$

We can roughly approximate

$MC\left(x\right)={C}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{C\left(x+h\right)-C\left(x\right)}{h}$

by choosing an appropriate value for $h.$ Since x represents objects, a reasonable and small value for $h$ is 1. Thus, by substituting $h=1,$ we get the approximation $MC\left(x\right)={C}^{\prime }\left(x\right)\approx C\left(x+1\right)-C\left(x\right).$ Consequently, ${C}^{\prime }\left(x\right)$ for a given value of $x$ can be thought of as the change in cost associated with producing one additional item. In a similar way, $MR\left(x\right)={R}^{\prime }\left(x\right)$ approximates the revenue obtained by selling one additional item, and $MP\left(x\right)={P}^{\prime }\left(x\right)$ approximates the profit obtained by producing and selling one additional item.

## Applying marginal revenue

Assume that the number of barbeque dinners that can be sold, $x,$ can be related to the price charged, $p,$ by the equation $p\left(x\right)=9-0.03x,0\le x\le 300.$

In this case, the revenue in dollars obtained by selling $x$ barbeque dinners is given by

$R\left(x\right)=xp\left(x\right)=x\left(9-0.03x\right)=-0.03{x}^{2}+9x\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}0\le x\le 300.$

Use the marginal revenue function to estimate the revenue obtained from selling the 101st barbeque dinner. Compare this to the actual revenue obtained from the sale of this dinner.

First, find the marginal revenue function: $MR\left(x\right)={R}^{\prime }\left(x\right)=-0.06x+9.$

Next, use ${R}^{\prime }\left(100\right)$ to approximate $R\left(101\right)-R\left(100\right),$ the revenue obtained from the sale of the 101st dinner. Since ${R}^{\prime }\left(100\right)=3,$ the revenue obtained from the sale of the 101st dinner is approximately \$3.

The actual revenue obtained from the sale of the 101st dinner is

$R\left(101\right)-R\left(100\right)=602.97-600=2.97,\text{or}\phantom{\rule{0.2em}{0ex}}2.97.$

The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.

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