2.14 Rates of Change

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In this section we interpret the derivative as an instantaneous rate of change.

Rates of Change

We begin by comparing the notion instantaneous rate of change to average rate of change. For two points $(x_1, y_1)$ and $(x_2, y_2)$ with $x_1 < x_2$ on the graph of a function $y = f(x)$ , the slope of the secant line connecting them is $m = \frac {\text {rise}}{\text {run}} = \frac {\Delta y}{\Delta x} = \frac {y_2 - y_1}{x_2 - x_1}.$ This slope can be interpreted as the average rate of change of $y$ with respect to $x$ over the interval $[x_1, x_2]$ .

The instantaneous rate of change of $y$ with respect to $x$ at $x = x_1$ is denoted by $\frac {dy}{dx}\Big {|}_{x=x_1},$ and it is obtained by letting $x_2$ approach $x_1$ in the formula for $\frac {\Delta y}{\Delta x}.$ Graphically, $\frac {dy}{dx}\Big {|}_{x=x_1}$ is the slope of the tangent line to the graph of $y = f(x)$ at $x=x_1$ . Generally speaking, $\frac {\Delta y}{\Delta x}$ represents the average rate of change of $y$ relative to $x$ whereas $\frac {dy}{dx}$ represents the instantaneous rate of change of $y$ relative to $x$ .

Examples of Rates of Change

If $y = f(t)$ is the distance traveled by an object as a function of time, then $\frac {\Delta y}{\Delta t}$ is the average speed of the object and $\frac {dy}{dt}$ is the instantaneous speed of the object.

If $V = f(t)$ is the volume of water in a tank as a function of time, then $\frac {\Delta V}{\Delta t}$ is the average rate that water is entering (or leaving, if this is negative) the tank and $\frac {dV}{dt}$ is the instantaneous rate that water is entering (or leaving) the tank.

If $C = f(x)$ is the cost of producing $x$ units then $\frac {dC}{dx}$ represents the instantaneous rate of change of cost with respect to the number of units produced. This is called the marginal cost of production.

If $P(t)$ represents population at time $t$ , then $\frac {\Delta P}{\Delta t}$ is the average growth rate of the population and $\frac {dP}{dt}$ is the instantaneous growth rate of the population. If the population of a certain bacteria is 50 (thousand) and it doubles every hour, then what is the population after 3 hours? How fast is the population growing at $t = 3$ hours? First, from the description of the population, we have $P(t) = 50 \cdot 2^t$ . Hence the population after 3 hours is given by $P(3) = 50\cdot 2^3 = 400$ (thousand) bacteria. The growth rate at time $t=3$ is given by $\frac {dP}{dt}\Big {|}_{t=3}.$ Computing the derivative, we have $\frac {dP}{dt}= 50 \cdot 2^t \cdot \ln (2),$ and hence, $\frac {dP}{dt}\Big {|}_{t=3} = 50 \cdot 2^3 \cdot \ln (2) = 400\ln (2) \approx 277.26 \mbox {(thousand) bacteria per hour}.$

If $Q(t)$ represents the quantity of a (medical) drug in the bloodstream at time $t$ , then $\frac {\Delta Q}{\Delta t}$ is the average rate of change of the quantity in the bloodstream and $\frac {dQ}{dt}$ is the instantaneous rate of change of the quantity in the bloodstream.