The instantaneous rate of change is the rate of change at a particular point $P$. You can find this rate of change by calculating the slope of the tangent at this point $P$. A tangent is a straight line that touches the graph at that exact point.

Theory

The instantaneous rate of change at a point $P=\phantom{\rule{-0.17em}{0ex}}\left({x}_{1},f({x}_{1})\right)$ on the graph of the function $f(x)$ is how much the graph is changing at this particular point. Let the tangent line to $f(x)$ at $x={x}_{1}$ be given by

$$y=ax+b$$ |

Then, the instantaneous rate of change at this point is equal to the slope of the tangent. Therefore,

$$\text{Instantaneousrateofchange}=a$$ |

For a function $f(x)$ and a point $P$ on the graph, you can choose two points $({x}_{1},{y}_{1})$ and $({x}_{2},{y}_{2})$ on the tangent at $P$. Then

$$\begin{array}{cc}\text{instantaneousrateofchange}& \\ =\frac{\text{changein}y\text{}}{\text{changein}x\text{}}& \\ =\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}& \end{array}$$

$$\begin{array}{llll}\hfill \text{instantaneousrateofchange}& =\frac{\text{changein}y\text{}}{\text{changein}x\text{}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

If $({x}_{1},{y}_{1})=P$, then

$$\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{{y}_{2}-f({x}_{1})}{{x}_{2}-{x}_{1}}$$ |

You can think of the difference between average rate of change and instantaneous rate of change as follows:

The average rate of change can be used to measure speed over a period of time. For example, how long it takes to drive a car $10\phantom{\rule{0.17em}{0ex}}\text{km}$.

Instantaneous rate of change, on the other hand, is comparable to what the car’s speedometer is showing at a specific point in time. Instantaneous rate of change means the speed you have at that exact moment.

Example 1

If you have the equation for the tangent at the point where you want to find the instantaneous rate of change, then just look at the slope of the line. But if you only have two points on the tangent, then the method presented here will help you find the slope. You will see that we actually calculate the average rate of change to find the slope.

In this figure, you see the function $f(x)$ (blue curve):

You draw the tangent (gray line) at the point $x={x}_{1}$ (where $x={x}_{1}$ means that the $x$-coordinate of the point has the value ${x}_{1}$, thus ${x}_{1}$ is a number). Note that one of the two points on the tangent line may be a point on the graph (the point of tangency).

You now choose a point you like on the tangent (gray line) and call it $({x}_{2},{y}_{2})$. Now, calculate the slope of the tangent to find the instantaneous rate of change.

You can use the expression for the average rate of change to find an approximation of the instantaneous rate of change at ${x}_{1}$. Then you select an ${x}_{2}$ value very close to the $x$-value you have been given in the exercise and enter both of them into the formula for the average rate of change.

Example 2

**Mark Zuckerberg has started a company and has found this function for the company’s revenue (given in millions of dollars): **

$$f(x)=\text{}0.5\text{}{x}^{2}-\text{}0.5\text{}x+2,$$ |

where $x$ is the number of weeks. Find the instantaneous rate of change after 4 weeks.

To be able to calculate this, you must draw the graph $f(x)$ in a coordinate system and then draw the tangent at $x=4$. To draw the graph you need a function table. Choose some $x$-values and calculate the $y$-values:

$x$ | $y=f(x)$ | $(x,y)$ |

$0$ | $f(0)=2$ | $(0,2)$ |

$1$ | $f(1)=2$ | $(1,2)$ |

$2$ | $f(2)=3$ | $(2,3)$ |

$3$ | $f(3)=5$ | $(3,5)$ |

$4$ | $f(4)=8$ | $(4,8)$ |

$5$ | $f(5)=12$ | $(5,12)$ |

$6$ | $f(6)=17$ | $(6,17)$ |

$7$ | $f(7)=23$ | $(7,23)$ |

$$ |

$x$ | $y=f(x)=0.5{x}^{2}-0.5x+2$ | $(x,y)$ |

$0$ | $f(0)=0.5\cdot {(0)}^{2}-0.5\cdot (0)+2=2$ | $(0,2)$ |

$1$ | $f(1)=0.5\cdot {(1)}^{2}-0.5\cdot (1)+2=2$ | $(1,2)$ |

$2$ | $f(2)=0.5\cdot {(2)}^{2}-0.5\cdot (2)+2=3$ | $(2,3)$ |

$3$ | $f(3)=0.5\cdot {(3)}^{2}-0.5\cdot (3)+2=5$ | $(3,5)$ |

$4$ | $f(4)=0.5\cdot {(4)}^{2}-0.5\cdot (4)+2=8$ | $(4,8)$ |

$5$ | $f(5)=0.5\cdot {(5)}^{2}-0.5\cdot (5)+2=12$ | $(5,12)$ |

$6$ | $f(6)=0.5\cdot {(6)}^{2}-0.5\cdot (6)+2=17$ | $(6,17)$ |

$7$ | $f(7)=0.5\cdot {(7)}^{2}-0.5\cdot (7)+2=23$ | $(7,23)$ |

$$ |

You then get this graph (blue curve):

Draw the tangent so that it touches only the point $(4,f(4))$. To find the instantaneous rate of change at $x=4$, you must find two points on the tangent line and insert into the formula. You see that the points $({x}_{1},{y}_{1})=(2,1)$ and $({x}_{2},{y}_{2})=(4,8)$ are on the tangent.

Then you get that

$$\begin{array}{cc}\text{instantaneousrateofchange}& \\ =\frac{8-1}{4-2}& \\ =\frac{7}{2}& \\ =3.5& \end{array}$$

$$\begin{array}{lll}\hfill \text{instantaneousrateofchange}=\frac{8-1}{4-2}=\frac{7}{2}=3.5& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$

You have to multiply this by $$1\phantom{\rule{0.17em}{0ex}}000\phantom{\rule{0.17em}{0ex}}000$ to get the correct answer since the revenue is in millions of dollars: $$3.5\cdot \text{\$}1\phantom{\rule{0.17em}{0ex}}000\phantom{\rule{0.17em}{0ex}}000=\text{\$}3\phantom{\rule{0.17em}{0ex}}500\phantom{\rule{0.17em}{0ex}}000$$ |

The instantaneous rate of change of the company’s revenue after four weeks is $$3\phantom{\rule{0.17em}{0ex}}500\phantom{\rule{0.17em}{0ex}}000$ per week.