Recall the following definitions:

A secant is a line that passes through a curve in at least two distinct points.

A secant through a single point is not unique.

The slope of the secant between two points represents the average rate of change.

\(m_{\text{secant}} = \text{average rate of change} = \dfrac{\Delta y}{\Delta x} = \dfrac{y_2 - y_1}{x_2 - x_1}\)

A tangent is a line which touches a curve at a point.

It does not cross the curve at that point if extended.

A tangent resembles the curve near that point.

A tangent to a single point is unique.

The slope of the tangent at a single point represents the instantaneous rate of change of the function at that point.

\(m_{\text{tangent}} = \text{instantaneous rate of change}\)

In this module, we will continue our discussion of average rate of change and instantaneous rate of change.

We will discuss possible methods which can be used to approximate the instantaneous rate of change at a particular point given data or a graph with discrete data points for which no algebraic model or formula is given.

We will complete this module by looking at examples of algebraic models which involve trigonometric functions and exponential functions.

The average monthly temperature in Waterloo over the last \(30\) years is summarized in the table. \(T\) is measured in degrees Celsius and \(n\) represents the month number (with \(1\) being January, with \(2\) being February, etc.)

a. Plot the data. Draw a secant through the points \((6,18.0)\) and \((7,20.2)\).

Solution

Plotting the data, a secant can be drawn between the points corresponding to the average temperature in June \((n = 6)\) and the average temperature in July \((n = 7)\).

The average monthly temperature in Waterloo over the last \(30\) years is summarized in the table. \(T\) is measured in degrees Celsius and \(n\) represents the month number (with \(1\) being January, with \(2\) being February, etc.)

b. Determine the average rate of change in the average monthly temperature over the interval \(6 \leq n \leq 7\). That is, determine the average rate of change in average monthly temperature from June to July.

Solution

Algebraically, the average rate of change in temperature between \(n = 6\) and \(n = 7\) can be calculated by finding the slope of the secant between the points \((6,18.0)\) and \((7,20.2)\).

The average monthly temperature in Waterloo over the last \(30\) years is summarized in the table. \(T\) is measured in degrees Celsius and \(n\) represents the month number (with \(1\) being January, with \(2\) being February, etc.)

b. Determine the average rate of change in the average monthly temperature over the interval \(6 \leq n \leq 7\). That is, determine the average rate of change in average monthly temperature from June to July.

Solution

If we were to draw all possible secants between points corresponding to consecutive months, we could determine which month to month interval has the greatest average monthly increase in average temperature.

We could also determine which month to month interval has the greatest average monthly decrease in average temperature.