A Brief Introduction to Calculus: Part 1.3 - Rate of Change

Math
Calculus
Further Math

Published on Tue Aug 02 2022. Views

An introduction to basic concepts in Calculus; mostly based on the IGCSE Further Math textbook.

Introduction

This is the third part of first series — Derivatives, focused on the concept of the rate of change of a function.

Note: the content introduced in this series is mostly based on the IGCSE Further Math textbook.

Quick Access

Rate of Change

Now, in calculus, the concept of rate of change is also important. It's simply a rate of how a dependent variable changes as the independent variable changes.

There are two types of rates of change we will look at here:

Instantaneous rate of change
Average rate of change

For example, for a linear function, the rate of change is constant, because there is a linear (straight line) correlation between the variables. And the rate also happens to be the derivative of the function, i.e. the gradient/slope of the line, as discussed above.

In contrast, a quadratic function, for example, would have a variable rate of change over an interval or a point. As you saw earlier, the derivative would produce an expression whose value actually depends on x; in other words, the derivative's value at different x coordinates may vary, unlike a linear function's derivative, which is a single constant that represents the gradient.

Average Rate of Change

The average rate of change over an interval $[a, b]$ is the gradient of the secant line that passes through the curve at $x_1 = a$ and $x_2 = b$ .

To calculate this, find the coordinates of the two points of intersection and calculate the gradient:

\begin{align*} m = \frac{y_2 - y_1}{x_2 - x_1} \end{align*}

And to form an equation for the secant line:

\begin{align*} y = m(x - x_1) + y_1 \end{align*}

Its Correlation with Derivatives

Suppose there exists a tangent to the blue curve, whose corresponding function is $f(x)$ , at point $P(x_0, f(x_0))$ :

Start by picking another point on the curve, whose $x$ -coordinate is displaced by $h$ units away from $x_0$ and $y$ -coordinate is hence $f(x_0 + h)$

By dragging the new point closer and closer towards point $P$ , $h$ gets smaller and smaller, and the gradient of the secant line approaches the tangent line:

At some point, when the second point reaches point $P$ , the gradient of the two lines would be equal.

Given that the gradient of a straight line can be calculated by:

\begin{aligned} \frac{\Delta y}{\Delta x} \end{aligned}

we can define the tangent line's gradient using limits:

\begin{aligned} \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \end{aligned}

Example

Let $f\left(x\right)\ =\ \frac{1}{20}x^{3}$ , find the average rate of change over the interval $[1, 4]$ , and hence the equation of the secant line.

\begin{align*} x_1 &= 1 \\ y_1 &= f(1) = 0.05 \\ \\ x_2 &= 4 \\ y_2 &= f(4) = 3.2 \\ \\ m &= \frac{3.15}{3} = 1.05 \\ \\ y &= 1.05(x - 1) + 0.05 = 1.05x - 1 \end{align*}

Thus the average rate of change required is $1.05$ and the equation of the secant line drawn is $y = 1.05x - 1$ .

Instantaneous Rate Of Change

The instantaneous rate of change is the value of the derivative at a specific point on the curve of the function, i.e. the gradient of the tangent line at a point.

To calculate its value for a curve $f(x)$ , is simply $f'(x)$ , where $x$ is the x-coordinate of a point on the curve.

Take this as an example:

The curve C is given by the equation $f(x) = 2x^2+4x-3$ , find the instantaneous rate of change at the point (7, 123).

\begin{align*} f'(x) &= 4x + 4 \\ f'(7) = 32 \end{align*}

In this case, $32$ represents the gradient of the line that's tangent to the curve at $(7, 123)$ .

Conclusion

Subsequent sections are accessible through this section at the top of the article.

Thanks for reading.