## Table of Contents

- Instantaneous Rate of Change vs. Average Rate of Change
- How to Calculate Instantaneous and Average Rate of Change
- When Instantaneous Rate of Change and Average Rate of Change are Equal
- Lesson Summary

- Instantaneous Rate of Change vs. Average Rate of Change
- How to Calculate Instantaneous and Average Rate of Change
- When Instantaneous Rate of Change and Average Rate of Change are Equal
- Lesson Summary

**Rate of change** explains how quick something is changing. In mathematics, it is characterized as a rate of one quantity related to another quantity. The most common definition of the rate of change is the change in *y* divided by the change in *x*. The variable *x* is an independent variable, the variable *y* is a dependent variable. Rate of change can be positive, negative, or zero.

The **instantaneous rate of change** definition is the rate of change at a specific point in time. On a graph, the instantaneous rate of change is represented as a tangent line against the main function at one point.

**Average rate of change** is the rate of change of a function during a certain interval, for example, from 0 to 10, or from 2 to 7.

Instantaneous rate of change vs. Average rate of change boils down to whether the measurement is taken over an interval or at a specific instant in time.

A great example of rate of change in real life is the speed of a vehicle. When driving somewhere, the speed of the vehicle is constantly changing. When driving on a highway, the moments where someone speeds up to pass by someone is considered their instantaneous rate of change. Looking at someone's everyday driving habits and taking the average speed they drive is considered their average rate of change.

Another example are runners and their speed. A runner's top speed could be consisted their instantaneous rate of change. A runner's average speed over a course of races could be considered their average rate of change of speed.

Using a graph or derivates are how to calculate the instantaneous and average rate of changes. The formula for calculating the average rate of change is:

{eq}(f(b) - f(a))/(b - a) {/eq}

The change in *y* values divided by the change in *x* values.

The formula for how to calculate the instantaneous rate of change is:

the limit as *h* approaches 0 of

{eq}(f(a +h) - f(a))/(a + h) {/eq}

In respect to *x* at *a*, the formula is now:

the limit as *h* approaches 0 of

{eq}(f(a +h) - f(a))/(h) {/eq}

This is otherwise known as the derivative of *f*(a). In conclusion, to find the instantaneous rate of change, find the derivate of the function at a certain time. This is using the equation of the curve to find the instantaneous rate of change.

Using a graph is another method of finding the rate of change. When using a graph to find instantaneous rate of change, instead of using the equation, use the actual curve. The instantaneous rate of change is a tangent line on a curve at a specific point. Then, find the slope of that tangent line.

Example:

Let the function be: {eq}f(x) = (x-1)^2 + 1 {/eq}

Graph it and draw a tangent line at the point {eq}(0,2) {/eq}

How to find the instantaneous rate of change from a graph? Find the slope of the tangent line.

Solution: Finding the instantaneous rate of change at the point {eq}(0,2) {/eq}

To find the slope of the tangent line, take the point that is tangent to the curve and the point where the tangent line crosses the x-axis: {eq}(1,0) {/eq}

Slope = change in *y* values divided by change in *x* values

{eq}(2 - 0)/(0 - 1) = 2/-1 = -2 {/eq}

So, the instantaneous rate of change of this function is -2.

This is is similar to estimating the instantaneous rate of change. To estimate the instantaneous rate of change, draw a line on graph between two points near the point that is being asked for. Then, find the slope of that line.

Example: Find the average rate of change from a graph with the function: {eq}f(x) = (x-1)^2 + 1 {/eq} over the interval {eq}[-2,5] {/eq}

Solution:

Graph the function:

{eq}= (f(5) - f(-2))/(5 - (-2)) {/eq}

{eq}= (17 - 10)/(5 - (-2)) {/eq}

{eq}= 7/7 {/eq}

{eq}= 1 {/eq}

The average rate of change from {eq}[-2,5] {/eq} is 1.

The derivative of a function {eq}dy/dx {/eq} represents the slope of {eq}y {/eq} with respect to {eq}x {/eq}.

The derivative is derived from the average rate of change formula:

{eq}(f(b) - f(a))/(b - a) {/eq}

Let {eq}b = a + h {/eq}

{eq}(f(a +h) - f(a))/(a + h - a) {/eq}

{eq}(f(a +h) - f(a))/(h) {/eq}

Using calculus, and taking the limit of that formula, gives the derivate of the function with respect to *x*. Therefore, providing the method to finding the instantaneous rate of change.

Example:

Let {eq}f(x) = x^3 + 2x^2 + 3x + 1 {/eq}. Find the instantaneous rate of change of the function at {eq}x = 2 {/eq}.

Solution:

Take the derivative of the function: {eq}f'(x) = 3x^2 + 4x + 3 {/eq}

Insert {eq}x = 2 {/eq}

{eq}f'(x) = (2)^2 + 4(2) + 3 {/eq}

{eq}f'(x) = 4 + 8 + 3 {/eq}

{eq}f'(x) = 15 {/eq}

The instantaneous rate of change of the function is 15.

There is an important theorem that states that on any interval of a continuous and differentiable function, there must be a point on the interval where the instantaneous rate of change equals the average rate of change. This is called the Mean Value Theorem. So, if {eq}''f'' {/eq} is a function that is continuous on in interval {eq}[a,b] {/eq}and differentiable on an interval {eq}(a,b) {/eq}, then there is at least a value, {eq}"c" {/eq}, that {eq}a < c < b {/eq}.

{eq}f'(c) = (f(b) - f(a))/(b - a) {/eq}

Example: Let the function be {eq}f(x) = x^3 + 4x^2 -2x +1 {/eq}. See if there is any place on the function between {eq}[-4,2] {/eq} where the instantaneous rate of change equals the average rate of change.

Solution:

First, check if the function is continuous and differentiable. Since, the function is a polynomial, it is continuous everywhere. Also, because the function is a polynomial, it is differentiable everywhere. So, in between the interval {eq}[-4,2] {/eq}, the function is continuous and differentiable. Therefore, using the Mean Value Theorem, there is at least one value "c" in the interval {eq}[-4,2] {/eq} where {eq}f'(c) = (f(b) - f(a))/(b - a) {/eq}.

Plug in -4 for "a" and 2 for "b":

{eq}f'(c) = (f(2) - f(-4))/(2 - (-4)) {/eq}

{eq}f'(c) = (21 - 9)/(2 - (-4)) {/eq}

{eq}f'(c) = 12/6 {/eq}

{eq}f'(c) = 2 {/eq}

The derivative of the function is: {eq}f'(x) = 3x^2 + 8x - 2 {/eq}

Insert "c" for "x" in the derivate function and set it equal to *f'(c)*. Solve for *c*.

{eq}3c^2 + 8c - 2 = 2 {/eq}

Solve using the quadratic formula:

{eq}3c^2 + 8c - 4 = 0 {/eq}

{eq}c = (\frac{-8 + \sqrt{8^2 - 4(3)(-4)}}{2(3)}) {/eq}

{eq}c = (\frac{-8 + \sqrt{64 - (-48)}}{6}) {/eq}

{eq}c = (\frac{-8 + \sqrt{112}}{6)}) {/eq}

{eq}c = (\frac{-8 + 4\sqrt{7}}{6}) {/eq}

{eq}c = (\frac{-8}{6})+(\frac{4\sqrt{7}}{6}) {/eq}

{eq}c = (\frac{-4}{3})+(\frac{2\sqrt{7}}{3}) {/eq}

{eq}c = 0.430501 {/eq}

{eq}c = (\frac{-8 - \sqrt{8^2 - 4(3)(-4)}}{2(3)}) {/eq}

{eq}c = (\frac{-8 - \sqrt{64 - (-48)}}{6}) {/eq}

{eq}c = (\frac{-8 - \sqrt{112}}{6)}) {/eq}

{eq}c = (\frac{-8 - 4\sqrt{7}}{6}) {/eq}

{eq}c = (\frac{-8}{6})-(\frac{4\sqrt{7}}{6}) {/eq}

{eq}c = (\frac{-4}{3})-(\frac{2\sqrt{7}}{3}) {/eq}

{eq}c = -3.09717 {/eq}

So, the two values of *c*, where the instantaneous and average rate of change are equal, are {eq}c = 0.430501 {/eq} and {eq}c = -3.09717 {/eq}.

The **rate of change** describes the relationship between between the rates of two quantities, one independent and one dependent. There are two rates of change, average and instantaneous. **Average rate of change** occurs during a specific interval. **Instantaneous rate** of change is the rate of change at a specific instant in time. A good example of rate of change is the speed of a moving vehicle. The instant a car speeds up to pass another is an example of instantaneous rate of change. Taking the average speed that a car goes through a trip from start to end, is an example of average rate of change. These rates of changes can be found graphically or numerically.

To find the average rate of change on a graph, locate the corresponding *y* values for each interval point on the graph. Then, divide the change in *y* values by the change in *x* values to find the average rate of change on a graph. Finding the average rate of change numerically is very similar to finding it graphically. To find the instantaneous rate of change on a graph, place a tangent line on a point and find the slope of the tangent line. The instantaneous rate of change can also be found numerically. Take the derivate of the function and insert the specific time in the function. Average and instantaneous rates of change can be equal under the Mean Value Theorem. It states that on any interval of a continuous and differentiable function, there must be a point on the interval where the instantaneous rate of change equals the average rate of change. Divide the change in *y* values by the change in *x* values and set that value equal to the derivative of the function to solve.

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Frequently Asked Questions

When given an interval, locate the corresponding *y* values for each interval point on the graph. Then, divide the change in *y* values by the change in *x* values to find the average rate of change on a graph.

You can find the instantaneous rate of change by taking the derivative of the function and inserting the instant of time being solved for. The derivate of the function is derived from the average rate of change formula because you are looking at a certain instant over the average interval.

Find the average rate of change by dividing the change in *y*, dependent variable, by the change in *x*, independent variable:

*(f(b) - f(a))/(b - a)*

On a graph, it is usually notated as "rise over run". Finding the average rate of change is similar to finding the slope of a line.

The average rate of change graphically represents how a function changed on average over an interval. It describes how the dependent variable, *y*, changed when the independent variable, *x*, changed.

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