# Rates of Change

## Conceptual Understanding of Rates of Change

A rate is a comparison between two quantities with different units.

We often judge performances by using rates. For example:

- The average speed of a motor vehicle is $90$ km per hour.
- The average typing rate is $55$ words per minute.
- The average score of a basketball player is $34$ points per game.

Speed is a commonly used rate. It is the rate of change in distance per unit of time. It is a well-known formula:

$$ \large \large \text{average speed}=\dfrac{\text{distance travelled}}{\text{time taken}}$$

However, if a car has an average speed of $80$ km/h for a trip, it does not mean that it travels at exactly $80$ km/h for the whole time.

The speed will probably vary continuously throughout the trip.

So, how can we calculate the car’s speed at any particular time?

A heavy particle falls from a tower, $50$ metres high, and crashes into the ground. How can we predict the damage the spanner causes when it hits the ground? We will begin by attempting to develop a mathematical model for the particle’s speed as it falls.

Speed is a rate – a measure of distance travelled per unit of time. Our modelling thus begins with a review of the rate and the concept of a constant rate.

## Constant Rates of Change

When the rate of change of one quantity with respect to another does not alter, the rate is constant.

For example, if petrol is $80$ cents per litre, then every litre of petrol purchased at this rate always costs $80$ cents. This means $10$ litres of petrol would cost $8.00$ dollars and $100$ litres of petrol would cost $80.00$ dollars.

## Variable Rates of Change

The particle travels $50$ metres in $3$ seconds and stops moving. In other words, it falls $50$ metres and then hits the ground. We also notice that the particle’s speed is not constant but variable.

If a rate is not constant but changing, it must be variable.

For example, the amount of electricity used per hour by a household is most likely to be a variable rate as the need for electricity will change throughout the day.

## Average Rates of Change

If a rate is variable, it is sometimes useful to know the average rate of change over a specific interval. For example, a tree grew from $5$ m last year to $6$ m now.

$$ \large \begin{align}

\text{average rate of growth} &= \dfrac{\text{change in height}}{\text{change in time}} \\

&= \dfrac{(6-5) \text{ m}}{1 \text{ year}} \\

&= 0.2 \text{ m/year} \\

\end{align} $$

This means that the tree grew $0.2$ metres over the past year but not necessarily constantly as that rate during the year.

## Instantaneous Rates of Change

If a rate is variable, it is often useful to know the rate of change at any given time or point, that is, the instantaneous rate.

For example, a police radar gun is designed to read a vehicle’s speed instantly. This enables the police to make an immediate decision as to whether a car is breaking the speed limit or not.

Instantaneous rates can be found from a curved graph by:

- drawing a tangent to the curve a the point in the equation,
- calculating the gradient of the tangent over an appropriate interval, that is, between two points whose coordinates are easily identified.

## Derivative in terms of Rates of Change

$$f^{\prime}(x) = \lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$$

The gradient of the tangent at the variable point $(x,f(x))$ is the limiting value of $\dfrac{f(x+h)-f(x)}{h}$ as $h$ approaches $0$.

This formula gives the gradient of the tangent to the curve $y=f(x)$ at the point $(x,f(x))$, for any value of the variable $x$ for which this limit exists. Since there is at most one value of the gradient for each value of $x$, the formula is a function.

## Algebraic vs Calculus in terms of Rates of Change

As stated above, *average rates of change* are often referred to as algebraic rates of change. Algebraic acceleration is the average rate of change, while acceleration is the term that is the instantaneous rate of change of speed of velocity obtained by differentiating velocity.

Assume $x(t)$ is a displacement, then $x^{\prime}(t)$ or $\dfrac{dx}{dt}$ is the instantaneous rate of change in displacement with respect to time, which is velocity.

An example is where quantities vary with time or with respect to another value.

- temperature changes
- the height of the surface of the water in a pond
- speed of a car

$\dfrac{dy}{dx}$ gives the rate of change in $y$ with respect to $x$.

We can therefore use the derivative of a function to inform us of the rate at which something is happening.

### Example 1

The number of people in a colony is approximated by the formula $P(t)=30000+72t^2-t^4$ where $t$ is the time in years after the year \( 2000 \).

(a) β Find the number of people in the colony in the year \( 2000 \).

\( \begin{align} \displaystyle

P(0) &= 30000 + 72 \times 0^2-0^4 \\

&= 30000

\end{align} \)

(b) β Find the number of people in the colony in the year \( 2008\).

\( \begin{align} \displaystyle

P(8) &= 30000 + 72 \times 8^2-8^4 \\

&= 30512

\end{align} \)

(c) β Find the rate at which the number of people in the colony changes when $t=2$.

\( \begin{align} \displaystyle

P^{\prime}(t) &= 144t^2-4t^3 \\

P^{\prime}(2) &= 144 \times 2-4 \times 2^3 \\

&= 256

\end{align} \)

(d) β Describe the significance of the result of part (c).

after two years, the number of people is increasing by \( 256 \) people per year

(e) β Find the rate at which the number of people in the colony changes when $t=8$.

\( \begin{align} \displaystyle

P^{\prime}(8) &= 144 \times 8-4 \times 8^3 \\

&= -896

\end{align} \)

(f) β Describe the significance of the result of part (e).

after eight years, the number of people is decreasing by $-896$ people per year.

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