## Rates of Change

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It is very useful to determine how fast (the rate at which) things are changing. Mathematically we can represent change in different ways. For example we can use algebraic formulae or graphs.

Graphs give a visual representation of the rate at which the function values change as the independent (input) variable changes. This rate of change is described by the gradient of the graph and can therefore be determined by calculating the derivative.

We have learnt how to determine the average gradient of a curve and how to determine the gradient of a curve at a given point. These concepts are also referred to as the average rate of change and the instantaneous rate of change.

\[\text{Average rate of change } = \cfrac{f(x+h)-f(x)}{(x + h) – x}\]\[\text{Instantaneous rate of change } = \lim_{h\to 0}\cfrac{f(x+h)-f(x)}{h}\]

When we mention rate of change, the instantaneous rate of change (the derivative) is implied. When average rate of change is required, it will be specifically referred to as average rate of change.

Velocity is one of the most common forms of rate of change:

Velocity refers to the change in distance (\(s\)) for a corresponding change in time (\(t\)).

\[v(t)=\cfrac{ds}{dt}={s}'(t)\]

Acceleration is the change in velocity for a corresponding change in time. Therefore, acceleration is the derivative of velocity

\[a(t)={v}'(t)\]

This implies that acceleration is the second derivative of the distance.

\[a(t)={s}”(t)\]

## Example

### Question

The height (in metres) of a golf ball \(t\) seconds after it has been hit into the air, is given by \(H(t)=20t-5{t}^{2}\). Determine the following:

The average vertical velocity of the ball during the first two seconds.

The vertical velocity of the ball after \(\text{1.5}\) \(\text{s}\).

The time at which the vertical velocity is zero.

The vertical velocity with which the ball hits the ground.

The acceleration of the ball.

### Determine the average vertical velocity during the first two seconds

\begin{align*} {v}_{\text{ave}} & = \cfrac{H(2)-H(0)}{2-0} \\ & = \cfrac{[20(2)-5{(2)}^{2}]-[20(0)-5{(0)}^{2}]}{2} \\ & = \cfrac{40-20}{2} \\ & = \text{10}\text{ m.s$^{-1}$}\end{align*}

### Calculate the instantaneous vertical velocity

\begin{align*} v(t) & = H'(t) \\ &= \cfrac{dH}{dt} \\ & = 20-10t \end{align*}

Velocity after \(\text{1.5}\) \(\text{s}\):

\begin{align*} v(\text{1.5}) & = 20-10(\text{1.5}) \\ & = \text{5}\text{ m.s$^{-1}$} \end{align*}

### Determine the time at which the vertical velocity is zero

\begin{align*} v(t) & = 0 \\ 20-10t & = 0 \\ 10t & = 20 \\ t & = 2 \end{align*}

Therefore, the velocity is zero after \(\text{2}\text{ s}\)

### Find the vertical velocity with which the ball hits the ground

The ball hits the ground when \(H(t)=0\)

\begin{align*} 20t-5{t}^{2} & = 0 \\ 5t(4-t) & = 0 \\ t=0& \text{ or } t=4 \end{align*}

The ball hits the ground after \(\text{4}\) \(\text{s}\). The velocity after \(\text{4}\) \(\text{s}\) will be:

\begin{align*} v(4) & = {H}'(4) \\ & = 20-10(4) \\ & = -\text{20}\text{ m.s$^{-1}$} \end{align*}

The ball hits the ground at a speed of \(\text{20}\text{ m.s$^{-1}$}\). Notice that the sign of the velocity is negative which means that the ball is moving downward (a positive velocity is used for upwards motion).

### Acceleration

\begin{align*} a=v'(t)&=H”(t) \\ &=-10 \end{align*}\begin{align*} \therefore a & = -\text{10}\text{ m.s$^{-2}$} \end{align*}

Just because gravity is constant does not mean we should necessarily think of acceleration as a constant. We should still consider it a function.