# Average and Instantaneous Rate of Change

We see changes around us everywhere. When we project a ball upwards, its position changes with respect to time, its velocity changes as its position changes. The height of a person changes with time. Prices of stock and options changes with time. Equilibrium price of goods changes with respect to demand and supply. The power radiated by a black body changes as its temperature changes. The surface area of a sphere changes as its radius changes. This list never ends. It is amazing to measure and study these changes.

These changes depend on many factors, such as power radiated by a black body depends on it's surface area as well as temperature. We shall be looking at cases where only one factor is varying, and all others are fixed. Then , we can model our system as \(y = f(x)\) , where \(y\) changes with regard to \(x\).

## Average rate of change

One way to measure changes is by looking at endpoints of a given interval.

If \(y_1 = f(x_1)\) and \(y_2 = f(x_2)\), the

averagerate of change of \(y\) wrt \(x\) in the interval from \(x_1\) to \(x_2\) is the average change in \(y\) for unit increase in \(x\). It is equal to:\[\dfrac{\Delta y}{\Delta x} = \dfrac{y_2 - y_1}{x_2 - x_1}\]

where \(\Delta x\) and \(\Delta y\) are the changes in \(x\) and \(y\) respectively.

Consider the following figure.

As \(x\) increased by \(\Delta x\), \(y\) increased by \(\Delta y\). So we can say,

on average, for every unit increase in \(x\), \(y\) increases by \(\dfrac{\Delta y}{\Delta x}\), therefore this is the average rate of change. \(_\square\)

A car is travelling on a straight road parallel to the \(x\)- axis. At \(t = 0\) seconds, the car is at \(x = 2\) meters, and at \(t = 6\) seconds, the car is at \(x = 14\) meters. Find the average rate of change of \(x\) coordinate of the car wrt time.

Using the formula, we get

\[ \text{Rate} = \dfrac{\Delta x}{\Delta t} = \dfrac{14 - 2}{6 - 0} = 2 \text{ ms}^{-1} ~~~~_ \square\]

## Instantaneous rate of change

Average rate of change tells us at what rate \(y\) increases in an interval. This just tells us the average, and no information of between. We have no idea how the function behaves in the interval. The following animation makes it clear. In all cases, the average rate of change is the same, but the function is very different in each case.

If we make \(\Delta x\) smaller, we get a more accurate representation of \(y\); as \(\Delta x \) tends to \(0\), the interval becomes smaller and smaller until it just becomes a point, an instant. Then the rate of change is not an average, but of an instant. It is the instantaneous rate of change of \(y\) wrt \(x\). We denote it as \(\dfrac{\text{d}y}{\text{d}x}\).

Mathematically,

\[\dfrac{\text{d}y}{\text{d}x} = \displaystyle \lim _{\Delta x\rightarrow 0} \dfrac{\Delta y}{\Delta x}\]

\(\dfrac{\text{d}y}{\text{d}x}\) is the instantaneous rate of change of \(y\) wrt \(x\). It is also called the derivative of \(y\) wrt \(x\).

Note 1: We can see that \(\dfrac{\text{d}y}{\text{d}x}\) will exist only when the limit exists. For example, in the green graph in the animation, \(\dfrac{\text{d}y}{\text{d}x}\) does not exist on some finite discrete points (the edges in the graph). It is not possible to find out the instantaneous rate of change at those points.

Note 2: At very small values of \(\Delta x\), we can see that \[\dfrac{\text{d}y}{\text{d}x} \approx \dfrac{\Delta y}{\Delta x}\]

## Examples

Let's solve some examples.

Let \(y = 2x \ln x \). What is the rate of change of \(y\) with respect to \(x\) when

(i) \(x = e\),(ii) \(y = 4e^2\)\(?\)

The question is asking to evaluate \(\frac{\text{d}y}{\text{d}x}\) at the given values of \(x\) and \(y\). To differentiate the expression, we must know product rule and differentiation of logarithmic functions.

\[\begin{align} \frac{\text{d}y}{\text{d}x} & = \frac{\text{d}}{\text{d}x}\left( 2x \ln x\right) \\ & = 2 \frac{\text{d}}{\text{d}x} \left(x \ln x\right) \\ & = 2\left(x \cdot \frac{\text{d}}{\text{d}x}\left(\ln x\right) + \ln x \cdot \frac{\text{d}}{\text{d}x} \left(x\right) \right) \\ & = 2\left(\frac{x}{x} + \ln x\right) \\ & = 2\left(1 + \ln x\right) \end{align}\]

(i) We now evaluate it at \(x = e\). When \(x = e\), \(\dfrac{\text{d}y}{\text{d}x} = 2(1 + \ln e) = 2 \cdot (1 + 1) = 4\)

(ii) When \(y = 4e^2, x = e^2\), \(\dfrac{\text{d}y}{\text{d}x} = 2(1 + \ln e^2) = 2 \cdot (1 + 2) = 6 \) \(_\square \)

A red cube has side length \(a\). \(a\) is changing with time such that \(a(t) = a_{\text{0}} t^2\). Find the instantaneous rate of change of volume of the red cube as a function of time.

Let the volume of the red cube be \(V\). We know that \(V = a^3 = (a_{\text{0}} t^2)^3.\)

We are asked to find out \(\dfrac{\text{d}V}{\text{d}t}\). We can solve this question in two ways:

Solution 1:First we find \(V\) and then \(\dfrac{\text{d}V}{\text{d}t}\).We know

\[ \begin{align} V & = a^3 \\ & = (a_{\text{0}} t^2)^3 \\ & = a_{\text{0}}^3 \cdot t^6 \\ \dfrac{\text{d}V}{\text{d}t} & = 6a_{\text{0}}^3 \cdot t^5 \end{align} \]

and we are done!

Solution 2:First we find \(\dfrac{\text{d}a}{\text{d}t}\) and then \(\dfrac{\text{d}V}{\text{d}t}\).Using the power rule, \(\dfrac{\text{d}a}{\text{d}t} = 2 a_{\text{0}} t\)

Using the chain rule to differentiate \(V = a^3\),

\[\dfrac{\text{d}V}{\text{d}t} = 3a^2 \cdot \dfrac{\text{d}a}{\text{d}t}\]

After subtistuting the values of \(3a^2\) and \(\dfrac{\text{d}a}{\text{d}t}\), we obtain the same result;

\[\dfrac{\text{d}V}{\text{d}t} = 3a_{\text{0}}^2 \cdot t^4 \cdot 2 a_{\text{0}} t = 6a_{\text{0}}^3 \cdot t^5 ~~~~ _\square\]

In a hollow inverted blue cone (vertex is downward) of radius \(r\) and height \(h\), water is being poured in at a constant rate of \(l\). Find the instantaneous rate of change of height of water in the cone at time \(t\). (assuming the cone isn't filled completely yet)

Solution to be added...

**Cite as:**Average and Instantaneous Rate of Change.

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