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 and its velocity changes as its position changes. The height of a person changes with time. The prices of stocks and options change with time. The equilibrium price of a good 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; for example, the power radiated by a black body depends on its 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 \(x\).

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 average rate of change of \(y\) with respect to \(x\) 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 \(\frac{\Delta y}{\Delta x}\), and 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; at \(t = 6\) seconds, the car is at \(x = 14\) meters. Find the average rate of change of the \(x\)-coordinate of the car with respect to time.

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Using the formula, we get

\[ \text{Rate} = \dfrac{\Delta x}{\Delta t} = \dfrac{14 - 2}{6 - 0} = 2 \text{ m/s}.\ _\square\]

The average rate of change tells us at what rate \(y\) increases in an interval. This just tells us the average and no information in-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 \(y\) with respect to \(x\). We denote it as \(\frac{\text{d}y}{\text{d}x}\).

Mathematically,

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

where \(\frac{\text{d}y}{\text{d}x}\) is the instantaneous rate of change of \(y\) with respect to \(x\). It is also called the derivative of \(y\) with respect to \(x\).

Note 1: We can see that \(\frac{\text{d}y}{\text{d}x}\) will exist only when the limit exists. For example, in the green graph in the animation, \(\frac{\text{d}y}{\text{d}x}\) 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 \(\frac{\text{d}y}{\text{d}x} \approx \frac{\Delta y}{\Delta x}.\)

Let's solve some examples.

Let \(y = 2x \ln x \).

What is the rate of change of \(y\) with respect to \(x\) when (i) \(x = e\) and (ii) \(y = 4e^2?\)

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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. We have

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

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

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

A red cube has side length \(a,\) and \(a\) is changing with time such that \(a(t) = a_{\text{0}} t^2\).

Find the instantaneous rate of change of the volume of the red cube as a function of time.

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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 \(\frac{\text{d}V}{\text{d}t}\). We can solve this question in the following two ways:

Solution 1: We first find \(V\) and then \(\frac{\text{d}V}{\text{d}t}\).

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 \(\frac{\text{d}a}{\text{d}t}\) and then \(\frac{\text{d}V}{\text{d}t}\).

Using the power rule, \(\frac{\text{d}a}{\text{d}t} = 2 a_{\text{0}} t.\)

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

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

After subtistuting the values of \(3a^2\) and \(\frac{\text{d}a}{\text{d}t}\), we obtain the same result as above:

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

In a hollow inverted blue cone (the 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 the height of water in the cone at time \(t\) (assuming the cone isn't filled completely yet).

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Solution to be added...