Examples of Average and Instantaneous Rate of Change

Example:

Let $$y = {x^2} – 2$$
(a) Find the average rate of change of $$y$$ with respect to $$x$$ over the interval $$[2,5]$$.
(b) Find the instantaneous rate of change of $$y$$ with respect to $$x$$ at point $$x = 4$$.

Solution:

(a) For Average Rate of Change:
We have
$$y = f(x) = {x^2} – 2$$

Put $$x = 2$$
$$\therefore f(2) = {(2)^2} – 2 = 4 – 2 = 2$$

Again put $$x = 5$$
$$\therefore f(5) = {(5)^2} – 2 = 25 – 2 = 23$$

The average rate of change over the interval $$[2,5]$$ is
$$\frac{{f(5) – f(2)}}{{5 – 2}} = \frac{{23 – 2}}{3} = \frac{{21}}{3} = 7$$

(b) For Instantaneous Rate of Change:
We have
$$y = f(x) = {x^2} – 2$$

Put $$x = 4$$
$$\therefore f(4) = {(4)^2} – 2 = 16 – 2 = 14$$

Now, putting $$x = {x_1}$$ then
$$\therefore f({x_1}) = {x_1}^2 – 2$$

The instantaneous rate of change at point $$x = 4$$ is

\[\begin{gathered} \mathop {\lim }\limits_{{x_1} \to 4} \frac{{f({x_1}) – f(4)}}{{{x_1} – 4}} = \mathop {\lim }\limits_{{x_1} \to 4} \frac{{{x_1}^2 – 2 – 14}}{{{x_1} – 4}} = \mathop {\lim }\limits_{{x_1} \to 4} \frac{{{x_1}^2 – 16}}{{{x_1} – 4}} \\ = \mathop {\lim }\limits_{{x_1} \to 4} \frac{{({x_1} + 4)({x_1} – 4)}}{{{x_1} – 4}} = \mathop {\lim }\limits_{{x_1} \to 4} ({x_1} + 4) = 4 + 4 = 8 \\ \end{gathered} \]

Example:
A particle moves on a line away from its initial position so that after $$t$$ seconds it is $$S = 2{t^2} – t$$ feet from its initial position.
(a) Find the average velocity of the particle over the interval$$[1,3]$$.
(b) Find the instantaneous velocity at $$t = 2$$.

Solution:

(a) For Average Velocity:
We have
$$S(t) = 2{t^2} – t$$

Put $$t = 1$$
$$\therefore S(1) = 2{(1)^2} – 1 = 2 – 1 = 1$$

Again put $$t = 3$$
$$\therefore S(3) = 2{(3)^2} – 3 = 18 – 3 = 15$$

The average velocity over the interval $$[1,3]$$ is
$${V_{ave}} = \frac{{S(3) – S(1)}}{{3 – 1}} = \frac{{15 – 1}}{2} = \frac{{14}}{2} = 7{\text{ }}ft/Sec$$

(b) For Instantaneous Velocity:
We have
$$S(t) = 2{t^2} – t$$

Put $$t = 2$$
$$\therefore S(2) = 2{(2)^2} – 2 = 8 – 2 = 6$$

Now putting $$t = {t_1}$$
$$\therefore S({t_1}) = 2{t_1}^2 – {t_1}$$

The instantaneous velocity at $$t = 2$$ is

\[\begin{gathered} {V_{ins}} = \mathop {\lim }\limits_{{t_1} \to 2} \frac{{S({t_1}) – S(2)}}{{{t_1} – 2}} = \mathop {\lim }\limits_{{t_1} \to 2} \frac{{2{t_1}^2 – {t_1} – 6}}{{{t_1} – 2}} = \mathop {\lim }\limits_{{t_1} \to 2} \frac{{2{t_1}^2 – 4{t_1} + 3{t_1} – 6}}{{{t_1} – 2}} \\ = \mathop {\lim }\limits_{{t_1} \to 2} \frac{{2{t_1}({t_1} – 2) + 3({t_1} – 2)}}{{{t_1} – 2}} = \mathop {\lim }\limits_{{t_1} \to 2} \frac{{(2{t_1} + 3)({t_1} – 2)}}{{{t_1} – 2}} \\ = \mathop {\lim }\limits_{{t_1} \to 2} (2{t_1} + 3) = 2(2) + 3 = 4 + 3 = 7{\text{ }}ft/Sec \\ \end{gathered} \]