# 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

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