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 the 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 the 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}

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