## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 2 - The Derivative - 2.1 Tangent Lines And Rates Of Change - Exercises Set 2.1 - Page 122: 29

#### Answer

(a) Average Velocity = $720ft/min$ (b) Instantaneous velocity at $t=2s$, $$v_{t=2}=192ft/min$$

#### Work Step by Step

Step-1: We know that the average velocity is equal to the ratio of the total displacement to time interval, that is, $$v_{avg} = \frac{\Delta s}{\Delta t}$$ $$\implies v_{avg} = \frac{s(4) - s(2)}{4-2} = \frac{6(4^4) - 6(2^4)}{2}$$ $$\therefore v_{avg} = 720ft/min$$ Step-2: We need to find instantaneous velocity of robot at $t=2$. Instantaneous velocity at time $t=t_0s$, $$v_{ins}(t_0)=\lim\limits_{t \to t_0}\frac{s(t)-s(t_0)}{t-t_0}$$ Let $h=t-t_0$. As $t\to t_0$, $h\to 0$. Thus, $$v_{ins}(t_0)=\lim\limits_{h\to 0}\frac{s(t_0+h)-s(t_0)}{h}$$ $$=\lim\limits_{h \to 0}\frac{6(t_0+h)^4-6(t_0)^4}{h}$$ $$=\lim\limits_{h \to 0}\frac{6(t_0^4+4t_0^3h+6t_0^2h^2+4t_0h^3+h^4)-6(t_0)^4}{h}$$ $$=\lim\limits_{h \to 0}6(4t_0^3+6t_0^2h+4t_0h^2+h^3)=24t_0^3$$ Step-3: Plug $2$ in place of $t_0$ to get $v_{ins}(t=2)$, $$v_{ins}(t=2)=24\times 2^3=192ft/min.$$

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