Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises: 16

Answer

a.) i.) $0 ft/sec$ $\quad$ ii.) $1 ft/sec$ $\quad$ iii.) $3 ft/sec$ $\quad$ iv.) $4ft/sec$ b.) $2 ft/sec$ c.) See attached file
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Work Step by Step

a.) The average velocity in the interval $[P,Q]$ is given by the slope of the secant line that goes through $P$ and $Q$, therefore: i.) The average velocity in the interval $[4,8]$ is: $\dfrac{f(8)-f(4)}{8-4} = \dfrac{\frac{1}{2}(8)^2-6\times8+23 -[\frac{1}{2}(4)^2-6\times4+23]}{4} = \dfrac{\frac{1}{2}64-48+23 -[\frac{1}{2}16-24+23]}{4} = \dfrac{32-25 -[8-1]}{4} = \dfrac{7 -7}{4} = \dfrac{0}{4} = 0 \\ \longrightarrow 0 ft/sec$ ii.) The average velocity in the interval $[6,8]$ is: $\dfrac{f(8)-f(6)}{8-6} = \dfrac{\frac{1}{2}(8)^2-6\times8+23 -[\frac{1}{2}(6)^2-6\times6+23]}{2} = \dfrac{\frac{1}{2}64-48+23 -[\frac{1}{2}36-36+23]}{2} = \dfrac{32-25 -[18-13]}{2} = \dfrac{7 -5}{2} = \dfrac{2}{2} = 1 \\ \longrightarrow 1 ft/sec $ iii.) The average velocity in the interval $[8,10]$ is: $\dfrac{f(10)-f(8)}{10-8} = \dfrac{\frac{1}{2}(10)^2-6\times10+23 -[\frac{1}{2}(8)^2-6\times8+23]}{2} = \dfrac{\frac{1}{2}100-60+23 -[\frac{1}{2}64-48+23]}{2} = \dfrac{50-37 -[32-25]}{2} = \dfrac{13 -7}{2} = \dfrac{6}{2} = 3 \longrightarrow 3 ft/sec$ iv.) The average velocity in the interval $[8,12]$ is: $\dfrac{f(12)-f(8)}{12-8} = \dfrac{\frac{1}{2}(12)^2-6\times12+23 -[\frac{1}{2}(8)^2-6\times8+23]}{4} = \dfrac{\frac{1}{2}144-72+23 -[\frac{1}{2}64-48+23]}{4} = \dfrac{72-49 -[32-25]}{4} = \dfrac{23 -7}{4} = \dfrac{16}{4} = 4 \longrightarrow 4ft/sec$ b.) The instantaneous velocity when $t=8$ is given by the equation: $\lim\limits_{h \to 0} \dfrac{f(8+h) - f(8)}{h} = \lim\limits_{h \to 0} \dfrac{\frac{1}{2}(8+h)^2-6(8+h) + 23 - [\frac{1}{2}(8)^2-6\times8+23]}{h} = \lim\limits_{h \to 0} \dfrac{\frac{1}{2}(64 + 16h + h^2)-48 -6h + 23 - [\frac{1}{2}64-48+23]}{h} = \lim\limits_{h \to 0} \dfrac{32 + 8h + \frac{1}{2}h^2 -6h -25 - [32-25]}{h} = \lim\limits_{h \to 0} \dfrac{\frac{1}{2}h^2 + 2h +7 - 7}{h} = \lim\limits_{h \to 0} \dfrac{\frac{1}{2}h^2 + 2h}{h} = \lim\limits_{h \to 0} \dfrac{h(\frac{1}{2}h + 2)}{h} = \lim\limits_{h \to 0} (\dfrac{1}{2}h + 2) = \dfrac{1}{2}0 +2=2 \longrightarrow 2 ft/sec$
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