Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 16

Answer

(i) $0 ft/s$ (ii) $ 1 ft/s$ (iii) $ 3ft/s$ (iv) $ 4ft/s$ b. $2 ft/s$

Work Step by Step

(i) $[4.8]$ $\frac{s(8)-s(4)}{8-4} = \frac{(7)-(7)}{4} = \frac{0}{4}= 0 ft/s$ (ii) $[6,8]$ $\frac{s(8)-s(6)}{8-6} = \frac{(7)-(5)}{2} = \frac{2}{2} = 1 ft/s$ (iii) $[8,10]$ $\frac{s(10)-s(8)}{10-8} = \frac{(13)-(7)}{2} = \frac{6}{2} = 3 ft/s$ (iv) $[8,12]$ $\frac{s(12)-s(8)}{12-8} = \frac{(23)-(7)}{4} = \frac{16}{4} = 4 ft/s$ b. Find the instantaneous velocity when $t = 8$. $s' = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$ $s' = \lim\limits_{h \to 0} \frac{1(a+h)^{2} - 6(a+h) + 23 - (\frac{1}{2} a^{2} - 6a + 23)}{h}$ $s' = \lim\limits_{h \to 0} \frac{\frac{1}{2}a^{2} + ah + \frac{1}{2} h^{2} - 6a +6h + 23 - \frac{1}{2}a^{2} +6a - 23}{h}$ $s' = \lim\limits_{h \to 0} \frac{\frac{1}{2}h^{2} + ah + 6h}{h}$ Simplify: $s' = \lim\limits_{h \to 0} \frac{h(\frac{1}{2}h + a - 6)}{h}$ $s' = \lim\limits_{h \to 0} \frac{1}{2}h + a - 6$ $s' = \frac{1}{2}(0) + a - 6$ $s' = a - 6$ $s'(8) = 8 - 6 = 2 ft/s$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.