Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.1 Tangent Lines And Rates Of Change - Exercises Set 2.1 - Page 121: 28

Answer

$$v_{avg}=\frac{4.5(12)^{2}-4.5(0)^{2}}{12-0}=54$$ $$v_{inst}=9t=9\cdot 6=54$$

Work Step by Step

The average velocity is: $$v_{avg}=\frac{s(12)-s(0)}{12-0}$$ $$v_{avg}=\frac{4.5(12)^{2}-4.5(0)^{2}}{12-0}=54$$ ---------------------------------------------------------------------------- $$v_{inst}=\lim\limits_{h \to 0}\frac{s(t+h)-s(t)}{h}$$ $$v_{inst}=\lim\limits_{h \to 0}\frac{4.5(t+h)^{2}-4.5t^{2}}{h}$$ $$v_{inst}=\lim\limits_{h \to 0}\frac{4.5t^{2}+9th+4.5h^{2}-4.5t^{2}}{h}$$ $$v_{inst}=\lim\limits_{h \to 0}\frac{+9th+4.5h^{2}}{h}$$ $$v_{inst}=\lim\limits_{h \to 0}(9t+4.5h)$$ $$v_{inst}=9t=9\cdot 6=54$$
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