University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.1 - Rates of Change and Tangents to Curves - Exercises - Page 56: 4

Answer

(a) $\frac{\Delta y}{\Delta t}=-\frac{2}{\pi}$ (b) $\frac{\Delta y}{\Delta t}=0$

Work Step by Step

*Average rates of change: The average rate of change of $y=f(x)$ with respect to $x$ over the interval $[x_1,x_2]$ is: $$\frac{\Delta y}{\Delta x}=\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ $$g(t)=2+\cos t$$ (a) $[0,\pi]$ The average rate of change of $y=h(t)$: $$\frac{\Delta y}{\Delta t}=\frac{(2+\cos\pi)-(2+\cos0)}{\pi-0}$$ $$\frac{\Delta y}{\Delta t}=\frac{\cos\pi-\cos0}{\pi}=\frac{-1-1}{\pi}=-\frac{2}{\pi}$$ (b) $[-\pi,\pi]$ The average rate of change of $y=h(t)$: $$\frac{\Delta y}{\Delta t}=\frac{(2+\cos\pi)-[2+\cos(-\pi)]}{\pi-(-\pi)}$$ $$\frac{\Delta y}{\Delta t}=\frac{\cos\pi-\cos(-\pi)}{2\pi}=\frac{-1-(-1)}{2\pi}=\frac{-1+1}{2\pi}=\frac{0}{2\pi}=0$$
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