## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 2 - Section 2.1 - Rates of Change and Tangents to Curves - Exercises - Page 56: 3

#### Answer

(a) $\frac{\Delta y}{\Delta t}=-\frac{4}{\pi}$ (b) $\frac{\Delta y}{\Delta t}=-\frac{3\sqrt3}{\pi}$

#### 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: $$\frac{\Delta y}{\Delta x}=\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ $$h(t)=\cot t$$ (a) $[\pi/4,3\pi/4]$ The average rate of change of $y=h(t)$: $$\frac{\Delta y}{\Delta t}=\frac{\cot(3\pi/4)-\cot(\pi/4)}{\frac{3\pi}{4}-\frac{\pi}{4}}$$ $$\frac{\Delta y}{\Delta t}=\frac{-1-1}{\frac{2\pi}{4}}=\frac{-2}{\frac{\pi}{2}}=-\frac{4}{\pi}$$ (b) $[\pi/6,\pi/2]$ The average rate of change of $y=h(t)$: $$\frac{\Delta y}{\Delta t}=\frac{\cot(\pi/2)-\cot(\pi/6)}{\frac{\pi}{2}-\frac{\pi}{6}}$$ $$\frac{\Delta y}{\Delta t}=\frac{0-\sqrt3}{\frac{2\pi}{6}}=\frac{-\sqrt3}{\frac{\pi}{3}}=-\frac{3\sqrt3}{\pi}$$

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