## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises: 57

#### Answer

$$\lim\limits_{\theta\to0^+}\frac{s}{d}=1$$

#### Work Step by Step

We call $r$ as the length of the radius of the circle. 1) Find the formula of $s$ A circular arc of length of $s$ would equal $$s=r\theta$$ $r$: the length of the radius of the circle $\theta$: the given angle of the arc 2) Find the formula of $d$ According to Law of Cosines, the length $d$ would be $$d^2=r^2+r^2-2rr\cos\theta$$ $$d^2=2r^2-2r^2\cos\theta$$ $$d^2=r^2\times2(1-\cos\theta)$$ Then, $$d=r\sqrt{2(1-\cos\theta)}$$ 3) Calculate $$\lim\limits_{\theta\to0^+}\frac{s}{d}=\lim\limits_{\theta\to0^+}\frac{r\theta}{r\sqrt{2(1-\cos\theta)}}=\lim\limits_{\theta\to0^+}\frac{\theta}{\sqrt{2(1-\cos\theta)}}$$ Multiply both numerator and denominator with $\sqrt{(1+\cos\theta)}$ $$\lim\limits_{\theta\to0^+}\frac{s}{d}=\lim\limits_{\theta\to0^+}\frac{\theta\sqrt{1+\cos\theta}}{\sqrt{2(1-\cos^2\theta)}}$$ $$=\lim\limits_{\theta\to0^+}\frac{\theta\sqrt{1+\cos\theta}}{\sqrt{2\sin^2\theta}}$$ $$=\lim\limits_{\theta\to0^+}\frac{\theta\sqrt{1+\cos\theta}}{|\sin\theta|\sqrt{2}}$$ $$=\lim\limits_{\theta\to0^+}\frac{\theta}{|\sin\theta|}\times\lim\limits_{\theta\to0^+}\frac{\sqrt{1+\cos\theta}}{\sqrt2}$$ Since $\theta\to0^+$, so here we only consider the cases for $\theta\gt0$. As $\theta\gt0$, $\sin\theta\gt0$, then $|\sin\theta|=\sin\theta$. Therefore, $$\lim\limits_{\theta\to0^+}\frac{s}{d}=\lim\limits_{\theta\to0^+}\frac{\theta}{\sin\theta}\times\frac{\sqrt{1+\cos\theta}}{\sqrt 2}$$ $$=1\times\frac{\sqrt{1+\cos0}}{\sqrt 2}$$ $$=1\times\frac{\sqrt{1+1}}{\sqrt 2}=1$$

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