Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.9 Linear Approximations and Differentials - 2.9 Exercises - Page 192: 2


$$\boxed{y-1/2 = \sqrt{3}/2*(x-\pi/6)}$$

Work Step by Step

To find the linearization we first need to calculate the slope by taking the derivative. $$f'(\pi/6) = cos(\pi/6) = \sqrt {3}/2$$ We also need to find the y-value at the given $a$ value, $$f(\pi/6) = sin(\pi/6) = 1/2$$ Since we have the point $(\pi/6, 1./2)$ and the slope $\sqrt {3}/2$, we can use point slope form. $$\boxed{y-1/2 = \sqrt{3}/2*(x-\pi/6)}$$
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