Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 33


$$y = 6 \pi x-\frac{9 \pi^{2}}{2}$$

Work Step by Step

Given $$y=x^{2}(1-\sin x), \quad x=\frac{3 \pi}{2}$$ Since $y(3\pi/2)= \frac{9 \pi^{2}}{2}$ and \begin{align*} y'&=\frac{d}{dx}\left(x^2\right)\left(1-\sin \left(x\right)\right)+\frac{d}{dx}\left(1-\sin \left(x\right)\right)x^2 \\ &= 2x\left(1-\sin \left(x\right)\right)-x^2\cos \left(x\right)\\ y'\bigg|_{x=3\pi/2}&= 6\pi \end{align*} Then the tangent line equation is given by \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-\frac{9 \pi^{2}}{2}}{x-3\pi/2} &= 6\pi \\ y- \frac{9 \pi^{2}}{2}&=6\pi [x-3\pi/2]\\ y&= 6 \pi x-\frac{9 \pi^{2}}{2} \end{align*}
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