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

Published by Pearson

# Chapter 3 - Section 3.7 - Implicit Differentiation - Exercises - Page 165: 38

#### Answer

a) The point $(\pi/4,\pi/2)$ is on the curve. b) The tangent line to the curve at $(\pi/4,\pi/2)$ is $$y=2x$$ c) The normal line to the curve at $(\pi/4,\pi/2)$ is $$y=-1/2x+5\pi/8$$

#### Work Step by Step

$$x\sin2y=y\cos2x$$ 1) Verify that the point $(\pi/4,\pi/2)$ is on the curve: We substitute the coordinates of the point $(\pi/4,\pi/2)$ into the formula of the function to see if 2 sides are equal or not: - Left side: $$\frac{\pi}{4}\sin\pi=\frac{\pi}{4}\times0=0$$ - Right side: $$\frac{\pi}{2}\cos\frac{\pi}{2}=\frac{\pi}{2}\times0=0$$ Since 2 sides are equal to each other, the point is on the curve. 2) Find the tangent at $(\pi/4,\pi/2)$: - Find the derivative of the function with implicit differentiation: $$\sin2y+x\cos2y(2y)'=y'\cos2x+y(-\sin2x)(2x)'$$ $$\sin2y+2x(\cos2y)y'=y'\cos2x-2y\sin2x$$ $$y'\cos2x-2x(\cos2y)y'=\sin2y+2y\sin2x$$ $$y'(\cos2x-2x\cos2y)=\sin2y+2y\sin2x$$ $$y'=\frac{\sin2y+2y\sin2x}{\cos2x-2x\cos2y}$$ - The slope of the tangent line at $(\pi/4,\pi/2)$ is $$y'=\frac{\sin\pi+\pi\sin\frac{\pi}{2}}{\cos\frac{\pi}{2}-\frac{\pi}{2}\cos\pi}=\frac{0+\pi\times1}{0-\frac{\pi}{2}(-1)}=\frac{\pi}{\frac{\pi}{2}}=2$$ - The tangent line to the given curve at $(\pi/4,\pi/2)$ is $$y-\frac{\pi}{2}=2(x-\frac{\pi}{4})$$ $$y-\frac{\pi}{2}=2x-\frac{\pi}{2}$$ $$y=2x$$ 3) Find the normal line at $(\pi/4,\pi/2)$: The normal line at $(\pi/4,\pi/2)$ would be perpendicular with the tangent line at $(\pi/4,\pi/2)$, so the product of the slopes of these two lines equals $−1$. So if we call the slope of the normal line at $(\pi/4,\pi/2)$ $k$, we would have $$k\times2=-1$$ $$k=-\frac{1}{2}$$ - The normal line to the given curve at $(\pi/4,\pi/2)$ therefore is $$y-\frac{\pi}{2}=-\frac{1}{2}(x-\frac{\pi}{4})$$ $$y-\frac{\pi}{2}=-\frac{1}{2}x+\frac{\pi}{8}$$ $$y=-\frac{1}{2}x+\frac{5\pi}{8}$$

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