University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 203: 115


The tangent is $$y=-\frac{5}{4}x+6$$ and the normal is $$y=\frac{4}{5}x-\frac{11}{5}$$

Work Step by Step

$$x+\sqrt{xy}=6\hspace{2cm}(4,1)$$ a) Find the derivative of the function: $$1+\frac{1}{2\sqrt{xy}}(xy)'=0$$ $$1+\frac{y+xy'}{2\sqrt{xy}}=0$$ $$1+\frac{y}{2\sqrt{xy}}+\frac{xy'}{2\sqrt{xy}}=0$$ $$\frac{2\sqrt{xy}+y}{2\sqrt{xy}}+\frac{xy'}{2\sqrt{xy}}=0$$ $$y'=-\frac{\frac{2\sqrt{xy}+y}{2\sqrt{xy}}}{\frac{x}{2\sqrt{xy}}}=-\frac{2\sqrt{xy}+y}{x}$$ b) The slope of the tangent at $(4,1)$ is $$y'=-\frac{2\sqrt{4\times1}+1}{4}=-\frac{2\times2+1}{4}$$ $$y'=-\frac{5}{4}$$ The tangent to the curve at $(4,1)$ is $$y-1=-\frac{5}{4}(x-4)$$ $$y-1=-\frac{5}{4}x+5$$ $$y=-\frac{5}{4}x+6$$ c) Since the normal is the perpendicular line to the tangent at a point, the product of their slopes equals $-1$. Therefore, if we call the slope of the normal at $(4,1)$ $s$, we would have $$s\times\Big(-\frac{5}{4}\Big)=-1$$ $$s=\frac{4}{5}$$ The normal to the curve at $(4,1)$ is $$y-1=\frac{4}{5}(x-4)$$ $$y-1=\frac{4}{5}x-\frac{16}{5}$$ $$y=\frac{4}{5}x-\frac{11}{5}$$
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