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: 116


The tangent is $$y=-\frac{1}{4}x+\frac{17}{4}$$ and the normal is $$y=4x$$

Work Step by Step

$$x^{3/2}+2y^{3/2}=17\hspace{2cm}(1,4)$$ a) Find the derivative of the function: $$\frac{3}{2}x^{1/2}+2\times\frac{3}{2}y^{1/2}y'=0$$ $$x^{1/2}+2y^{1/2}y'=0$$ $$y'=-\frac{x^{1/2}}{2y^{1/2}}=-\frac{1}{2}\sqrt{\frac{x}{y}}$$ b) The slope of the tangent at $(1,4)$ is $$y'=-\frac{1}{2}\sqrt{\frac{1}{4}}=-\frac{1}{2}\times\frac{1}{2}=-\frac{1}{4}$$ The tangent to the curve at $(1,4)$ is $$y-4=-\frac{1}{4}(x-1)$$ $$y-4=-\frac{1}{4}x+\frac{1}{4}$$ $$y=-\frac{1}{4}x+\frac{17}{4}$$ 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 $(1,4)$ $s$, we would have $$s\times\Big(-\frac{1}{4}\Big)=-1$$ $$s=4$$ The normal to the curve at $(1,4)$ is $$y-4=4(x-1)$$ $$y-4=4x-4$$ $$y=4x$$
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