Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 72


The two points are $(6, -8)$ and $(-6, 8).$

Work Step by Step

$\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2)=\dfrac{d}{dx}(100)\rightarrow$ $2x+\dfrac{dy}{dx}(2y)=0\rightarrow\dfrac{dy}{dx}=-\dfrac{x}{y}.$ Slope $= \dfrac{3}{4}\rightarrow \dfrac{dy}{dx}=\dfrac{3}{4}\rightarrow-\dfrac{x}{y}=\dfrac{3}{4}\rightarrow y=-\frac{4}{3}x$ Substituting $y=-\frac{4}{3}x$ in the original equation gives: $x^2+(-\frac{4}{3}x)^2=100\rightarrow\frac{25}{9}x^2=100\rightarrow x^2=36\rightarrow x=6$ or $x=-6$ When $x=6\rightarrow y=-\frac{4}{3}(6)=-8.$ When $x=-6\rightarrow y=-\frac{4}{3}(-6)=8.$ The two points are $(6, -8)$ and $(-6, 8).$
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