Calculus (3rd Edition)

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

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 604: 58

Answer

The answers: $y=f(x)=-x^2+x+1$. $dy/dx=-2x+1$.

Work Step by Step

First way: Using Eq. (8) we have $dy/dx=(y'(\theta))/(x'(\theta))=(-\sin {\theta} + 2\cos \theta \sin \theta)/(-\sin \theta)=1-2 \cos\theta$. Second way: Since $x=\cos \theta$ and so $\sin^2 \theta=1-\cos^2 \theta=1-x^2$. Substituting these into $y=\cos \theta+\sin^2 \theta$ gives $y=x+1-x^2$, $y=f(x)=-x^2+x+1$. Differentiating $f(x)$ we get $dy/dx=-2x+1$. Since $x=\cos \theta$, this is the same as $1-2\cos \theta$.
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