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: 56


The answers: $y=f(x)=x^2-2x$. $\frac{\text{dy}}{\text{dx}}=2 x-2$.

Work Step by Step

First way: Using Eq. (8) we have $\frac{\text{dy}}{\text{dx}}=\frac{y'(t)}{x'(t)}=\frac{\frac{t}{2}-1}{\frac{1}{2}}=t-2$. Second way: Since $x=(1/2) t$, so $t=2x$. Substituting $t$ into $y=(1/4) t^2-t$ gives $y=(1/4) (2x)^2-2x=x^2-2x$, $y=f(x)=x^2-2x$. Differentiating $f(x)$ we get $\frac{\text{dy}}{\text{dx}}=2 x-2$. Since $t=2x$, this is the same as $t-2$.
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