Calculus (3rd Edition)

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

Chapter 14 - Calculus of Vector-Valued Functions - 14.1 Vector-Valued Functions - Exercises - Page 712: 42


$$r(t)=(\frac{1}{t}-t, t,\frac{1}{t^2}-2).$$

Work Step by Step

We have $z=x^2-y^2$ and $z=x^2+xy-1$. Hence $$x^2-y^2=x^2+xy-1\Longrightarrow xy=1-y^2\Longrightarrow x=\frac{1}{y}-y.$$ Also, $$z=x^2-y^2\Longrightarrow z=(\frac{1}{y}-y)^2-y^2=\frac{1}{y^2}-2.$$ Now let $y=t$; then we have $x=\frac{1}{t}-t$ and $z=\frac{1}{t^2}-2$. So we have the parametrization $$r(t)=(\frac{1}{t}-t, t,\frac{1}{t^2}-2).$$
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