Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 13 - Vector Functions - 13.2 Exercises - Page 876: 12

Answer

$\displaystyle \mathrm{r}^{\prime}(t)=-\frac{1}{(1+t)^{2}}\mathrm{i}+\frac{1}{(1+t)^{2}}\mathrm{j}+\frac{t^{2}+2t}{(1+t)^{2}}\mathrm{k}$

Work Step by Step

Parametric equations: $\mathrm{r}(t):\quad\left\{\begin{array}{ll} x=(1+t)^{-1} & \\ y=t(1+t)^{-1} & \\ y=t^{2}(1+t)^{-1} & \end{array}\right.$ Differentiate $(\displaystyle \frac{d}{dt})$ each component function. x: chain rule y,z: quotient rule $\mathrm{r}^{\prime}(t):\quad\left\{\begin{array}{ll} x=-1(1+t)^{-2}(1) & =-\dfrac{1}{(1+t)^{2}}\\ \\ y=\dfrac{(1+t)\cdot 1-t(1)}{(1+t)^{2}} & =\dfrac{1}{(1+t)^{2}}\\\\ z=\dfrac{(1+t)\cdot 2t-t^{2}(1)}{(1+t)^{2}} & =\dfrac{t^{2}+2t}{(1+t)^{2}} \end{array}\right.\\ $ $\displaystyle \mathrm{r}^{\prime}(t)=-\frac{1}{(1+t)^{2}}\mathrm{i}+\frac{1}{(1+t)^{2}}\mathrm{j}+\frac{t^{2}+2t}{(1+t)^{2}}\mathrm{k}$
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