Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Review - Exercises: 26

Answer

$\frac{dy}{dx}=\frac{1-3t^{2}}{2t}$ and $\frac{d^{2}y}{dx^{2}}=\frac{-1-3t^{2}}{4t^{3}}$

Work Step by Step

Given: $x=1+t^{2}$ and $y=t-t^{3}$ $\frac{dx}{dt}=2t$ $\frac{dy}{dt}=1-3t^{2}$ $\frac{dy}{dx}=\frac{{dy}/{dt}}{{dx}/{dt}}=\frac{1-3t^{2}}{2t}$ $\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}({dy}/{dx})}{dx/dt}$ $=\frac{\frac{d}{dt}(\frac{1-3t^{2}}{2t})}{2t}$ $=\frac{-\frac{1}{2}t^{-2}-\frac{3}{2}}{2t}\times \frac{2t^{2}}{2t^{2}}$ $=\frac{-1-3t^{2}}{4t^{3}}$ Hence, $\frac{dy}{dx}=\frac{1-3t^{2}}{2t}$ and $\frac{d^{2}y}{dx^{2}}=\frac{-1-3t^{2}}{4t^{3}}$
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