Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Review - Exercises: 25

Answer

$\frac{dy}{dx}=\frac{1+sint}{1+cost}$ and $\frac{d^{2}y}{dx^{2}}=\frac{1+cost+sint}{(1+cost)^{3}}$

Work Step by Step

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