Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Calculus with Parametric Curves - 10.2 Exercises - Page 695: 15

Answer

$ 0

Work Step by Step

Given $$x=t-\ln t, \ \ \ \ y=t+\ln t $$ Since \begin{aligned} \frac{dy}{dt}& = 1+\frac{1}{t}\\ \frac{dx}{dt}& = 1-\frac{1}{t}\\ \end{aligned} Then \begin{aligned} \frac{d y}{d x}&=\frac{d y / d t}{d x / d t}\\ &=\frac{1+1 / t}{1-1 / t}\\ &=\frac{t+1}{t-1} \end{aligned} and \begin{aligned} \frac{d^2 y}{d x^2}&=\frac{\frac{d}{d t}\left(\frac{d y}{d x}\right)}{d x / d t}\\ &=\frac{\frac{(t-1)(1)-(t+1)(1)}{(t-1)^2}}{(t-1) / t}\\ &=\frac{-2 t}{(t-1)^3} \end{aligned} Since the curve is concave up when $ \frac{d^2y}{dx^2}>0 $ ,then $ \frac{d^2y}{dx^2}>0 $ for $ 0
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