Multivariable Calculus, 7th Edition

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

Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Exercises - Page 675: 13

Answer

The curve is concave up when $\frac{d^{2}y}{dx^{2}} \gt 0 $, when $t \gt \frac{3}{2}$

Work Step by Step

$x=e^{t}$, $y=te^{-t}$ Step 1 $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{-te^{-t}+e^{-t}}{e^{t}}=e^{-2t}(1-t)$ Step 2 Find second derivative $\frac{d^{2}y}{dx^{2}}= \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}= \frac{e^{-2t}(-1)+(1-t)(-2e^{-2t})}{e^{t}}=\frac{e^{-2t}(-1-2+2t)}{e^{t}}= e^{-2t}(2t-3)$ The curve is concave up for $\frac{d^{2}y}{dx^{2}} \gt 0$. This happens when $(2t-3)\gt 0$ or when $t \gt \frac{3}{2}$
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