Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 687: 10


$y=-3x+\dfrac{13}{4}$ and $6$

Work Step by Step

Here, we have $\dfrac{dy}{dx}=(\dfrac{-3}{2})t$ Now, $ \dfrac{dy}{dx}(t=2)=(\dfrac{-3}{2})(2)=-3$ Now, $y+\dfrac{1}{2}=(-3)(x-\dfrac{5}{4})$ or, $y=-3x+\dfrac{13}{4}$ Now, $\dfrac{d^2y}{dx^2}=\dfrac{dy'/dt}{dx/dt}=(\dfrac{3}{4})t^3$ and $\dfrac{d^2y}{dx^2}(t=2)=\dfrac{3}{4} (2)^3=(\dfrac{3}{4}) (8)=6$
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