University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 8

Answer

Tangent line: $\quad y=-2x-1$ $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=3}=-\frac{1}{3}$

Work Step by Step

The slope of the tangent line at $t=3$ is $\left. \displaystyle \frac{dy}{dx} \right|_{t=3}$ $\displaystyle \left\{\begin{array}{l} \frac{dx}{dt}=-\frac{1}{2}(t+1)^{-1/2},\\ \\ \frac{dy}{dt}=\frac{3}{2}(3t)^{-1/2} \end{array}\right.\quad, \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{(\frac{3}{2})(3t)^{-1/2}}{(-\frac{1}{2})(t+1)^{-1/2}}=-\frac{3\sqrt{t+1}}{\sqrt{3t}}$ $\left. \displaystyle \frac{dy}{dx} \right|_{t=3}=\frac{-3\sqrt{3+1}}{\sqrt{3(3)}}=\frac{-3(2)}{3}=-2$ With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=3}$, $\displaystyle \frac{dy'}{dt}=\frac{d}{dt}[-\frac{3\sqrt{t+1}}{\sqrt{3t}}]$ $=-\displaystyle \frac{3\cdot\frac{1}{2\sqrt{t+1}}\cdot\sqrt{3t}-(3\sqrt{t+1})\cdot\frac{1}{2\sqrt{3t}}\cdot 3}{3t}$ $=-\displaystyle \frac{3\cdot 3t-(3\sqrt{t+1})\cdot 3\sqrt{t+1}}{(3t)(2\sqrt{t+1})(\sqrt{3t})}$ $=-\displaystyle \frac{3t-(3\sqrt{t+1})\sqrt{t+1}}{2t\sqrt{t+1})(\sqrt{3t})}$ $=-\displaystyle \frac{3t-3(t+1)}{2t\sqrt{t+1}\cdot\sqrt{3t}}$ $=\displaystyle \frac{3}{2t\sqrt{t+1}\cdot\sqrt{3t}}$ $\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{3}{2t\sqrt{t+1}\cdot\sqrt{3t}[(-\frac{1}{2})(t+1)^{-1/2}]}=\frac{-3}{t\cdot\sqrt{3t}}$ $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=3}=\frac{-3}{3\cdot 3}=-\frac{1}{3}$ The point $P$ on the curve corresponding to $t=3$ is $\left\{\begin{array}{l} x=-\sqrt{3+1}=-2\\ \\ y=\sqrt{3(3)}=3 \end{array}\right.\quad, P(-2,3)$ Use the point-slope equation for the tangent line. $y-3=-2(x+2)$ $y=-2x-4+3$ $y=-2x-1$
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