Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 680: 5

Answer

The slope is $\ln 2 -\frac{1}{4}$.

Work Step by Step

Find $\frac{dx}{dt}$: $\frac{dx}{dt}=\frac{d}{dt}(t^2+2t)=2t+2$ Find $\frac{dy}{dt}$: $\frac{dy}{dt}=\frac{d}{dt}(2^t-2t)=2^t\ln 2-2$ Find $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2^t\ln 2-2}{2t+2}$ Find $t$ such that $(x,y)=(15,2)$: $t^2+2t=15$ and $2^t-2t=2$ $t^2+2t-15=0$ and $2^t-2t=2$ $(t+5)(t-3)=0$ and $2^t-2t=2$ $t=-5,t=3$ and $2^t-2t=2$ For $t=-5$, $2^{-5}-2\cdot (-5)=\frac{1}{32}+10\neq 2$ (not satisfied) For $t=3$, $2^3-2\cdot 3=8-6=2$ (satisfied) Find the slope at $(15,2)$: $m=\frac{dy}{dx}=\frac{2^3\ln 2-2}{2\cdot 3+2}=\frac{8\ln 2-2}{8}=\ln 2 -\frac{1}{4}$
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