Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 655: 1



Work Step by Step

Given: $$x=\frac{t}{1+t}$$ and $$y=\sqrt 1+t$$ We can solve for dy/dx: $$\frac{dy}{dt}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ So, we find the derivatives of each of the parametric equations with respect to t. $$\frac{dx}{dt}=\frac{dt}{dt}[\frac{1\times(1+t)-t\times(1)}{(1+t)^2}]=1[\frac{1}{(1+t)^2}]=\frac{1}{(1+t)^2}$$ and $$\frac{dy}{dt}=\frac{dt}{dt}\frac{1}{2}(1+t)^{-\frac{1}{2}}=1\frac{1}{2}(1+t)^{-\frac{1}{2}}=\frac{1}{2}(1+t)^{-\frac{1}{2}}$$ Then, $$\frac{dy}{dt}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{1}{2}(1+t)^{-\frac{1}{2}}}{\frac{1}{(1+t)^2}}=\frac{1}{2}(1+t)^{-\frac{1}{2}}\times(1+t)^{2}=\frac{1}{2}(1+t)^\frac{3}{2}$$
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