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: 7

Answer

(a) The equation of the tangent line is $~~y = 2x+1$ (b) The equation of the tangent line is $~~y = 2x+1$

Work Step by Step

(a) The equation of the tangent line is $\frac{dy}{dx}$ We can find $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2t}{1/t} = 2t^2$ When $x=1$: $x = 1+ln~t = 1$ $ln~t = 0$ $t=1$ The slope of the tangent line is $2(1)^2 = 2$ We can find the equation of the tangent line: $y-3 = 2(x-1)$ $y = 2x+1$ (b) $x = 1+ln~t$ $ln~t = x-1$ $t = e^{x-1}$ We can replace $t$ in the expression for $y$: $y = t^2+2$ $y = (e^{x-1})^2+2$ $y = e^{2x-2}+2$ We can find $\frac{dy}{dx}$: $\frac{dy}{dx} = 2e^{2x-2}$ When $x=1$: $x = 1+ln~t = 1$ $ln~t = 0$ $t=1$ The slope of the tangent line is $2e^{2(1)-2} = 2$ We can find the equation of the tangent line: $y-3 = 2(x-1)$ $y = 2x+1$
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