Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Calculus with Parametric Curves - 10.2 Exercises - Page 695: 6


$\displaystyle y=\frac{2}{\pi}x+1$

Work Step by Step

$\qquad\qquad\qquad\qquad\qquad\qquad$Find the point at which $t = 0$ $x|_{t=0}=e^0\sin(0)=1(0)=0$ $y|_{t=0}=e^{2(0)}=1$ Point: $(0,1)$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad$Now find $\displaystyle \frac{dy}{dx}$ $\displaystyle \frac{dy}{dt}=2e^{2t}$ $\displaystyle \frac{dx}{dt}=e^t[π\cos(πt)]+e^t\sin(πt)=e^t[\pi\cos(\pi t)+\sin(πt)]$ $\displaystyle \frac{dy}{dx}=\frac{2e^{2t}}{e^t[\pi\cos(\pi t)+\sin(πt)]}=\frac{2e^t}{\pi\cos(\pi t)+\sin(πt)}$ $\qquad\qquad\qquad\qquad$Now plug $t=0$ in for $t$ and find the slope of the line: $\displaystyle \frac{dy}{dx}|_{t=0}=\frac{2e^0}{\pi\cos(0)+\sin(0)}=\frac{2(1)}{\pi(1)+0}=\frac{2}{\pi}$ $\qquad\qquad\qquad\qquad$Now use the point-slope form of a line to $\qquad\qquad\qquad\qquad$find the equation tangent to the parametric curve. $y-y_1=m(x_1-x)\quad$ $\displaystyle y-1=\frac{2}{π}(x-0)$ $\displaystyle y=\frac{2}{\pi}x+1$
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