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: 3



Work Step by Step

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$Find the point at which $t = -1$ $x|_{t=-1}=(-1)^3+1=0$ $y|_{t=-1}=(-1)^4+(-1)=0$ Point: $(0,0)$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$Now find $\displaystyle \frac{dy}{dx}$ $dx=3t^2$ $dy=4t^3+1$ $\displaystyle \frac{dy}{dx}=\frac{4t^3+1}{3t^2}$ $\qquad\qquad\qquad\qquad\qquad$Now plug $t=-1$ in for $t$ and find the slope of the line: $\displaystyle \frac{dy}{dx}|_{t=-1}=\frac{-4+1}{3}=\frac{-3}{3}=-1$ $\qquad\qquad\qquad\qquad\qquad$Now use the point-slope form of a line to $\qquad\qquad\qquad\qquad\qquad$find the equation tangent to the parametric curve. $y-y_1=m(x_1-x)\quad$ $y-0=(-1)(x-0)$ $y=-x$
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