Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 604: 59

Answer

The points are (0,0) and (96,180).

Work Step by Step

Slope of the tangent line= $c'(t)=\frac{(t^{3}-6t)'}{(3t^{2}-2t)'}=\frac{3t^{2}-6}{6t-2}=3$ $\implies 3t^{2}-6=3(6t-2)$ $\implies 3t^{2}-6=18t-6$ $\implies 3t^{2}=18t\implies t^{2}=6t$ $\implies t= 0, 6$ When $t=0$, the point is $(x,y)= (3(0)^{2}-2(0), (0)^{3}-6(0))=(0,0)$ When $t=6$, the point is $(3(6)^{2}-2(6), (6)^{3}-6(6))=(96,180)$
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