Answer
(55, 0)
Work Step by Step
Slope of the tangent line= $c'(t)=\frac{(t^{2}-8t)'}{(t^{2}-9)'}=\frac{2t-8}{2t}=\frac{1}{2}$
$\implies 2t-8=t$
$\implies t=8$
When $t=8$, the point is $ ((8)^{2}-9, (8)^{2}-8(8))=(55,0)$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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: 63
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
