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: 9

Answer

$y=3x+3$ Graph:
1559253328

Work Step by Step

$x=t^{2}-t,y=t^{2}+t+1;(0,3).\quad $ First, find the t for which $x=0,y=3,$ $x=t^{2}-t$ $0=t(t-1)\quad \Rightarrow t=0$ or $t=1.$ When$ t=0,\quad y=0+0+1,$ When $t=1, \quad y=1+1+1=3$ so, t=1. Now, the slope of the tangent at the given point. $\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t+1}{2t-1}$ For $t=1,$ $\displaystyle \frac{dy}{dx}=\frac{2+1}{2-1}=3$ With m=3, P(0,3), an equation of the tangent is $y-y_{1}=m(x-x_{1})$ $y-3=3x$ $y=3x+3$ To graph the curve, build a table of coordinates (x,y) $x=t^{2}-t,\quad y=t^{2}+t+1$ Plot the points and join with a smooth curve. To graph the line, we have (0,3), move 1 unit to the right, go up 3 units (slope=3), and plot the second point on the tangent, (1,6)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.