University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 12 - Section 12.1 - Curves in Space and Their Tangents - Exercises - Page 648: 3


$y=\dfrac{2}{9}x^2$; $v(\ln 3)=3i+4j$ and $a(\ln 3)=3i+8j$

Work Step by Step

Since, we have $x=e^t; y=\dfrac{2}{9}e^{2t}$ This can be written as: $y=\dfrac{2}{9}x^2$ Now, velocity $v(t)=r'(t)=e^ti+\dfrac{4}{9}e^{2t}j$ and $v(\ln 3)=3i+4j$ As we know acceleration $a(t)=v'(t)=e^ti+\dfrac{8}{9}e^{2t}j$ and $a(\ln 3)=3i+8j$ Hence, we have $y=\dfrac{2}{9}x^2$; $v(\ln 3)=3i+4j$ and $a(\ln 3)=3i+8j$
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.