Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.1 - Curves in Space and Their Tangents - Exercises 13.1 - Page 745: 3


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

Work Step by Step

We know that the velocity and acceleration is defined as: velocity is given as: $v(t)=r'(t)$ and acceleration is given as: $a(t)=v'(t)$ In the given problem, we have $x=e^t; y=\dfrac{2}{9}e^{2t}$ Re-write as follows: $y=\dfrac{2}{9}x^2$ $v(t)=r'(t)=e^ti+\dfrac{4}{9}e^{2t}j\\ \implies v(\ln 3)=3i+4j$ Next, we have $a(t)=v'(t)=e^ti+\dfrac{8}{9}e^{2t}j\\ \implies a(\ln 3)=3i+8j$
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