Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 13 - Section 13.4 - Motion in Space: Velocity and Acceleration - 13.4 Exercise - Page 879: 37



Work Step by Step

Formula to calculate the tangential acceleration component is: $a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}$ ....(1) and Formula to calculate the normal acceleration component is: $a_N=\dfrac{r'(t) \times r''(t)}{|r'(t)|}$ ....(2) Thus, $r'(t)= 2ti+3t^2 j$ and $r''(t)=2i+6tj$ $|r'(t)|=\sqrt{(2t)^2+(3t^2)}=t\sqrt{4+9t^2}$ $r'(t) \cdot r''(t)=[2ti+3t^2 j] \cdot [2i+6tj]=4+18t^2$ and $r'(t) \times r''(t)=[2ti+3t^2 j] \times [2i+6tj]=6t^2 k$ From equation (1), we have $a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{4+18t^2}{\sqrt{4+9t^2}}$ From equation (2), we have $a_N=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|}=\dfrac{6t^2}{\sqrt{4+9t^2}}$
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.