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: 40



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, $r'(t)= i+2e^tj+2e^{2t}k$ and $r''(t)=2e^tj+4e^{2t}k$ $|r'(t)|=\sqrt{(1)^2+(2e^t)^2+(2e^{2t})^2}=1+2e^{2t}$ $r'(t) \cdot r''(t)=[i+2e^tj+2e^{2t}k] \cdot [2e^tj+4e^{2t}k]=4e^{2t}(1+2e^t)$ and $|r'(t) \times r''(t)|=[i+2e^tj+2e^{2t}k] \times [2e^tj+4e^{2t}k]=2e^{t}(1+2e^t)$ From equation (1), we have $a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{4e^{2t}(1+2e^t)}{1+2e^{2t}}=4e^{2t}$ From equation (2), we have $a_N=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|}=\dfrac{2e^{t}(1+2e^t)}{1+2e^{2t}}=2e^{t}$
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