## Calculus 8th Edition

Published by Cengage

# Chapter 13 - Vector Functions - 13.4 Motion in Space: Velocity and Acceleration - 13.4 Exercises - Page 919: 37

#### Answer

$\dfrac{4+18t^2}{\sqrt{4+9t^2}},\dfrac{6t^2}{\sqrt{4+9t^2}}$

#### Work Step by Step

Tangential acceleration component can be calculated as: $a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}$ Normal acceleration component can be calculated as: $a_N=\dfrac{r'(t) \times r''(t)}{|r'(t)|}$ From the question, $r'(t)= 2ti+3t^2 j$ and $r''(t)=2i+6tj$ This yields, $|r'(t)|=\sqrt{(2t)^2+(3t^2)}=t\sqrt{4+9t^2}$ Also, $r'(t) \cdot r''(t)=[2ti+3t^2 j] \cdot [2i+6tj]=4+18t^2$ Also, $r'(t) \times r''(t)=[2ti+3t^2 j] \times [2i+6tj]=6t^2 k$ Plug all the calculated in their respective formulas. Therefore, $a_T=\dfrac{r'(t) \cdot r''(t)}{|r'(t)|}=\dfrac{4+18t^2}{\sqrt{4+9t^2}}$ Plug all the calculated in their respective formulas. $a_N=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|}=\dfrac{6t^2}{\sqrt{4+9t^2}}$

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