University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 12 - Practice Exercises - Page 673: 5

Answer

$$\dfrac{1}{5}$$

Work Step by Step

The vector formula for curvature can be written as: $k=\dfrac{|v \times a(t)|}{|v^3|} ...(1)$ But $v \times a(t)=25 k$ So, $|v \times a(t)|=25 $ and $|v|=\sqrt {(3)^2+(4)^2}=5$ Thus, equation (1) becomes: $$k=\dfrac{|v \times a(t)|}{|v^3|}=\dfrac{25}{(5)^3}=\dfrac{1}{125}=\dfrac{1}{5}$$
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