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: 44


$L'(t)=\tau(t)$ When $\tau(t)=0$, then $L'(t)=\tau(t)=0$. This means that $L$ is constant.

Work Step by Step

Since, $L(t)=mr(t) \times v(t)$ Apply differentiation rule, we get $L'(t)=mr'(t) \times v(t)+mr(t) \times v'(t)$ $\implies L'(t)=mv(t) \times v(t)+mr(t) \times a(t)$ $\implies L'(t)=0+r(t) \times (ma(t))$ $\implies L'(t)=r(t) \times F(t)$ As we know, Torque,$\tau(t)=r(t) \times F(t)$ Hence, $L'(t)=\tau(t)$ When $\tau(t)=0$, then $L'(t)=\tau(t)=0$. This means that $L$ is constant.
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