## University Calculus: Early Transcendentals (3rd Edition)

$|r|$ is constant.
Since, $r \dfrac{dr}{dt}=0$ or, $2r \dfrac{dr}{dt}=0$ or, $r \dfrac{dr}{dt}+r \dfrac{dr}{dt}=0$ This implies $\dfrac{d(r\cdot r)}{dt}=0$ or, $\dfrac{d(|r^2|)}{dt}=0$ This means that $|r^2|$ is constant , so $|r|$ is also constant.