## Thomas' Calculus 13th Edition

$|r|$ is also constant.
As we are given that $r \dfrac{dr}{dt}=0 \\ \implies (2r) \dfrac{dr}{dt}=0 \\ \implies r \dfrac{dr}{dt}+r \dfrac{dr}{dt}=0$ Thus, we have $\dfrac{d(r\cdot r)}{dt}=0\\ \implies \dfrac{d(|r^2|)}{dt}=0$ This implies that $|r^2|$ is constant . Hence, $|r|$ is also constant.