## Calculus 8th Edition

Given; $|r(t)|=1$ for all the value of $t$ $|r(t)|$ can be written as; $|r(t)|=r(t).r(t)$ Thus, $r(t).r(t)=1$ Let us find the derivative. $0=|r(t)|=r'(t).r(t)+r(t).r'(t)$ $2r'(t).r(t)=0$ This implies that $r'(t).r(t)=0$ Therefore, $r'(t)$ is orthogonal to $r(t)$ for any value of $t$. Hence, the given statement is true.