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

$r(t)=sec t -t-1$
Here, the derivative is given as: $r'(t)=sec t \tan t-1$, thus, the general form of the function is: $r(t)=sec t -t+C$ The graph of the given function passes through $(0,0)$ Therefore, $f(0)=0 \implies 0=sec(0)-0+C$ Thus, $C=-1$ Hence, $r(t)=sec t -t-1$