Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 198: 40

$r(t)=\sec t -t-1$

Since, the derivative of the function $r'(t)=sec t \tan t-1$ is $r(t)=sec t -t+C$ Here, the graph of the given function passes through $(0,0)$. Thus, we have $r(t)=sec t -t+C$ So, $r(0)=0$ or, $0=\sec(0)-(0)+C \implies C=-1$ Hence, $r(t)=\sec t -t-1$