Centrifugal acceleration in a hyperbola. The following is a question in Tom M. Apostol Calculus book. > A particle moves along a hyperbola according to the equation $$r(t)=a\cosh\omega t \hat{i} + b\sinh\omega t \hat{j},$$where $\omega$ is a constant. Prove that acceleration is centrifugal. I tried to show that $r'\cdot r''=0$ for all $t$. But $r'\cdot r''=(a^2+b^2)\omega^3\sinh \omega t \cosh \omega t$, which isn't zero. What was my mistake?

You have $$\underline{r}=\left(\begin {matrix}a\cosh\omega t\\\b\sinh\omega t\end{matrix}\right)\Rightarrow\underline{\dot{r}}=\left(\begin {matrix}a\omega\sinh\omega t\\\b\omega\cosh\omega t\end{matrix}\right)$$ $$\Rightarrow\underline{\ddot{r}}=\left(\begin {matrix}a\omega^2\cosh\omega t\\\b\omega^2\sinh\omega t\end{matrix}\right)=\omega^2\underline{r}$$

Thus the acceleration acts in a direction along the radius vector and away from the origin, i.e. is centrifugal.