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?
 A: 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.
A: 
The term centrifugal here is not precise physically, $\omega^2 \boldsymbol{r}$ is the net force which is central and repulsive (or outward radially).

Let $S$ be the inertial frame and accelerated frame $S'$:
$$\boldsymbol{a}=
\boldsymbol{a}'+2\boldsymbol{\omega} \times \boldsymbol{v}'+\boldsymbol{\alpha} \times \boldsymbol{r}'+\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r}')$$
The centrifugal force is given by $-m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r}')$ or alternatively,
\begin{align*}
  \mathbf{F}'_{\perp} &= -m\kappa v^2 \mathbf{N} \\
  &=-\frac{m}{a^2b^2 \left(
               \frac{\cosh^2 \omega t}{a^2}+\frac{\sinh^2 \omega t}{b^2}
             \right)^{3/2}} \times
      \omega^{2} \left( a^2\sinh^2 \omega t+b^2\cosh^2 \omega t
                 \right) \mathbf{N} \\
  &= -\frac{m\omega^{2}ab}
           {\sqrt{a^2\sinh^2 \omega t+b^2\cosh^2 \omega t
                 }} \mathbf{N}
\end{align*}
