## Thomas' Calculus 13th Edition

$\dfrac{\pi}{2}$
$v(t)=\dfrac{dr}{dt}=i[e^t(\cos t -\sin t) ]+j [e^t(\sin t +\cos t) ]$ and $a(t)=\dfrac{dv(t)}{dt}=-2 e^t ( \sin t i- \cos t j)$ Let us suppose that $\theta$ be the angle between $r$ and $a(t)$ Thus, $\theta =\cos^{-1} [\dfrac{r \cdot a(t)}{|r| |a(t)|}]$ or, $=arccos [\dfrac{ e^t \cos t \times (-2 e^t \sin t) +(e^t \sin t) (2e^t \cos t)}{\sqrt {(e^t \cos t )^2+(e^t \cos t )^2 \cdot \sqrt {(-2 e^t \sin t)^2 +( e^t \cos t)^2}}}]$ or, $\theta=\dfrac{\pi}{2}$