## Calculus 8th Edition

$\theta =0^\circ$ at $(0,0)$ and $\theta \approx 8.1^\circ$ at $(1,1)$
$f(x)=x^2$ and $g(x)=x^3$ $x^2=x^3$ $\implies$ $x=0,1$ $f'(x)=2x$ and $g'(x)=3x^2$ $f'(0)=0$ and $g'(0)=0$ $f'(1)=2$ and $g'(1)=3$ Thus, $a= \lt 1,2\gt$ and $b =\lt 1,3 \gt$ $\theta =0 ^\circ$ at $x=0$ $\theta = cos^{-1}\dfrac{a \cdot b}{|a||b|}=cos^{-1}\dfrac{1(1)+2(3)}{\sqrt {5}\sqrt {10}}$ $\theta = cos^{-1}\dfrac{a \cdot b}{|a||b|}=cos^{-1}\dfrac{7}{\sqrt {50}}\approx 8.13^\circ$ at $x=1$ Hence, $\theta =0^\circ$ at $(0,0)$ and $\theta \approx 8.1^\circ$ at $(1,1)$