Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 55

The angle of inclination of the tangent line $\ell$ to the parabola $y=x^2$ at the point $(1,1)$
is the angle $\phi$ that $\ell$ makes with the positive direction of the $x$-axis. Calculate $\phi$
correct to the nearest degree.





$
\begin{equation}
\begin{aligned}
f'(x) &= \lim\limits_{h \to 0} \frac{(x+h)^2-(x)^2}{h}\\
f'(x) &= \lim\limits_{h \to 0} \frac{\cancel{x^2}+2xh+h^2-\cancel{x^2}}{h}\\
f'(x) &= \lim\limits_{h \to 0} \frac{\cancel{h}(2x+h)}{\cancel{h}}\\
f'(x) &= \lim\limits_{h \to 0} (2x+h)\\
f'(x) &= 2x+0\\
f'(x) &= 2x
\end{aligned}
\end{equation}
$


@ point $(1,1)$
$f'(x) = 2(1) = 2$
$f'(x) = \text{slope} = 2$

Based from the graph, we can compute the angle $\phi$ by taking the tangent of the slope of the line


$
\begin{equation}
\begin{aligned}
\tan \phi &= 2\\
\phi &= \tan^{-1}[2]\\
\phi &= 63.43^\circ
\end{aligned}
\end{equation}
$