Answer
$\phi = 63^{\circ}$
Work Step by Step
$y = x^2$
We can find an expression for $y'$:
$y' = \lim\limits_{h \to 0}\frac{(x+h)^2-x^2}{h}$
$= \lim\limits_{h \to 0}\frac{(x^2+2xh+h^2)-x^2}{h}$
$= \lim\limits_{h \to 0}\frac{2xh+h^2}{h}$
$= \lim\limits_{h \to 0}2x+h$
$= 2x$
The slope of the tangent line is equal to $y'$.
We can find the slope of the tangent line at the point $(1,1)$:
$m = 2x = 2(1) = 2$
The slope of a straight line is $\frac{\Delta y}{\Delta x}$
Then:
$tan ~\phi = \frac{\Delta y}{\Delta x} = 2$
$\phi = tan^{-1}(2)$
$\phi = 63^{\circ}$
