Angle of intersection

Recall

Recall that the graph of a linear function $f(x)=ax+b$ is a straight line with slope $a$ . The (smaller) angle $\alpha$ between this straight line and the $x$ -axis is the arc tangent of the slope: $\alpha = \arctan(a)$ . This is a direct consequence of the trigonometric relations (SOHCAHTOA) for the slope triangle, as is shown in the figure below.

Indeed, we have seen

Equation 1

\tan(\alpha)=\frac{O}{A}=\frac{\Delta y}{\Delta x} = a \implies \alpha=\arctan(a)

Convert slope to angles.

Also remember, that:

The arc tangent is also called inverse tangent, written $\tan^{-1}$ .
If the calculator is in degree mode, the angle $\alpha=\arctan(a)$ is in degrees; if the calculator is in radian mode, the angle $\alpha=\arctan(a)$ is in radians.
For a positive slope $a$ , $\alpha=\arctan(a)$ is positive (below, left); for a negative slope $a$ , $\alpha=\arctan(a)$ is negative (below, right). A negative angle means you measure downwards or clockwise.

Exercise 1

In the left figure above, assume that $\Delta x=2$ and $\Delta y=2.8$ . In the right figure above, assume that $\Delta x=2$ and $\Delta y=-3.2$ . Calculate the angle $\alpha$ for both cases.
Consider the two linear functions $f(x)=0.5x+1$ and $g(x)=-1.5x+3$ . Draw the two graphs, and calculate their smallest angle of intersection with the $x$ -axis. Also, indicate these angles on the picture.

Solution

Left: $a=\frac{\Delta y}{\Delta x}=\frac{2.8}{2}=1.4 \implies \alpha=\arctan(1.4)=54.46^\circ$ . Right: Note that $\Delta y<0$ because we go down on the slope triangle. $a=\frac{\Delta y}{\Delta x}=\frac{-3.2}{2}=-1.6 \implies \alpha=\arctan(-1.6)=-58^\circ$ . Thus the angle is $58^\circ$ , measured downwards or clockwise.
Figures are not shown. $\alpha_f=\arctan(0.5)=26.57^\circ$ and $\alpha_g=\arctan(-1.5)=-56.31^\circ$ (thus the angle is $56.31^\circ$ , measured downwards or clockwise).

Definition 1

Consider a function $f$ that intersects the $x$ -axis at some point $x$ . The angle between the graph of $f$ and the $x$ -axis at this point is defined as the angle between the tangent to the graph of $f$ at this point and the $x$ -axis. We call this angle the angle between the graph and the x-axis.

Theorem 1

Consider a function $f$ and the tangent to the graph of $f$ at $x=u$ . The angle $\alpha$ between the tangent and the $x$ -axis is

Equation 2

\alpha = \arctan(f^\prime(u))

Angle between the graph and the x-axis

Proof

The proof should be clear. The tangent has the slope $a=f^\prime(u)$ .

Exercise 2

The graph above is given by the function $f(x)=x^2-1$ . Find both angles of intersection of the graph with the $x$ -axis.

Solution

Find the $x$ -intercepts: $f(x)=0 \rightarrow x^2-1=0 \rightarrow x_1=-1$ and $x_2=1$ .
The derivative for calculating the slope of the tangent at $x$ is $f'(x)=2x$ . Thus the slope of the tangent at $x_1=-1$ is $a_1=f'(-1)=-2$ , and the slope of the tangent at $x_2=1$ is $a_2=f'(1)=2$ .
The angle at $x_1$ is $\alpha_1 =\arctan(-2)=\underline{-63.44^\circ}$ . Thus the angle is $63.44^\circ$ if measured clockwise.
The angle at $x_2$ is $\alpha_2 =\arctan(2)=\underline{63.44^\circ}$ .

Definition 2

Consider two functions $f$ and $g$ whose graphs intersect at $x=u$ (see figure below). The angle between graphs is defined as the (smaller) angle between the tangents to the graphs of $f$ and $g$ at $x=u$ .

Recipe 1: Finding the angle of intersection

To determine $\alpha$ , it is helpful to first find the angles $\alpha_f$ and $\alpha_g$ between the $x$ -axis and the tangents (see figure below).

Exercise 3

Assume that in the figure above, the slopes of the tangents $t_f$ and $t_g$ are given by $a_f=2.5$ and $a_g=-1$ . Determine the angle of intersection $\alpha$ .

Solution

We have $\alpha_f = \arctan(a_f)=\arctan(2.5) = 68.19^\circ$ and $\alpha_g = \arctan(a_g)=\arctan(-1) = -45^\circ$ . Thus, $\alpha=180^\circ - 68.19^\circ - 45^\circ=66.8^\circ$ .

Exercise 4

Determine the angles of intersection between the $x$ -axis and the graph of $f$ :
1. $f(x)=5-3x^2$
2. $f(x)=2(x+1)(x-2)(x-3)$
Determine all the angles of intersection between the two graphs $f$ and $g$ :
1. $f(x)=x^2$ and $g(x)=3-2x^2$
2. $f(x)=x^2$ and $g(x)=2x+3$
Consider the function $f(x)=3\sqrt[3]{x}-6$ .
1. Find the angle of intersection with the $x$ -axis.
2. Find all points on the graph where the tangent intersects the $x$ -axis at an angle of $30^\circ$ .

Solution

1. At $x_1=-\sqrt{\frac{5}{3}}$ , the angle is $\alpha_1=82.64^\circ$ , and at $x_2=\sqrt{\frac{5}{3}}$ , the angle is $\alpha_2=-82.64^\circ$ .
2. The derivative is $f'(x)=6x^2-16x+2$ . At $x_1=-1$ , the slope is $a_1=24$ and $\alpha_1=87.61^\circ$ ; at $x_2=2$ , the slope is $a_2=-6$ and $\alpha_2=-80.54^\circ$ ; and at $x_3=3$ , the slope is $a_3=8$ and $\alpha_3=82.87^\circ$ .
1. At $P_1=(1\vert 1)$ , it is $\alpha_f=63.43^\circ, \alpha_g=-75.95^\circ, \alpha=40.91^\circ$ . At $P_2=(-1\vert 1)$ , it is $\alpha=40.91^\circ$ .
2. At $P_1(3\vert 9)$ , it is $\alpha = 17.1^\circ$ , and at $P_2(-1\vert 1)$ , it is $\alpha = 53.13^\circ$ .
1. At $x=8$ , the angle is $\alpha=14.04^\circ$ .
2. Find $x$ such that $f'(x)=x^{-2/3}=\tan(30^\circ) \implies x \approx 2.2795 \implies P(2.2795\vert -2.051)$ .