Remember, orthogonality is the idea of being "perpendicular". In this example, we want to find the the value of the slope (a.k.a. derivative) of the function at the specified pint so we can determine one point on the line which will be used to construct the unit vectors. In addition, since we are in 2-space, the unit vectors will be negative directions of one another.
y = e2x at x = 0.
1st point can be found by plugging in the value of x into the function --> (x,y) = (0,1)
∂/∂x (e2x) = 2e2x
So the slope at x = 0 --> m = 2
Now we have to find the perpendicular slope which is -1/m ( by definition ). So the unit vectors perpendicular at connected to the point (0,1) and with slope of -1/2.
y = -1/2(x) + b
using the point provided gives --> 1 = -1/2(0) + b --> b = 1
So the line perpendicular to the derivative at point x = 0 is y = -x/2 + 1
Using the unit_vector idea of (x2 + y2 = 1). and the known representation for y we get an equation for x to solve...
x2 + (-x/2 + 1)2 = 1
x2 + x2/4 - x + 1 = 1
5x2/4 -x = 0
x (5x/4 - 1) = 0
x = 0 or 4/5
Since we used the point x=0 for generation of the slope at the point of interest, we should use x = 4/5
This leads to y = -3/5. Futhermore, selecting x = -4/5 leads to y = 3/5 (opposite direction -- our second unit vector perpendicular to the derivative at x = 0)
So v1 = (0,1) -> (4/5, -3/5) and v2 = (0,1) -> (-4/5, 3/5)
Both vectors are unit vectors!