Find the unit vectors parallel to the tangent line of the parabola y = x^2 at the point (2, 4).

To find the unit vector(s) parallel to the tangent line at a point on a curve, we first need to find the slope of the tangent line at that point. The slope of the tangent is given by the derivative of the function at that point.

For the parabola \( y = x^2 \):

1. Find the derivative of \( y \) with respect to \( x \) to get the slope of the tangent line: \( \frac{dy}{dx} = 2x \)

2. Evaluate the derivative at the point \( x = 2 \) to find the slope of the tangent line at \( (2, 4) \): \( m = 2(2) = 4 \)

Now we have the slope of the tangent line, we know that the direction of the tangent line at \( (2, 4) \) has a direction vector with components proportional to \( (1, m) \), which would give us \( (1, 4) \) in this case.

3. To find the unit vector, we must normalize this direction vector by dividing it by its magnitude. The magnitude of the vector \( (1, 4) \) is given by: \( \|v\| = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \)

4. Now, to find the unit vector (let's call it \( u \)), divide the vector \( (1, 4) \) by its magnitude: \( u = \frac{1}{\sqrt{17}}(1, 4) \)

Hence, the unit vector parallel to the tangent line at \( (2, 4) \) is: \( u = \left(\frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}}\right) \)

Since a unit vector pointing in the exact opposite direction would also be parallel to the tangent, there is another solution which is the negation of the above vector: \( u' = -\left(\frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}}\right) \)

The two unit vectors parallel to the tangent line are: \( u = \left(\frac{1}{\sqrt{17}}, \frac{4}{\sqrt{17}}\right) \) and \( u' = \left(-\frac{1}{\sqrt{17}}, -\frac{4}{\sqrt{17}}\right) \)