# How do you find the unit vector, in the first quadrant, that is parallel to the line tangent to the parabola y = x^2 at the point (4, 16)?

Jul 6, 2016

(1/sqrt 65, 8/sqrt 65)

#### Explanation:

The slope of the tangent $\tan \psi$ = y', at x=4, = 2 x, at x=4, =8 > 0..

So, the principal value of $\psi$ is in $\left(0. \frac{\pi}{2}\right)$.

And so,$\cos \psi = \frac{1}{\sqrt{65}} \mathmr{and} \sin \psi = \frac{8}{\sqrt{65}}$.

For the opposite direction, $\psi = \psi + \pi$, and both sin and cos will

be < 0... .

The unit vector for inclination $\psi$ to the x-axis is

$\left(\cos \psi , \sin \psi\right) = \left(\frac{1}{\sqrt{65}} , \frac{8}{\sqrt{65}}\right)$

The unit vector in this direction could be anywhere in the plane.

The reference to first quadrant, for its presence, is not necessary..