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)?

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Answer 1 · 2016-07-06T07:30:37

(1/sqrt 65, 8/sqrt 65)

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 $(0. pi/2)$ .

And so, $cos psi=1/sqrt 65 and sin psi =8/sqrt 65$ .

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

$(cos psi, sin psi)=(1/sqrt 65, 8/sqrt 65)$

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

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