Answer:
Find a unit vector that is parallel to the line tangent to the parabola y = x2 at the point (2, 4).
Summary:
The unit vector that is parallel to the line tangent to the parabola y = x2 at the point (2, 4) is ±(i + 4j)/√17.
Explanation:
Find a unit vector that is parallel to the line tangent to the parabola y = x2 at the point (2, 4).
Solution:
Given parabola y = x2
Point (2, 4)
The slope of the tangent line to the parabola at (2,4) can be written as
(dy/dx) at (2,4) = 2x at (2,4) =4
So, any line parallel to the tangent line has slope ‘4’
Let us assume the unit vector is ±(i + 4j)
The length of the vector is √(12 + 42) = √17
So, the required unit vectors are ±(i + 4j)/√17