Question

Find a vector (in 2 space) of length 2 that is parallel to the curve y=x^(2/3) at the point (-8,4)

Answer 1

The vector v should be parallel to the tangent vector tothe curve at point A.

Let us find that tangent vector. y' = 2/3 *x^(2/3-1) = 2/3 * x^(-1/3).
Hence at x=-8 we have that y'(-8) = 2/3 *8^(-1/3) = 2/3 * 1/2 = 1/3
Thus the tangent vector to the curve at point A is w = (1, 1/3)
Its length is
|w| =sqrt(1+1/3^2) = sqrt(1+1/9) = sqrt(10/9) = sqrt(10)/3.
We should multiply this vector by a number
t = 2 / (sqrt(10)/3)= 6/sqrt(10)
then its length will be equal to 2:
So
v = 6/sqrt(10) * w
= (6, 6/3) /sqrt(10)
= (6/sqrt(10), 2/sqrt(10) ).