# Answer to Question #22356 in Vector Calculus for Jon

Question #22356

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

Expert's answer

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) ).

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) ).

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