68 784
Assignments Done
98,8%
Successfully Done
In January 2019

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) ).

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions