(3) This problem will walk you through the steps of finding a directional derivative through thepoint (−1, 12) in the direction ~a = 〈6, 8〉 for the function g(x, y) = x4y.(a) Our goal is to find a directional derivative in the direction of ~a. To do this, we first needto find a unit vector ~u pointing in this direction.~u =(b) Next, we need to find the gradient vector for g, which contains the information about therate of change in the x-direction and the rate of change in the y-direction.∇g(x, y) = 〈gx, gy〉=(c) Since we want the directional derivative through the point (−1, 12), we need to know thegradient vector specific to this point:∇g(−1, 12) =(d) Lastly, we can assemble the directional derivative using the dot product of the gradientvector and the unit vector.D~u(−1, 12) = ∇g(−1, 12) ·~u =please write you process cleary (3) This problem will walk you through the steps of finding a directional derivative through thepoint (–1, ) in the direction ä = (6, 8) for the function g(x, y) = r*y.(a) Our goal is to find a directional derivative in the direction of a. To do this, we first needto find a unit vector i pointing in this direction.i =(b) Next, we need to find the gradient vector for g, which contains the information about therate of change in the r-direction and the rate of change in the y-direction.Vg(r, y) = (92, 9y) =%3D(c) Since we want the directional derivative through the point (-1,), we need to know thegradient vector specific to this point:Vg(-1, ¿) =,(d) Lastly, we can assemble the directional derivative using the dot product of the gradientvector and the unit vector.DE(-1, ) = Vg(-1, ) · i =

Answered: Image /qna-images/answer/9414b231-af92-45a4-8cb3-3a53af12e7b5.jpg

Answered: (3) This problem will walk you through…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

See similar textbooks

Related questions

Question

(3) This problem will walk you through the steps of finding a directional derivative through the
point (−1, 1
2) in the direction ~a = 〈6, 8〉 for the function g(x, y) = x4y.
(a) Our goal is to find a directional derivative in the direction of ~a. To do this, we first need
to find a unit vector ~u pointing in this direction.
~u =
(b) Next, we need to find the gradient vector for g, which contains the information about the
rate of change in the x-direction and the rate of change in the y-direction.
∇g(x, y) = 〈gx, gy〉=
(c) Since we want the directional derivative through the point (−1, 1
2), we need to know the
gradient vector specific to this point:
∇g(−1, 1
2) =
(d) Lastly, we can assemble the directional derivative using the dot product of the gradient
vector and the unit vector.
D~u(−1, 1
2) = ∇g(−1, 1
2) ·~u =