Thread: Local linearization and directional derivatives

Here's another question:

""Let f(x,y)=2(x^2)y-(y^3)+x-4.

a) Find the local linearization of f at (3,1)

b) Compute the directional derivative of f at (3,1) in the direction towards the point (-2,3). At (3,1), is the function f increasing or decreasing in the direction towards the point (-2,3)? Give a reason for your answer.

c) What direction, given as a unit vector, gives the maximum instantaneous rate of change in the function at (3,1)? In what direction, given as a unit vector, is the instantaneous rate of change at (3,1) equal to 0?"

My work:

a) Local linearization: f(x,y) + fx(x,y)(x-a) + fy(x,y)(y-b)

fx(x,y)= 4xy+1 ; fx(3,1) = 13
fy(x,y) = 2x^2-3y^2 ; fy(3,1) = 15

So, the local linearization is: f(x,y) = 16 + 13(x-3) + 15(y-1)

b) u = (-2,3) ; u = 1/sqrt(2^2+3^2) = -2/sqrt(13), 3/sqrt(13) or -1/sqrt(13)

Directional derivative: (-2/sqrt(13))(13) + (3/sqrt(13))(15) = -3.05
So, the function in the direction of (3,1) is decreasing, since the rate of change is negative.

c) gradf(3,1) = 13i+15j = (13, 15) [From what I understand, the greatest increase of instantaneous change is in the direction of the gradient, so I'm assuming this is what's being looked for in terms of an answer.]

For the rate to be zero: gradf(3,1) (dot product) u = 0, so 13(x-3)+15(y-1)=0.

13x+15y=54, so anywhere on this plane is where the instantaneous rate is 0.

Did I make any mistakes?

Originally Posted by Rumor

Here's another question:

""Let f(x,y)=2(x^2)y-(y^3)+x-4.

a) Find the local linearization of f at (3,1)

b) Compute the directional derivative of f at (3,1) in the direction towards the point (-2,3). At (3,1), is the function f increasing or decreasing in the direction towards the point (-2,3)? Give a reason for your answer.

c) What direction, given as a unit vector, gives the maximum instantaneous rate of change in the function at (3,1)? In what direction, given as a unit vector, is the instantaneous rate of change at (3,1) equal to 0?"

My work:

a) Local linearization: f(x,y) + fx(x,y)(x-a) + fy(x,y)(y-b)

fx(x,y)= 4xy+1 ; fx(3,1) = 13
fy(x,y) = 2x^2-3y^2 ; fy(3,1) = 15

So, the local linearization is: f(x,y) = 16 + 13(x-3) + 15(y-1)

Very good!

b) u = (-2,3) ; u = 1/sqrt(2^2+3^2) = -2/sqrt(13), 3/sqrt(13) or -1/sqrt(13)

No, the problem asked for "the directional derivative of f at (3,1) in the direction towards the point (-2,3)". The vector u is <-2- 3, 1- 3>= <-5, -2>.

Directional derivative: (-2/sqrt(13))(13) + (3/sqrt(13))(15) = -3.05
So, the function in the direction of (3,1) is decreasing, since the rate of change is negative.

c) gradf(3,1) = 13i+15j = (13, 15) [From what I understand, the greatest increase of instantaneous change is in the direction of the gradient, so I'm assuming this is what's being looked for in terms of an answer.]

For the rate to be zero: gradf(3,1) (dot product) u = 0, so 13(x-3)+15(y-1)=0.

13x+15y=54, so anywhere on this plane is where the instantaneous rate is 0.

Did I make any mistakes?

You did not answer either part of (c)! Yes, the greatest increase is in the direction of the gradient. The gradient at (3, 1) is, as you say, <13, 15>. What is a unit vector in that direction?

The rate will be 0 in a direction perpendicular to the gradient but your answer should not be a plane, but a unit vector <a, b> such that 13a+ 15b= 0. (There are actually two correct answers to this, one being the negative of the other.)