"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?"
Thanks for your help!
a) To find the local linearization of f at (3,1), we need to use the first-order Taylor expansion. The local linearization is given by:
L(x, y) = f(a, b) + ∇f(a, b) · (x - a, y - b)
where (a, b) is the point of interest, ∇f is the gradient of f, and · denotes the dot product.
In this case, (a, b) = (3, 1). Let's find ∇f:
∂f/∂x = 4xy + 1
∂f/∂y = 2x^2 - 3y^2
Hence, ∇f = (4xy + 1, 2x^2 - 3y^2). Evaluating ∇f at (3, 1):
∇f(3, 1) = (4(3)(1) + 1, 2(3)^2 - 3(1)^2) = (13, 15)
Now we can write the local linearization:
L(x, y) = f(3, 1) + (13, 15) · (x - 3, y - 1)
Substituting in the given values, we have:
L(x, y) = (2(3^2)(1) - (1^3) + 3 - 4) + (13, 15) · (x - 3, y - 1)
Simplifying further:
L(x, y) = 6 - 1 + 3 - 4 + (13x - 39, 15y - 15) = (13x - 37, 15y - 13)
Therefore, the local linearization of f at (3,1) is L(x, y) = 13x - 37 + 15y - 13.
b) The directional derivative of f at (3, 1) in the direction towards the point (-2, 3) can be computed using the gradient of f. The directional derivative is given by the dot product of the gradient and the unit vector in the direction of (-2, 3).
First, let's find the unit vector in the direction towards (-2, 3):
u = (-2, 3) / ||(-2, 3)|| = (-2, 3) / √((-2)^2 + 3^2) = (-2/√13, 3/√13)
The gradient of f at (3, 1) is ∇f(3, 1) = (13, 15), which we found in part (a).
The directional derivative is then given by:
D_uf(3, 1) = ∇f(3, 1) · u = (13, 15) · (-2/√13, 3/√13)
Evaluating this dot product:
D_uf(3, 1) = 13(-2/√13) + 15(3/√13) = 4/√13 = (4√13)/13
The fact that the directional derivative is positive indicates that the function f is increasing in the direction towards the point (-2, 3) at (3, 1).
c) To find the direction that gives the maximum instantaneous rate of change in the function at (3, 1), we need to find the direction of the gradient vector ∇f(3, 1).
From part (a), we found that ∇f(3, 1) = (13, 15). To find the unit vector in this direction, we divide it by its magnitude:
||∇f(3, 1)|| = √(13^2 + 15^2) = √(169 + 225) = √394
The unit vector in the direction of ∇f(3, 1) is:
v = ∇f(3, 1) / ||∇f(3, 1)|| = (13/√394, 15/√394)
Therefore, the direction that gives the maximum instantaneous rate of change in the function at (3, 1) is the unit vector v = (13/√394, 15/√394).
To find the direction in which the instantaneous rate of change at (3, 1) is equal to 0, we are looking for the direction perpendicular to the gradient vector ∇f(3, 1).
The direction perpendicular to ∇f(3, 1) can be found by taking the negative reciprocal of the gradient vector's components and then normalizing:
w = (-15/√394, 13/√394)
Thus, the unit vector w = (-15/√394, 13/√394) gives the direction in which the instantaneous rate of change at (3, 1) is equal to 0.