Consider the function F(x, y)=exp(x2/6−y2) and the point P(3,−3).

(a) Find the unit vectors that give the direction of steepest ascent and steepest descent at P.
(b) Find a vector that points in a direction of no change in the function at P.

Step-by-step explanation:

We find the unit vector in the direction of P, we are going to find the Grad of the function ∇F

Then, ∇F is defined as

∇F= ifx + jfy + kfz

fx, ft and fz are partial derivatives with respect to x (taking y and z as constant), y (taking x and z as constant ) and z ( taking x and y as constant) respectively.

Given that F(x, y)=exp(x^2/6−y2)

Then Fx = (x/3) exp(x^2/6-y^2)

Then Fx at x=3 and y=-3

Fx=(3/3)exp(3^2/6-(-3)^2)

Fx=exp(9/6-9)

Fx=exp(-15/2)

Fx=5.53×10^-4

Also Fy= 2yexp(x^2/6-y^2)

Then Fy at x=3 and y=-3

Fy=2(-3)exp(3^2/6-(-3)^2)

Fy=-6exp(9/6-9)

Fy=-6exp(-15/2)

Fy=-3.32×10^-3

And Fz=0

There is no function of z, then we will discard it

Therefore

∇F=5.53E-4 i - 3.32E-3 j

Then unit vector is given as

Vector / magnitude of vector

Magnitude of ∇F

|∇F|=√( 5.53E-4)^2 +(-3.32E-3)^2

|∇F|=3.364×10^-3

Then,

Unit vector = ∇F/|∇F|

Unit vector =(5.53E-4 i - 3.32E-3 j) / 3.364×10^-3

Unit vector= 0.16i - 0.99j

b. The direction of no change is a unit vector orthogonal to u. Noting that the unit vectors orthogonal to a unit vector (a, b) are (−b, a) and (b, −a), we have that a direction of no change at the point

Then, the unit vector orthogonal to 0.16i - 0.99j is -0.16i + 0.99j

The answer is -016i+0.99j

ok

step-by-step explanation: