Question

Consider the function f(x,y)=3x⁴ −x²y+y² +5 and the point P(−1,3).

a. Find the unit vectors that give the direction of steepest ascent and steepest descent at P.

Answer 1

The unit vectors that give the direction of steepest ascent and steepest descent at point P(-1,3) are (-6/√61, 5/√61) and (6/√61, -5/√61) respectively.

Step-by-step explanation:

To find the unit vectors that give the direction of steepest ascent and steepest descent at point P(-1,3), we first need to calculate the gradient vector at P. The gradient vector is a vector that gives the direction of the steepest ascent. The steepest ascent and descent unit vectors can be found by normalizing the gradient vector.

First, we calculate the gradient vector:

∇f(x, y) = (∂f/∂x, ∂f/∂y) = (12x³ - 2xy, -x² + 2y)

Substituting the coordinates of point P into the gradient vector, we have:

∇f(-1, 3) = (12(-1)³ - 2(-1)(3), -(-1)² + 2(3)) = (-12 + 6, -1 + 6) = (-6, 5)

To find the unit vectors, we divide the gradient vector by its magnitude:

||∇f(-1, 3)|| = √((-6)² + 5²) = √(36 + 25) = √61

So, the unit vectors that give the direction of steepest ascent and steepest descent at point P are:

Steepest Ascent Unit Vector: (-6/√61, 5/√61)

Steepest Descent Unit Vector: (-(-6)/√61, -5/√61) = (6/√61, -5/√61)