Find the directions in which the function increases and decreases most rapidly at P0. Then find the derivatives of the function in those directions.

Answer and Step-by-step explanation:

SOLUTION:

Given function:

h (x, y, z) = ln (x2 + y2 – 1) + y +6z,  p0(1, 1, 0)

Derivatives of the function:

∂h / ∂x = (1 / x2 + y2 -1). (2x)

          = 2x / x2 + y2 -1

∂h/∂y = (1 / x2 + y2 – 1) (2y + 1)

          = (2y / x2 + y2 -1) + 1

∂h / ∂z = 6

Now compute the partial derivatives:

∂h (1, 1, 0) / ∂x = 2(1) / 1 +1 -1

                            = 2

∂h(1, 1, 0) / ∂y =( 2(1) / 1 +1 -1) + 1

                           = 3

∂h (1, 1, 0) / ∂z = 6

Δh (1,1,0) = ( 2, 3, 6)

The definition of gradient is:

Δf (x, y, z) = (∂f / ∂x , ∂f / ∂y)

Put the coordinate values to arrive at solution:

Umax = Δh (1,1,0) / | Δh (1,1,0) |

      = (2, 3, 6) / |2, 3, 6|

     = (2, 3, 6) / √(22+33+62)

   = (2, 3, 6) / √49

    = (2, 3, 6) / 7

= (2/7 , 3/7, 6/7)

Umin = - U max = ( -2/7, -3/7, -6/7)

Compute direction by finding unit vector:

Du h max = Δh (1,1,0)  . u max

                 = (2, 3, 6) . (2/7, 3/7, 6/7)

                 = 2(2/7) + 3(3/7) + 6(6/7)

                 = 4/7 +  9/7 +36/ 7

                  = 49 / 7

                  = 7

Du h min = -Du h max

                = -7

im trinna find out'

step-by-step explanation: