219k views
1 vote
Find the directions in which the function increases and decreases most rapidly at P0. Then find the derivatives of the function in those directions.

User HackAfro
by
6.5k points

1 Answer

6 votes

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

User Rahul Katariya
by
6.6k points