Answer:
a) 12/√3
b)<36, 12, 36>/√97.
c) 3
Step-by-step explanation:
(a) Normalize (make length 1) the vector v = <1, 1, -1>.
||v|| = √(1^2 + 1^2 + (-1)^2) = √3.
So, the unit vector u is given by u = v/||v|| = <1, 1, -1>/√3.
To find the rate of change of the potential in the direction of the vector v, you should calculate the
∇f(6, 6, 5) and multiply by u.
=<df/dx, df/dy, df/dz>*u
= <4x-3y+yz, -3x+xz, xy> {at (6, 6, 5)} * u
= <36, 12, 36> · <1, 1, -1>/√3
= 12/√3.
b) The direction which V change more rapidly is in the direction of the gradient at the given point:
u_max = ∇f(6, 6, 5)/||f(6, 6, 5)|| = <36, 12, 36>/√97.
c) The maximum rate of change is |∇F (1,1,−1)|= |(2,−2,1)|= 3.
