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Suppose that over a certain region of space the electrical potential V is given by the following equation. V(x, y, z) = 2x2 − 3xy + xyz (a) Find the rate of change of the potential at P(6, 6, 5) in the direction of the vector v = i + j − k. (b) In which direction does V change most rapidly at P? (c) What is the maximum rate of change at P?

Answer :

debrubs97

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.

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