Find a unit vector in the direction in which f increases most rapidly at P and give the rate of chance of f in that direction; find a unit vector in the direction in which f decreases most rapidly at P and give the rate of change of f in that direction.

There's a part of the question missing and it is:

f (x, y) = 4{x (^3) }{y^ (2) }; P (-1,1)

Answer:

A) Unit vector = 4 (3i - 2j) / (√13)

B) The rate of change;

|Δf (1, - 1) | = 4 / (√13)

Explanation:

First of all, f increases rapidly in the positive direction of Δf (x, y)

Now;

[differentiation of the x item alone] to get;

fx (x, y) = 12{x (^2) }{y^ (2) }

So at (1,-1), fx (x, y) = 12

Similarly, [differentiation of the y item alone] to get; fy (x, y) =

8{x (^3) }{y}

At (1,-1), fy (x, y) = - 8

Therefore, Δf (1, - 1) = 12i - 8j

Simplifying this, vector along gradient = 4 (3i - 2j)

Unit vector = 4 (3i - 2j) / (√ (3^2) + (-2^2) = 4 (3i - 2j) / (√13)

Therefore, the rate of change;

|Δf (1, - 1) | = 4 / (√13)