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.

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Question

Find a unit vector in the direction in which

f increases most rapidly at P and give the rate of change 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.

f (x, y, z) = x²z e^y + xz²; P(1, ln 2, 2).

Explanation:

The function, z = f(x, y,z), increases most rapidly at (a, b,c) in the

direction of the gradient and decreases

most rapidly in the opposite direction

Given that

F=x²ze^y+xz² at P(1, In2, 2)

1. F increases most rapidly in the positive direction of ∇f

∇f= df/dx i + df/dy j +df/dz k

∇f=(2xze^y+z²)i + (x²ze^y) j + (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= (2×1×2×e^In2+2²)i +(1²×2×e^In2)j +(1²e^In2+2×1×2)

∇f=12i + 4j + 6k

Then, unit vector

V= ∇f/|∇f|

Then, |∇f|= √ 12²+4²+6²

|∇f|= 14

Then,

Unit vector

V=(12i+4j+6k)/14

V=6/7 i + 2/7 j + 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

2. F increases most rapidly in the positive direction of -∇f

∇f=- (df/dx i + df/dy j +df/dz k)

∇f=-(2xze^y+z²)i - (x²ze^y) j - (x²e^y + 2xz)k

At the point P(1, In2, 2)

Then,

∇f= -(2×1×2×e^In2+2²)i -(1²×2×e^In2)j -(1²e^In2+2×1×2)

∇f=-12i -4j - 6k

Then, unit vector

V= -∇f/|∇f|

Then, |∇f|= √ 12²+4²+6²

|∇f|= 14

Then,

Unit vector

V=-(12i+4j+6k)/14

V= - 6/7 i - 2/7 j - 3/7 k

This is the increasing unit vector

The rate of change of f at point P is.

|∇f|= √ 12²+4²+6²

|∇f|= 14

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

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

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)

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.