Question

If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k. SOLUTION (a) The gradient of f is ∇f(x, y, z) = fx(x, y, z), fy(x, y, z), fz(x, y, z) = . (b) At (2, 4, 0) we have ∇f(2, 4, 0) = <0,0,8> . The unit vector in the direction of v = i + 3j − k is u = < 1 √11​, 3 √11​,− 1 √11​> . Therefore this equation gives Duf(2, 4, 0) = ∇f(2, 4, 0) · u = 8k · 1 √11​, 3 √11​,− 1 √11​ = − 8 √11​ .

Answer 1

a)
`\\abla f(x,y,z) = sin(yz)\mathbf{i} + xzcos(yz)\mathbf{j} + xy cos(yz)\mathbf{k}`.

b)
`Du_(f)(2,4,0) = -(8)/(√(11))`

Explanation:

Given a function
`f(x,y,z)`, this function has the following gradient:

`\\abla f(x,y,z) = f_(x)(x,y,z)\mathbf{i} + f_(y)(x,y,z)\mathbf{j} + f_(z)(x,y,z)\mathbf{k}`.

(a) find the gradient of f

We have that
`f(x,y,z) = xsin(yz)`. So

`f_(x)(x,y,z) = sin(yz)`

`f_(y)(x,y,z) = xzcos(yz)`

`f_(z)(x,y,z) = xy cos(yz)`.

`\\abla f(x,y,z) = f_(x)(x,y,z)\mathbf{i} + f_(y)(x,y,z)\mathbf{j} + f_(z)(x,y,z)\mathbf{k}`.

`\\abla f(x,y,z) = sin(yz)\mathbf{i} + xzcos(yz)\mathbf{j} + xy cos(yz)\mathbf{k}`

(b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k.

The directional derivate is the scalar product between the gradient at (2,4,0) and the unit vector of v.

We have that:

`\\abla f(x,y,z) = sin(yz)\mathbf{i} + xzcos(yz)\mathbf{j} + xy cos(yz)\mathbf{k}`

`\\abla f(2,4,0) = sin(0)\mathbf{i} + 0cos(0)\mathbf{j} + 8 cos(0)\mathbf{k}`.

`\\abla f(2,4,0) = 0i+0j+8k=(0,0,8)`

The vector is
`v = i + 3j - k = (1,3,-1)`

To use v as an unitary vector, we divide each component of v by the norm of v.

`|v| = \sqrt{1^(2) + 3^(2) + (-1)^(2)} = √(11)`

So

`v_(u) = ((1)/(√(11)), (3)/(√(11)), (-1)/(√(11)))`

Now, we can calculate the scalar product that is the directional derivative.

`Du_(f)(2,4,0) = (0,0,8).((1)/(√(11)), (3)/(√(11)), (-1)/(√(11))) = -(8)/(√(11))`