Question

Consider the following equation. f(x, y) = sin(4x + 3y), P(−6, 8), u = 1 2 3 i − j (a) Find the gradient of f. ∇f(x, y) = (b) Evaluate the gradient at the point P. ∇f(−6, 8) = (c) Find the rate of change of f at P in the direction of the vector u. Duf(−6, 8) =

Answer 1

Explanation:

a) Find the gradient

`\\abla f(x,y) =f_x(x,y)\bold{i} +f_y (x,y)\bold{j}`

`Also, sin(4x+3y)= sin(4x)cos(3y)+sin(3y)cos (4x)`

`f_x(x,y) = 4cos(4x)cos(3y)-4sin(3y)sin(4x) = 4 cos(4x-3y)`

`f_y(x,y) = -3sin(4x)sin(3y)+3cos(3y)cos(4x)=3cos(4x-3y)`

Hence

`\\abla f(x,y) = 4cos(4x-3y)\bold{i}+3cos(4x-3y)\bold{j}`

b) At point (-6, 8) just replace the values in gradient to find it out. You can do it.

c) directional derivative in the direction u.

`D_uf(x,y)=\\abla f(x,y). u`

I stuck with your 1 2 3 i -j . What does 1 2 3 mean? is it not 123 or 1 2/3 or else?