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siniylev [52]
2 years ago
13

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) Ev

aluate 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) =
Mathematics
1 answer:
Gennadij [26K]2 years ago
7 0

Answer:

Step-by-step explanation:

a) Find the gradient

\nabla 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

\nabla 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)=\nabla 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?

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