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Chapter 12: Q 69, (page 965)

Prove that

∇(αf(x,y)+βg(x,y))=α∇f(x,y)+β∇g(x,y)

α,βare constants.

Short Answer

Expert verified

The relation can be proved using∇(f(x,y)+g(x,y))=i∂(f(x,y)+g(x,y))∂x+j∂(f(x,y)+g(x,y))∂y

Step by step solution

01

Given Information

Let the function be

f(x,y),g(x,y)

We need to show that∇(f(x,y)+g(x,y))=∇f(x,y)+∇g(x,y)

02

Simplification

Solving ∇(f(x,y)+g(x,y))=i∂(f(x,y)+g(x,y))∂x+j∂(f(x,y)+g(x,y))∂y, we get

∇(f(x,y)+g(x,y))=i∂(f(x,y))+∂(g(x,y))∂x+j∂(f(x,y))+∂(g(x,y))∂y

∇(f(x,y)+g(x,y))=i∂(f(x,y))∂x+j∂(f(x,y))∂y+i∂(g(x,y))∂x+j∂(g(x,y))∂y

∇(f(x,y)+g(x,y))=i∂(f(x,y))∂x+j∂(f(x,y))∂y+i∂(g(x,y))∂x+j∂(g(x,y))∂y

∇(f(x,y)+g(x,y))=∇f(x,y)+∇g(x,y)

Hence proved.

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