Chapter 12: Q 69, (page 965)

Prove that
$\nabla (α f (x, y) + β g (x, y)) = α \nabla f (x, y) + β \nabla g (x, y)$
$α, β$ are constants.

Short Answer

Expert verified

The relation can be proved using $\nabla (f (x, y) + g (x, y)) = i \frac{\partial (f (x, y) + g (x, y))}{\partial x} + j \frac{\partial (f (x, y) + g (x, y))}{\partial y}$

Step by step solution

Given Information

Let the function be

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

We need to show that $\nabla (f (x, y) + g (x, y)) = \nabla f (x, y) + \nabla g (x, y)$

Simplification

Solving $\nabla (f (x, y) + g (x, y)) = i \frac{\partial (f (x, y) + g (x, y))}{\partial x} + j \frac{\partial (f (x, y) + g (x, y))}{\partial y}$ , we get

$\nabla (f (x, y) + g (x, y)) = i \frac{\partial (f (x, y)) + \partial (g (x, y))}{\partial x} + j \frac{\partial (f (x, y)) + \partial (g (x, y))}{\partial y}$

$\nabla (f (x, y) + g (x, y)) = [i \frac{\partial (f (x, y))}{\partial x} + j \frac{\partial (f (x, y))}{\partial y}] + [i \frac{\partial (g (x, y))}{\partial x} + j \frac{\partial (g (x, y))}{\partial y}]$

$\nabla (f (x, y) + g (x, y)) = \nabla f (x, y) + \nabla g (x, y)$

Hence proved.

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Prove that
$\nabla (α f (x, y) + β g (x, y)) = α \nabla f (x, y) + β \nabla g (x, y)$
$α, β$ are constants.

Short Answer

Step by step solution

Given Information

Simplification

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Prove that∇(αf(x,y)+βg(x,y))=α∇f(x,y)+β∇g(x,y)α,βare constants.

Short Answer

Step by step solution

Given Information

Simplification

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Decision Maths

Geometry

Theoretical and Mathematical Physics

Discrete Mathematics

Calculus

Study anywhere. Anytime. Across all devices.

Prove that
$\nabla (α f (x, y) + β g (x, y)) = α \nabla f (x, y) + β \nabla g (x, y)$
$α, β$ are constants.