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Chapter 12: Q 67. (page 965)

Prove that $\nabla f = 0$ if $f$ is the constant function $f (x, y) = c$ .

Short Answer

Expert verified

Solve for $\nabla f = i \frac{\partial f}{\partial x} + j \frac{\partial f}{\partial y}$ to prove the above result.

Step by step solution

01

Given Information

It is given that $f (x, y) = c$ .

02

Simplification

Using $\nabla f = i \frac{\partial f}{\partial x} + j \frac{\partial f}{\partial y}$

$\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$

Hence

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

$\nabla f (x, y) = i (0) + j (0)$

role="math" localid="1653834651698" $\nabla f (x, y) = 0$

Therefore $\nabla f (x, y) = 0$

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Most popular questions from this chapter

In Exercises, find the maximum and minimum of the function f subject to the given constraint. In each case explain why the maximum and minimum must both exist.

$f (x, y) = x + y when x^{2} + 4 y^{2} = 16$

Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors.

$f (x, y) = \frac{x y}{x^{2} + y^{2}}, P = (- 2, 1), v = <\frac{3}{5}, - \frac{4}{5}>$

Let $f (x, y)$ be a differentiable function such that $\nabla f (x, y) \neq 0$ for every point in the domain of f, and let $R$ be a closed, bounded subset of role="math" localid="1649887954022" $ℝ^{2} .$ Explain why the maximum and minimum of f restricted to $R$ occur on the boundary ofrole="math" localid="1649888770915" $R .$

Evaluate the following limits, or explain why the limit does not exist.

$\lim_{(x, y) \to (1, 2)} \frac{x^{2} + y^{2}}{x^{2} - y^{2}}$

Explain the steps you would take to find the extrema of a function of two variables $f (x, y),$ if $(x, y)$ is a point in a triangle role="math" localid="1649884242530" $T$ in the xy-plane.

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