Q. 39 Find the gradient of the given f... [FREE SOLUTION]

Chapter 12: Q. 39 (page 961)

Find the gradient of the given functions in Exercises 37鈥�42.
$f (x, y) = \sqrt{x^{2} + y^{2}}$

Short Answer

Expert verified

The gradient of the given function is $鈭� f (x, y) = (\frac{x}{\sqrt{x^{2} + y^{2}}}, \frac{y}{\sqrt{x^{2} + y^{2}}})$ .

Step by step solution

Step 1. Given Information.

The given function is:

$f (x, y) = \sqrt{x^{2} + y^{2}}$

Step 2. Calculation.

The gradient of the given function is:

$z = f (x, y) = \sqrt{x^{2} + y^{2}} 鈭� f (x, y) = [f_{x} (x, y), f_{y} (x, y)] - - - - - - - (1)$

Now find

$f_{x} = \frac{鈭� f (x, y)}{鈭� x} = \frac{1 路 2 x}{2 \sqrt{x^{2} + y^{2}}} = \frac{x}{\sqrt{x^{2} + y^{2}}} f_{y} = \frac{鈭� f (x, y)}{鈭� y} = \frac{1 路 2 y}{2 \sqrt{x^{2} + y^{2}}} = \frac{y}{\sqrt{x^{2} + y^{2}}}$

Use these above values in (1) we get

$鈭� f (x, y) = (\frac{x}{\sqrt{x^{2} + y^{2}}}, \frac{y}{\sqrt{x^{2} + y^{2}}})$

Step 3. Conclusion.

The gradient of the given function is $鈭� f (x, y) = (\frac{x}{\sqrt{x^{2} + y^{2}}}, \frac{y}{\sqrt{x^{2} + y^{2}}})$ .

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91影视

Find the gradient of the given functions in Exercises 37鈥�42.
$f (x, y) = \sqrt{x^{2} + y^{2}}$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Calculation.

Step 3. Conclusion.

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91影视

Find the gradient of the given functions in Exercises 37鈥�42.fx,y=x2+y2

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Calculation.

Step 3. Conclusion.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Statistics

Probability and Statistics

Mechanics Maths

Pure Maths

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Find the gradient of the given functions in Exercises 37鈥�42.
$f (x, y) = \sqrt{x^{2} + y^{2}}$