Chapter 12: Q. 38 (page 961)

Find the gradient of the given functions in Exercises 37–42.
$z = \tan^{- 1} (\frac{y}{x})$

Short Answer

Expert verified

The gradient of the given function is $(- \frac{y}{x^{2} + y^{2}}, \frac{x}{x^{2} + y^{2}})$ .

Step by step solution

Step 1. Given Information.

The given function is:

$z = \tan^{- 1} (\frac{y}{x})$

Step 2. Calculation

The gradient of the given function is:

$z = f (x, y) = \tan^{- 1} (\frac{y}{x}) \nabla f (x, y) = [f_{x} (x, y), f_{y} (x, y)] - - - - - - - (1)$

Now find

$f_{x} = \frac{\partial f (x, y)}{\partial x} = \frac{1 \cdot (- \frac{y}{x^{2}})}{1 + {(\frac{y}{x})}^{2}} = - \frac{y}{x^{2} + y^{2}} f_{y} = \frac{\partial f (x, y)}{\partial y} = \frac{1 \cdot (\frac{1}{x})}{1 + {(\frac{y}{x})}^{2}} = \frac{x}{x^{2} + y^{2}}$

Use these above values in (1) we get

$\nabla f (x, y) = (- \frac{y}{x^{2} + y^{2}}, \frac{x}{x^{2} + y^{2}})$

Step 3. Conclusion.

The gradient of the given function is $(- \frac{y}{x^{2} + y^{2}}, \frac{x}{x^{2} + y^{2}})$ .

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Find the gradient of the given functions in Exercises 37–42.
$z = \tan^{- 1} (\frac{y}{x})$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Calculation

Step 3. Conclusion.

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Find the gradient of the given functions in Exercises 37–42.z=tan-1yx

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

Geometry

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Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Find the gradient of the given functions in Exercises 37–42.
$z = \tan^{- 1} (\frac{y}{x})$