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Chapter 12: Q. 38 (page 961)

Find the gradient of the given functions in Exercises 37鈥�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

01

Step 1. Given Information.

The given function is:

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

02

Step 2. Calculation

The gradient of the given function is:

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

Now find

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

Use these above values in (1) we get

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

03

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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Most popular questions from this chapter

In Exercises 24鈥�32, find the maximum and minimum of the functionf subject to the given constraint. In each case explain why the maximum and minimum must both exist.

$f (x, y, z) = x y z when x^{2} + 4 y^{2} + 16 z^{2} = 64$

Gradients: Find the gradient of the given function, and find the direction in which the function increases most rapidly at the specified point P.

f(x, y ,z) = ln(x + y + z), P = (e, 0, 鈭�1) .

Use Theorem 12.32 to find the indicated derivatives in Exercises 21鈥�26. Express your answers as functions of a single variable. $\frac{d x}{d t} w h e n x = r \cos 胃, r = t^{2} 鈭� 5, a n d 胃 = t^{3} + 1$

$A chain rule for functions of two variables : Let f (x, y) = x^{2} y^{3}, x = s \cos t, and y = s \sin t . Find \frac{鈭� f}{鈭� x} and \frac{鈭� f}{鈭� y} .$

Solve the exact differential equations in Exercises 63鈥�66. $e^{y} + (x e^{y} 鈭� 7) \frac{d y}{d x} = 0$

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