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Chapter 12: Q. 40 (page 961)

Find the gradient of the given functions in Exercises 37鈥�42.
$f (x, y) = y s e c^{- 1} (\frac{1}{x})$

Short Answer

Expert verified

The gradient of the given function is $(- \frac{y}{\sqrt{1 - x^{2}}}, s e c^{- 1} (\frac{1}{x})) .$

Step by step solution

01

Step 1. Given Information.

The given function is:

$f (x, y) = y s e c^{- 1} (\frac{1}{x})$

02

Step 2. Calculation.

The gradient of the given function is:

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

Now find

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

Use these above values in (1) we get

$鈭� f (x, y) = (- \frac{y}{\sqrt{1 - x^{2}}}, s e c^{- 1} (\frac{1}{x}))$

03

Step 3. Conclusion.

The gradient of the given function is $(- \frac{y}{\sqrt{1 - x^{2}}}, s e c^{- 1} (\frac{1}{x}))$ .

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Most popular questions from this chapter

In Exercises, by considering the function $f (x, y) = x^{2} y$ subject to the constraint $x + y = 0,$ you will explore a situation in which the method of Lagrange multipliers does not provide an extremum of a function.

Why does the method of Lagrange multipliers fail with this function?

Use Theorem 12.32 to find the indicated derivatives in Exercises 21鈥�26. Express your answers as functions of a single variable.

$\frac{d r}{d t} w h e n r = \sqrt{x^{2} + y^{2}}, x = \sqrt{t} a n d y = t^{2}$

$Recall that the volume, V, of a cylinder is given by V (r, h) = 蟺 r^{2} h . Find \frac{鈭� V}{鈭� r} and \frac{鈭� V}{鈭� h} . What are the physical interpretations of these partial derivatives ?$

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.

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

Use Theorem 12.32 to find the indicated derivatives in Exercises 21鈥�26. Express your answers as functions of a single variable.

$\frac{d y}{d t} w h e n y = r \sin 胃, r = t^{3}, a n d 胃 = \sqrt{t}$

See all solutions

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