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Chapter 12: Q. 24 (page 953)

In Exercises 21鈥�28, find the directional derivative of the given
function at the specified point P and in the direction of the
given unit vector u.
$f (x, y) = \frac{x}{y^{2}} at P = (鈭� 2, 1), u = 鈭� \frac{5}{13} i + \frac{12}{13} j$

Short Answer

Expert verified

The directional derivative of the

$f (x, y) = \frac{x}{y^{2}}$ function is

Step by step solution

01

Given data

The given function is $f (x, y) = \frac{x}{y^{2}}$

$P = (x_{0}, y_{0}) = (鈭� 2, 1) u = (伪 i + 尾 j) = (\frac{鈭� 5}{13} i + \frac{12}{13} j)$

02

Solution

Consider directional derivative

$D_{k} f (x_{e}, y_{e}) = {Lim}_{h 鈫� 0} 鈦� \frac{f (x_{e} + 伪 h + y_{0} + 尾 h) 鈭� f (x_{e +} y_{e})}{h}$

$D_{k} f (鈭� 2, 1) = {Lim}_{k 鈫� 0} 鈦� \frac{f (鈭� 2 鈭� \frac{5}{13} h, 1 + \frac{12}{13} h) 鈭� f (鈭� 2, 1)}{h}$

Therefore

$f (鈭� 2 鈭� \frac{5}{13} h, 1 + \frac{12}{13} h) = \frac{鈭� 2 鈭� \frac{5}{13} h}{{(1 + \frac{12}{13} h)}^{2}}$

$= \frac{鈭� 2 鈭� \frac{5}{13} h}{(1 + \frac{24}{13} h + \frac{144}{169} h^{2})}$

$= \frac{鈭� 26 鈭� 5 h}{(1 + \frac{24}{13} h + \frac{144}{169} h^{2})}$

$= \frac{鈭� 26 鈭� 5 h}{13 鈭� (1 + \frac{24}{13} h + \frac{144}{169} h^{2})}$

03

Step 3

$= \frac{鈭� 26 鈭� 5 h}{13 鈭� (\frac{169 + 312 h + 144 h^{2}}{169})}$

$= \frac{鈭� 338 鈭� 65 h}{169 + 312 h + 144 h^{2}}$

and

$f (x_{0}, y_{0}) = f (鈭� 2, 1) = \frac{鈭� 2}{1^{2}} = 鈭� 2$

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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.

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

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.

$f (x, y, z) = \sqrt{\frac{x y}{z}}, P = (2, 3, 1), v = 鉄� 2, 1, 鈭� 2 鉄�$

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.

$f (x, y, z) = x + y + z when x^{2} + y^{2} + z^{2} = 9$

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.

Show that the only point given by the method of Lagrange multipliers for the function $f (x, y) = x^{2} y$ subject to the constraint $x + y = 0 is (0, 0) .$

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

See all solutions

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