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Chapter 12: Q. 22 (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 $u .$
$f (x, y) = x^{2} 鈭� y^{2} at P = (2, 3), u = \frac{3}{5} i 鈭� \frac{4}{5} j$

Short Answer

Expert verified

The directional derivative of the $f (x, y) = x^{2} 鈭� y^{2}$ function is $\frac{36}{5}$

Step by step solution

01

Given data

The objective to determine the directional derivative function

$f (x, y) = x^{2} 鈭� y^{2}$

$P = (2, 3), u = \frac{3}{5} i 鈭� \frac{4}{5} j$

02

Solution

Consider directional derivative

$D_{n} f (x_{0}, y_{6}) = {Lim}_{h 鈫� 0} 鈦� \frac{f (x_{0} + 伪 h, y_{0} + 尾 h) 鈭� f (x_{0}, y_{0})}{h}$

$D_{n} f (2, 3) = {Lim}_{a 鈫� 0} 鈦� \frac{f (2 + \frac{3}{5} h, 3 鈭� \frac{4}{5} h) 鈭� f (2, 3)}{h}$ Equation $1$

Therefore

$f (2 + \frac{3}{5} h, 3 鈭� \frac{4}{5} h) = {(2 + \frac{3}{5} h)}^{2} 鈭� {(3 鈭� \frac{4}{5} h)}^{2}$

$= (4 + \frac{12}{5} h + \frac{9}{25} h^{2}) 鈭� (9 鈭� \frac{24}{5} h + \frac{16}{25} h^{2})$

$= 鈭� 5 + \frac{36}{5} h 鈭� \frac{7}{25} h^{2}$ Equation $2$

and

$f (x_{0}, y_{0}) = 2^{2} 鈭� 3^{2} = 鈭� 5$ Equation $3$

03

Substitute

Substituting

$D_{N} f (2, 3) = {Lim}_{h 鈫� 0} 鈦� \frac{鈭� 5 + \frac{36}{5} h 鈭� \frac{7}{25} h^{2} + 5}{h}$

$= {Lim}_{h 鈫� 0} 鈦� \frac{36}{5} 鈭� \frac{7}{25} h$

$D_{N} f (2, 3) = \frac{36}{5}$

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Most popular questions from this chapter

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

$f (x, y, z) = \frac{x^{2} y}{z}, P = (0, 3, 鈭� 1)$

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

$\frac{d z}{d t} w h e n z = x^{3} e^{y}, x = \sin t, a n d y = \cos t$

$Let p (x) be a polynomial of degree n, q (y) be any function of y, and f (x, y) = p (x) q (y) . Prove that \frac{{鈭�}^{n + 1} f}{鈭� x^{n + 1}} = 0$

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?

Let T be a triangle with side lengths a, b, and c. The semi-perimeter of T is defined to be $s = \frac{1}{2} (a + b + c)$ Heron鈥檚 formula for the area A of a triangle is

$A = \sqrt{s (s - a) (s - b) (s - c)}$

Use Heron鈥檚 formula and the method of Lagrange multipliers to prove that, for a triangle with perimeter P, the equilateral triangle maximizes the area.

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