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Chapter 12: Q. 36 (page 953)

In Exercises $35 - 38$ , find the directional derivative of the given
function at the specified point $P$ and in the direction of the
given vector $v$ .
$f (x, y) = \frac{x}{y^{2}} at P = (9, 鈭� 3), v = 2 i + 7 j$

Short Answer

Expert verified

The directional derivative function is $\frac{44}{9 \sqrt{53}}$

Step by step solution

01

Given data

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

$P = (x_{0}, y_{0}) = (9, 鈭� 3) and v = 2 i + 7 j$

02

Solution

Therefore

$鈭� v 鈭� = \sqrt{2^{2} + 7^{2}} = \sqrt{53}$

$u = (伪, 尾) = (\frac{2}{\sqrt{53}}, \frac{7}{\sqrt{53}})$

Direction derivative function of point is given by

$鈭� f (P) 鈰� u = 鈭� f (9, 鈭� 3) 鈰� u = ({\frac{d f}{d x}|}_{(9, 鈭� 3)} i + {\frac{d f}{d y}|}_{(9, 鈭� 3)} j) 鈰� (\frac{2}{\sqrt{53}} i + \frac{7}{\sqrt{53}} j)$ $= [{(\frac{1}{y^{2}})}_{(9, 鈭� 3)} i + {(\frac{鈭� 2 x}{y^{3}})}_{(9, 鈭� 3)} j] 鈰� (\frac{2}{\sqrt{53}} i + \frac{7}{\sqrt{53}} j)$

$= (\frac{1}{9} i + \frac{18}{27} j) 鈰� (\frac{2}{\sqrt{53}} i + \frac{7}{\sqrt{53}} j)$

$= (\frac{1 鈰� 2}{9 鈰� \sqrt{53}} + \frac{18 鈰� 7}{27 鈰� \sqrt{53}})$

$= (\frac{1 脳 2}{9 脳 \sqrt{53}} + \frac{18 脳 7}{27 - \sqrt{53}})$

$= (\frac{2}{9 脳 \sqrt{53}} + \frac{126}{27 脳 \sqrt{53}})$

$= \frac{2}{9 脳 \sqrt{53}} (1 + \frac{63}{3})$

$鈭� (P) 脳 u = \frac{44}{9 \sqrt{53}}$

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