Chapter 12: Q. 25 (page 953)

In Exercises 21–28, find the directional derivative of the given
function at the specified point P and in the direction of the
given unit vector u.
$f (x, y) = \sqrt{\frac{y}{x}} at P = (4, 9), u = (- \frac{\sqrt{17}}{17}, - \frac{4 \sqrt{17}}{17})$

Short Answer

Expert verified

The directional derivative of the given

function is $- \frac{7}{816} \sqrt{17}$

Step by step solution

Given data

To find the directional derivative function $f (x, y) = \sqrt{\frac{y}{x}}$

$P = (x_{0}, y_{0}) = (4, 9) u = (α, β) = (- \frac{\sqrt{17}}{17}, - \frac{4 \sqrt{17}}{17})$

Solution

the directional derivative of

point P and in the direction of the given unit vector u given by

$\nabla f (P) \cdot u = \nabla f (4, 9) \cdot u$

$= ({\frac{d f}{d x}|}_{(4, 9)} i + {\frac{d f}{d y}|}_{(4, 9)} j) \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$

$= [{(\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x}))}_{(4, 9)} i + {(\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x}))}_{(4, 9)} j] \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$

$= [(\frac{\frac{- y}{x^{2}}}{{2 \sqrt{\frac{y}{x}})}_{(4, 9)}} i + (\frac{\frac{1}{x}}{{2 \sqrt{\frac{y}{x}})}_{(4, 9)}} j] \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$

$= [(\frac{\frac{- 9}{4^{2}}}{2 \sqrt{\frac{9}{4}}}) i + (\frac{\frac{1}{4}}{2 \sqrt{\frac{9}{4}}})] \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$

$= [(\frac{- \frac{9}{16}}{2 \cdot \frac{3}{2}}) i + (\frac{\frac{1}{4}}{2 \cdot \frac{3}{2}})] \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$

$= (- \frac{9}{48} i + \frac{1}{12} j) \cdot (- \frac{\sqrt{17}}{17} i - \frac{4 \sqrt{17}}{17} j)$

Step 3

$= (\frac{9 \cdot \sqrt{17}}{48 \cdot 17} - \frac{4 \sqrt{17}}{12 \cdot 17})$

$= (\frac{9 \cdot \sqrt{17}}{16 \cdot 3 \cdot 17} - \frac{\sqrt{17}}{3 \cdot 17})$

$= \frac{1}{3 \cdot 17} (\frac{9 \cdot \sqrt{17}}{16} - \frac{\sqrt{17}}{1})$

$= \frac{1}{3 \cdot 17} (\frac{9 \cdot \sqrt{17} - 16 \sqrt{17}}{16})$

$\nabla f (P) \cdot u = - \frac{7}{816} \sqrt{17}$

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In Exercises 21–28, find the directional derivative of the given
function at the specified point P and in the direction of the
given unit vector u.
$f (x, y) = \sqrt{\frac{y}{x}} at P = (4, 9), u = (- \frac{\sqrt{17}}{17}, - \frac{4 \sqrt{17}}{17})$

Short Answer

Step by step solution

Given data

Solution

Step 3

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In Exercises 21–28, find the directional derivative of the givenfunction at the specified point P and in the direction of thegiven unit vector u.f(x,y)=yxatP=(4,9),u=−1717,−41717

Short Answer

Step by step solution

Given data

Solution

Step 3

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Discrete Mathematics

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Probability and Statistics

Study anywhere. Anytime. Across all devices.

In Exercises 21–28, find the directional derivative of the given
function at the specified point P and in the direction of the
given unit vector u.
$f (x, y) = \sqrt{\frac{y}{x}} at P = (4, 9), u = (- \frac{\sqrt{17}}{17}, - \frac{4 \sqrt{17}}{17})$