Q. 37 In Exercises 35-38, find the dir... [FREE SOLUTION]

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Chapter 12: Q. 37 (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 $v$ .
$f (x, y) = \sqrt{\frac{y}{x}} at P = (1, 16), v = 鉄� 2, 鈭� 1 鉄�$

Short Answer

Expert verified

The directional derivative function is $\frac{鈭� 33}{40} \sqrt{5}$

Step by step solution

Given data

$f (x, y) = \sqrt{\frac{y}{x}}$

$P = (x_{0}, y_{0}) = (1, 16) and v = (2, 鈭� 1)$

Solution

Therefore

$鈭� v 鈭� = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$

$u = (伪, 尾) = (\frac{2 \sqrt{5}}{5}, \frac{鈭� \sqrt{5}}{5})$

Directional derivative of function is given by

$鈭� f (P) 鈰� u = 鈭� f (1, 16) 鈰� u = ({\frac{d f}{d x}|}_{(1, 16)} i + {\frac{d f}{d y}|}_{(1, 16)} j) 鈰� (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

$= [{(\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x})]}_{(1, 16)} i + {(\frac{1}{2 \sqrt{\frac{y}{x}} \frac{d}{d x} (\frac{y}{x})})}_{(1, 16)}^{j} 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

role="math" $= [{(\frac{\frac{鈭� y}{x^{2}}}{2 \sqrt{\frac{y}{x}}})}_{(1, 16)} i + {(\frac{\frac{1}{x}}{2 \sqrt{\frac{y}{x}}})}_{(1, 16)} j] 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

role="math" $= [(\frac{\frac{鈭� 16}{1^{2}}}{2 \sqrt{\frac{16}{1}}}) i + (\frac{\frac{1}{1}}{2 \sqrt{\frac{16}{1}}}) j] 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

$= [(\frac{鈭� 16}{8}) i + (\frac{1}{8}) j] 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

$= (鈭� 2 i + \frac{1}{8} j) 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

Step 3

$= (\frac{鈭� 2 脳 2 \sqrt{5}}{5} 鈭� \frac{\sqrt{5}}{8 脳 5})$

localid="1650377777679" $= \frac{\sqrt{5}}{5} (鈭� 4 鈭� \frac{1}{8})$

$= \frac{\sqrt{5}}{5} (鈭� \frac{33}{8})$

$鈭� f (P) 鈰� u = \frac{鈭� 33}{40} \sqrt{5}$

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91影视

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 $v$ .
$f (x, y) = \sqrt{\frac{y}{x}} at P = (1, 16), v = 鉄� 2, 鈭� 1 鉄�$

Short Answer

Step by step solution

Given data

Solution

Step 3

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91影视

In Exercises 35-38, find the directional derivative of the givenfunction at the specified point Pand in the direction of thegiven vector v.f(x,y)=yxatP=(1,16),v=鉄�2,鈭�1鉄�

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

Pure Maths

Calculus

Logic and Functions

Mechanics Maths

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

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 $v$ .
$f (x, y) = \sqrt{\frac{y}{x}} at P = (1, 16), v = 鉄� 2, 鈭� 1 鉄�$