Q. 37 In Exercise 35-38, find the dire... [FREE SOLUTION]

Chapter 12: Q. 37 (page 953)

In Exercise $35 - 38$ , find the directional derivative of the given function at the specified point $P$ and in the direction of the given vector $v$ .
$f (x, y) = \sqrt{\frac{y}{x}} a t P = (1, 16), v = (2, 1)$

Short Answer

Expert verified

The directional derivative is $鈭� f (P) . u = \frac{- 33}{40} \sqrt{5}$ .

Step by step solution

Directional derivative.

For a given function $P = (x_{0}, y_{0}) = (1, 16)$ and $v = (2, - 1)$ , we must find the directional derivative $f (x, y) = \sqrt{\frac{y}{x}}$ .

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

Therefore;

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

Directional unit vector.

The function's directional derivative at point $P$ with directional unit vector $u$ is,

localid="1650642585491" $鈭� f (P) 鈰� u = 鈭� f (1, 16) 脳 u = ({\frac{d f}{d x}|}_{(1, 16)} i + {\frac{d f}{d y}|}_{(1, 16)} j) 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

localid="1650642610945" $= [{(\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x}))}_{(1, 16)} i + {(\frac{1}{2 \sqrt{\frac{y}{x}}} \frac{d}{d x} (\frac{y}{x}))}_{(1, 16)} j] 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

localid="1650642627783" $= {[\frac{\frac{鈭� y}{x^{2}}}{2 \sqrt{\frac{y}{x}}})}_{(1, 16)} i + (\frac{\frac{1}{x}}{{2 \sqrt{\frac{y}{x}})}_{(1, 16)}} j] 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

localid="1650642643659" $= [(\frac{\frac{鈭� 16}{1^{2}}}{2 \sqrt{\frac{16}{1}}}) i + (\frac{\frac{1}{1}}{2 \sqrt{\frac{16}{1}}}) j] 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

localid="1650642674743" $= [(\frac{鈭� 16}{8}) i + (\frac{1}{8}) j] 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

localid="1650642705305" $= (鈭� 2 i + \frac{1}{8} j) 脳 (\frac{2 \sqrt{5}}{5} i 鈭� \frac{\sqrt{5}}{5} j)$

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

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

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

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

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

In Exercise $35 - 38$ , find the directional derivative of the given function at the specified point $P$ and in the direction of the given vector $v$ .
$f (x, y) = \sqrt{\frac{y}{x}} a t P = (1, 16), v = (2, 1)$

Short Answer

Step by step solution

Directional derivative.

Directional unit vector.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Theoretical and Mathematical Physics

Probability and Statistics

Geometry

Applied Mathematics

Study anywhere. Anytime. Across all devices.

91影视

In Exercise 35-38, find the directional derivative of the given function at the specified point Pand in the direction of the given vector v.f(x,y)=yxatP=(1,16),v=(2,1)

Short Answer

Step by step solution

Directional derivative.

Directional unit vector.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Theoretical and Mathematical Physics

Probability and Statistics

Geometry

Applied Mathematics

Study anywhere. Anytime. Across all devices.

In Exercise $35 - 38$ , find the directional derivative of the given function at the specified point $P$ and in the direction of the given vector $v$ .
$f (x, y) = \sqrt{\frac{y}{x}} a t P = (1, 16), v = (2, 1)$