Q. 51 Find all points where the first-... [FREE SOLUTION]

Chapter 12: Q. 51 (page 954)

Find all points where the first-order partial derivatives of the functions in Exercises 43鈥�54 are continuous. Then use Theorems 12.28 and 12.31 to determine the sets in which the functions are differentiable. $f (x, y) = \sqrt{\frac{y}{x}}$

Short Answer

Expert verified

points where the first-order partial derivatives of the functions in $f (x, y) = \sqrt{\frac{y}{x}}$ is $9 x - y + 6 z = 18$

Step by step solution

Given.

Given $f (x, y) = \sqrt{\frac{y}{x}}$ at point $P = (x_{0}, y_{0}) = (1, 9)$

The equation of line of tangent is

$f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0}) = z - f (x_{0}, y_{0})$

$f_{x} (1, 9) (x - 1) + f_{y} (1, 9) (y - 9) = z - f (1, 9)$ $(1)$

Considering.

$f_{x} (x_{0}, y_{0}) = {\frac{d}{d x} f (x, y)|}_{(x_{0}, y_{0})} = {\frac{d}{d x} \sqrt{\frac{y}{x}}|}_{(x_{0}, y_{0})}$

$f_{x} (x_{0}, y_{0}) = \frac{d}{d x} ({\sqrt{\frac{y}{x})} \frac{d}{d x} (\frac{y}{x})|}_{1}$

$f_{x} (x_{0}, y_{0}) = {\frac{1}{2 \sqrt{\frac{y}{x}}} (- \frac{y}{x^{2}})|}_{(x_{0}, y_{0})}$

$f_{x} (1, 9) = {\frac{1}{2 \sqrt{\frac{y}{x}}} (- \frac{y}{x^{2}})|}_{(1, 9)}$

$f_{x} (1, 9) = \frac{1}{2 \sqrt{\frac{9}{1}}} (- \frac{9}{1^{2}})$

$f_{x} (1, 9) = - \frac{9}{6}$ $(2)$

Considering.

$f_{y} (x_{0}, y_{0}) = {\frac{d}{d y} f (x, y)|}_{(x_{0}, y_{0})} = {\frac{d}{d y} \sqrt{\frac{y}{x}}|}_{(x_{0}, y_{0})}$

$f_{y} (x_{0}, y_{0}) = {\frac{d}{d y} (\sqrt{\frac{y}{x}}) \frac{d}{d y} (\frac{y}{x})|}_{(x_{0}, y_{0})}$

$f_{y} (x_{0}, y_{0}) = {\frac{1}{2 \sqrt{\frac{y}{x}}} 路 \frac{1}{x}|}_{(x_{0}, y_{0})}$

$f_{y} (1, 9) = {\frac{1}{2 x \sqrt{\frac{y}{x}}}|}_{(1, 9)}$

localid="1650642163935" $f_{y} (1, 9) = \frac{1}{2 脳 1 脳 \sqrt{\frac{9}{1}}}$

$f_{y} (1, 9) = \frac{1}{6}$ $(3)$

Equation (4).

$f (x_{0}, y_{0}) = f (1, 9) = \sqrt{\frac{9}{1}}$

$$ f (1, 9) = 3 $$ $(4)$

Substituting (2),(3) and (4) in equation (1)to get.

$- \frac{9}{6} (x - 1) + \frac{1}{6} (y - 9) = z - 3 - \frac{9}{6} x + \frac{9}{6} + \frac{1}{6} y - \frac{9}{6} = z - 3 - \frac{9}{6} x + \frac{1}{6} y - z = - 3$

Multiply by 6 on both sides, get

$- 9 x + y - 6 z = - 18 9 x - y + 6 z = 18$

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91影视

Find all points where the first-order partial derivatives of the functions in Exercises 43鈥�54 are continuous. Then use Theorems 12.28 and 12.31 to determine the sets in which the functions are differentiable. $f (x, y) = \sqrt{\frac{y}{x}}$

Short Answer

Step by step solution

Given.

Considering.

Considering.

Equation (4).

Substituting (2),(3) and (4) in equation (1)to get.

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91影视

Find all points where the first-order partial derivatives of the functions in Exercises 43鈥�54 are continuous. Then use Theorems 12.28 and 12.31 to determine the sets in which the functions are differentiable.f(x,y)=yx

Short Answer

Step by step solution

Given.

Considering.

Considering.

Equation (4).

Substituting (2),(3) and (4) in equation (1)to get.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Applied Mathematics

Statistics

Logic and Functions

Discrete Mathematics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

Find all points where the first-order partial derivatives of the functions in Exercises 43鈥�54 are continuous. Then use Theorems 12.28 and 12.31 to determine the sets in which the functions are differentiable. $f (x, y) = \sqrt{\frac{y}{x}}$