Find a function of two variables with the given gradient.∇f(x,y)=1y-yx2i+1x-xy2j

Q 60. - Calculus | StudyQuestionHub

Q 60.

Question

Find a function of two variables with the given gradient.

$\nabla f (x, y) = (\frac{1}{y} - \frac{y}{x^{2}}) i + (\frac{1}{x} - \frac{x}{y^{2}}) j$

Step-by-Step Solution

Verified

Answer

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

1Step 1: Given information

$\nabla f (x, y) = (\frac{1}{y} - \frac{y}{x^{2}}) i + (\frac{1}{x} - \frac{x}{y^{2}}) j$

2Step 2: Calculation

Consider the gradient

$\nabla f (x, y) = (\frac{1}{y} - \frac{y}{x^{2}}) i + (\frac{1}{x} - \frac{x}{y^{2}}) j$

The goal is to deduce the function from the gradient.

Rewrite $(1)$ as

$\nabla f (x, y) = (\frac{1}{y} - \frac{y}{x^{2}}) i + (\frac{1}{x} - \frac{x}{y^{2}}) j f_{x} (x, y) i + f_{y} (x, y) j = (\frac{1}{y} - \frac{y}{x^{2}}) i + (\frac{1}{x} - \frac{x}{y^{2}}) j$

Equate both sides

$f_{x} (x, y) = (\frac{1}{y} - \frac{y}{x^{2}}) \dots \dots (2)$

And

$f_{y} (x, y) = (\frac{1}{x} - \frac{x}{y^{2}}) \dots \dots (3)$

Check to see if the function is available. If there is a function, it exists.

$f_{x y} (x, y) = f_{y x} (x, y)$

Now, find $f_{x y} (x, y)$ by differentiating $f_{x} (x, y)$ partially with respect to $y$

f_{x y} (x, y) = \frac{\partial}{\partial y} (\frac{1}{y} - \frac{y}{x^{2}}) = \frac{\partial}{\partial y} \frac{1}{y} - \frac{\partial}{\partial y} \frac{y}{x^{2}} = (- 1) y^{- 2} - \frac{1}{x^{2}} \frac{\partial}{\partial y} y = - \frac{1}{y^{2}} - \frac{1}{x^{2}}

Also, find $f_{y x} (x, y)$ by differentiating $f_{y} (x, y)$ partially with respect to $x$

f_{y x} (x, y) = \frac{\partial}{\partial x} (\frac{1}{x} - \frac{x}{y^{2}}) = \frac{\partial}{\partial x} \frac{1}{x} - \frac{\partial}{\partial x} \frac{x}{y^{2}} = (- 1) x^{- 2} - \frac{1}{y^{2}} \frac{\partial}{\partial x} x = - \frac{1}{x^{2}} - \frac{1}{y^{2}}

Since $f_{x y} (x, y) = - \frac{1}{x^{2}} - \frac{1}{y^{2}} = f_{y x} (x, y)$ so the function exists.

3Step 3: Calculation

Integrate $(3)$ with respect to $y$

$f (x, y) = \int (\frac{1}{x} - \frac{x}{y^{2}}) d y + q (x)$ where $q (x)$ is an arbitrary function

$= \int \frac{1}{x} d y - \int \frac{x}{y^{2}} d y + q (x) = \frac{1}{x} \int d y - x \int \frac{1}{y^{2}} d y + q (x) = \frac{y}{x} - x \cdot \frac{y^{- 2 + 1}}{(- 2 + 1)} + q (x) = \frac{y}{x} - x \cdot \frac{y^{- 1}}{(- 1)} + q (x) = \frac{y}{x} + \frac{x}{y} + q (x) \dots . . (4)$

Next, find $q (x)$ to partially differentiate $(4)$ with regard to $x$

$\frac{\partial}{\partial x} f (x, y) = \frac{\partial}{\partial x} (\frac{y}{x} + \frac{x}{y}) + \frac{\partial}{\partial x} q (x) f_{x} (x, y) = (\frac{\partial}{\partial x} \frac{y}{x} + \frac{\partial}{\partial x} \frac{x}{y}) + q^{'} (x) = (y \frac{\partial}{\partial x} \frac{1}{x} + \frac{1}{y} \frac{\partial}{\partial x} x) + q^{'} (x)$

$= (y (- 1) x^{- 1 - 1} + \frac{1}{y}) + q^{'} (x) = (- y x^{- 2} + \frac{1}{y}) + q^{'} (x) = - \frac{y}{x^{2}} + \frac{1}{y} + q^{'} (x) Comment \int q^{'} (x) d x = \int 0 d x + C Integrate (5) with respect to x . \frac{1}{y} - \frac{y}{x^{2}} = - \frac{y}{x^{2}} + \frac{1}{y} + q^{'} (x) [From (2); f_{x} (x, y) = (\frac{1}{y} - \frac{y}{x^{2}})] \frac{1}{y} - \frac{y}{x^{2}} + \frac{y}{x^{2}} - \frac{1}{y} = q^{'} (x) q^{'} (x) = 0$