Q 57. - Calculus

Question

Find a function of two variables with the given gradient. 

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

Step-by-Step Solution

Answer

$f(x,y)={\tan }^{-1}\frac{y}{x}+C$

Step 1: Given information

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

Step 2: Calculation

Consider the gradient

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

The goal is to deduce the function from the gradient.

Rewrite (1) as

$$\begin{gathered} \nabla f(x,y)=-\frac{y}{x^2+y^2}\mathrm{i}+\frac{x}{x^2+y^2}\mathrm{j} \\ {f}_x(x,y)\mathrm{i}+f_y(x,y)\mathrm{j}=-\frac{y}{x^2+y^2}\mathrm{i}+\frac{x}{x^2+y^2}\mathrm{j} \end{gathered}$$

Equate both sides

$f_x(x,y)=-\frac{y}{x^2+y^2}\dots \dots \text{ (2) }$

And

$f_y(x,y)=\frac{x}{x^2+y^2}\dots \dots \text{ (3) }$

First, check whether the function exists or not. Function exists if

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

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

$$\begin{gathered} f_{xy}(x,y)=\frac{\partial }{\partial y}\left(-\frac{y}{x^2+y^2}\right) \\ =-\frac{\partial }{\partial y}\left(\frac{y}{x^2+y^2}\right) \\ =-\left\{\frac{\left(x^2+y^2\right)\frac{\partial }{\partial y}y-y\frac{\partial }{\partial y}\left(x^2+y^2\right)}{{\left(x^2+y^2\right)}^2}\right\} \\ =-\left\{\frac{\left(x^2+y^2\right)-y\left(\frac{\partial }{\partial y}{x}^2+\frac{\partial }{\partial y}{y}^2\right)}{{\left(x^2+y^2\right)}^2}\right\} \end{gathered}$$

$$\begin{gathered} =-\left\{\frac{\left(x^2+y^2\right)-y(0+2y)}{{\left(x^2+y^2\right)}^2}\right\} \\ =-\left\{\frac{x^2+y^2-2y^2}{{\left(x^2+y^2\right)}^2}\right\} \\ =-\left\{\frac{x^2-y^2}{{\left(x^2+y^2\right)}^2}\right\} \end{gathered}$$

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

$$\begin{gathered} f_{yx}(x,y)=\frac{\partial }{\partial x}\left(\frac{x}{x^2+y^2}\right) \\ =\frac{\partial }{\partial x}\left(\frac{x}{x^2+y^2}\right) \\ =\left\{\frac{\left(x^2+y^2\right)\frac{\partial }{\partial x}x-x\frac{\partial }{\partial x}\left(x^2+y^2\right)}{{\left(x^2+y^2\right)}^2}\right\} \end{gathered}$$

$$\begin{gathered} =\left\{\frac{\left(x^2+y^2\right)-x\left(\frac{\partial }{\partial x}{x}^2+\frac{\partial }{\partial x}{y}^2\right)}{{\left(x^2+y^2\right)}^2}\right\} \\ =\left\{\frac{\left(x^2+y^2\right)-x(2x+0)}{{\left(x^2+y^2\right)}^2}\right\} \\ =\left\{\frac{x^2+y^2-2x^2}{{\left(x^2+y^2\right)}^2}\right\} \\ =\left\{\frac{-x^2+y^2}{{\left(x^2+y^2\right)}^2}\right\} \\ =-\left\{\frac{x^2-y^2}{{\left(x^2+y^2\right)}^2}\right\} \end{gathered}$$

Since

$f_{xy}(x,y)=\left\{\frac{x^2-y^2}{{\left(x^2+y^2\right)}^2}\right\}=f_{yx}(x,y)$ so the function exists.

Step 3: Calculation

Integrate $\left(3\right)$ with respect to $y$

$$\begin{gathered} f(x,y)=\int \frac{x}{x^2+y^2}dy+q\left(x\right) \\ =\int \frac{x}{x^2+y^2}dy+q\left(x\right) \\ =x·\frac{1}{x}{\tan }^{-1}\frac{y}{x}+q\left(x\right) \\ ={\tan }^{-1}\frac{y}{x}+q\left(x\right)\dots \dots \left(4\right) \end{gathered}$$

where $q\left(x\right)$ is an arbitrary function?

The following step is to locate $q\left(x\right)$differentiating $\left(4\right)$ with respect to $x$ but only in part

$$\begin{gathered} \frac{\partial }{\partial x}f(x,y)=\frac{\partial }{\partial x}{\tan }^{-1}\frac{y}{x}+\frac{\partial }{\partial x}q\left(x\right) \\ {f}_x(x,y)=\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\frac{\partial }{\partial x}\left(\frac{y}{x}\right)+q^{\text{'}}\left(x\right) \\ =\frac{1}{1+{\left(\frac{y}{x}\right)}^2}·y\frac{\partial }{\partial x}\left(\frac{1}{x}\right)+q^{\text{'}}\left(x\right) \\ =\frac{1}{\left(1+\frac{y^2}{x^2}\right)}\left(-\frac{y}{x^2}\right)+q^{\text{'}}\left(x\right) \\ =\frac{1}{\left(\frac{x^2+y^2}{x^2}\right)}\left(-\frac{y}{x^2}\right)+q^{\text{'}}\left(x\right) \\ =\frac{1}{\left(\frac{x^2+y^2}{x^2}\right)}\left(-\frac{y}{x^2}\right)+q^{\text{'}}\left(x\right) \\ {x}^2+y^2\left(-\frac{y}{x^2}\right)+q^{\text{'}}\left(x\right) \end{gathered}$$

$$\begin{gathered} =-\frac{y}{x^2+y^2}+q^{\text{'}}\left(x\right) \\ -\frac{y}{x^2+y^2}=-\frac{y}{x^2+y^2}+q^{\text{'}}\left(x\right)\left[\text{ From }\left(2\right);f_x(x,y)=-\frac{y}{x^2+y^2}\right] \\ -\frac{y}{x^2+y^2}+\frac{y}{x^2+y^2}=q^{\text{'}}\left(x\right) \\ 0=q^{\text{'}}\left(x\right) \\ {q}^{\text{'}}\left(x\right)=0 \end{gathered}$$

Integrate $\left(5\right)$ with respect to $y$

$$\begin{gathered} \int q^{\text{'}}\left(x\right)dx=\int 0dx+C \\ q\left(x\right)=0+C \\ =C \end{gathered}$$

where $C$ is the constant of integration?

Put $q\left(x\right)=C$ in (4)

$f(x,y)={\tan }^{-1}\frac{y}{x}+C$

Therefore, the required function is

$f(x,y)={\tan }^{-1}\frac{y}{x}+C\text{. }$