Q 59. - Calculus

Question

Find a function of two variables with the given gradient.

Step-by-Step Solution

Answer

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

Step 1: Given information

Step 2: Calculation

Consider the gradient

The goal is to deduce the function from the gradient.

Rewrite (1) as

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

Equate both sides

$f_x(x,y)=\frac{y}{(x+y{)}^2}\dots \dots .\left(2\right)$

And

$f_y(x,y)=-\frac{x}{(x+y{)}^2}\dots \dots .\left(3\right)$

Open with -

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

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

Find $f_{xy}(x,y)$ by partially differentiating $f_x(x,y)$ with regard to $y$

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

$$\begin{gathered} =\frac{(x+y{)}^2\frac{\partial }{\partial y}y-y\frac{\partial }{\partial y}(x+y{)}^2}{{\left\{(x+y{)}^2\right\}}^2} \\ =\frac{(x+y{)}^2-2y(x+y\left)\frac{\partial }{\partial y}\right(x+y)}{(x+y{)}^4} \end{gathered}$$

$=\frac{(x+y{)}^2-2y(x+y)\left(\frac{\partial }{\partial y}x+\frac{\partial }{\partial y}y\right)}{(x+y{)}^4}$

$=\frac{(x+y{)}^2-2y(x+y\left)\right(0+1)}{(x+y{)}^4}$

$=\frac{(x+y{)}^2-2y(x+y)}{(x+y{)}^4}$

$$\begin{gathered} =\frac{(x+y)\left\{\right(x+y)-2y\}}{(x+y{)}^4} \\ =\frac{x+y-2y}{(x+y{)}^3} \\ =\frac{x-y}{(x+y{)}^3} \end{gathered}$$

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

$$\begin{gathered} f_{yx}(x,y)=\frac{\partial }{\partial x}\left(-\frac{x}{(x+y{)}^2}\right) \\ =-\frac{(x+y{)}^2\frac{\partial }{\partial x}x-x\frac{\partial }{\partial x}(x+y{)}^2}{{\left\{(x+y{)}^2\right\}}^2} \\ =-\frac{(x+y{)}^2-2x(x+y\left)\frac{\partial }{\partial x}\right(x+y)}{(x+y{)}^4} \\ =-\frac{(x+y{)}^2-2x(x+y)\left(\frac{\partial }{\partial x}x+\frac{\partial }{\partial x}y\right)}{(x+y{)}^4} \end{gathered}$$

$$\begin{gathered} =-\frac{(x+y{)}^2-2x(x+y\left)\right(1+0)}{(x+y{)}^4} \\ =-\frac{(x+y)\left\{\right(x+y)-2x\}}{(x+y{)}^4} \\ =-\frac{(x+y-2x)}{(x+y{)}^3} \\ =-\frac{y-x}{(x+y{)}^3} \\ =\frac{x-y}{(x+y{)}^3} \end{gathered}$$

Since $f_{xy}(x,y)=\frac{x-y}{(x+y{)}^3}=f_{yx}(x,y)$ so the function exists.

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

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

Where

$I=\int \frac{1}{(x+y{)}^2}dy$

Take $t=x+y$ then $d t=d y$ (Since $x$ is a constant, so $d x=0$ )

Thus, 

$$\begin{gathered} I=\int \frac{1}{(x+y{)}^2}dy \\ =\int \frac{1}{t^2}dt \end{gathered}$$

$$\begin{gathered} =\int t^{-2}dt \\ =\frac{t^{-2+1}}{(-2+1)} \\ =\frac{t^{-1}}{(-1)} \\ =-\frac{1}{t} \\ =-\frac{1}{x+y}[\text{ Since }t=x+y] \\ \text{ From }\left(4\right) \\ f(x,y)=-xI+q\left(x\right) \\ =-x\left(-\frac{1}{x+y}\right)+q\left(x\right) \\ =\frac{x}{x+y}+q\left(x\right)\dots \dots \left(5\right) \end{gathered}$$

Step 3: Calculation

Next, find $q\left(x\right)$1q to partially differentiate$\left(5\right)$ with regard to $x$

$$\begin{gathered} \frac{\partial }{\partial x}f(x,y)=\frac{\partial }{\partial x}\left(\frac{x}{x+y}\right)+\frac{\partial }{\partial x}q\left(x\right) \\ {f}_x(x,y)=\frac{(x+y)\frac{\partial }{\partial y}x-x\frac{\partial }{\partial y}(x+y)}{(x+y{)}^2}+q^{\text{'}}\left(x\right) \\ =\frac{(x+y)-x\left(\frac{\partial }{\partial y}x+\frac{\partial }{\partial y}y\right)}{(x+y{)}^2}+q^{\text{'}}\left(x\right) \\ =\frac{(x+y)-x(0+1)}{(x+y{)}^2}+q^{\text{'}}\left(x\right) \\ =\frac{x+y-x}{(x+y{)}^2}+q^{\text{'}}\left(x\right) \\ =\frac{y}{(x+y{)}^2}+q^{\text{'}}\left(x\right) \\ \frac{y}{(x+y{)}^2}=\frac{y}{(x+y{)}^2}+q^{\text{'}}\left(x\right)\left[From\left(2\right);f_x(x,y)=\frac{y}{(x+y{)}^2}\right] \end{gathered}$$

$$\begin{gathered} \frac{x}{(x+y{)}^2}-\frac{x}{(x+y{)}^2}=q^{\text{'}}\left(x\right) \\ {q}^{\text{'}}\left(x\right)=0\dots \dots .\left(6\right) \end{gathered}$$

Integrate ( 6 ) with respect to $x$

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

where $C$ is the constant of integration.

Put $q(x)=C$ in (5)

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

Therefore, the required function is

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