Q 58. - Calculus

Question

Find a function of two variables with the given gradient. 

Step-by-Step Solution

Answer

$f(x,y)=e^{x^2}+C$

Step 1: Given information

Step 2: Calculation

Consider the gradient

The goal is to deduce the function from the gradient.

Rewrite ( 1 ) as 

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

Equate both sides

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

And

$f_y(x,y)=2xy{e}^{x{y}^2}\dots \dots .\left(3\right)$

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

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

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

$$\begin{gathered} f_{xy}(x,y)=\frac{\partial }{\partial y}\left(y^2{e}^{x{y}^2}\right) \\ =y^2\frac{\partial }{\partial y}\left(e^{x{y}^2}\right)+e^{x{y}^2}\frac{\partial }{\partial y}{y}^2 \\ =y^2{e}^{x{y}^2}\frac{\partial }{\partial y}x{y}^2+e^{x{y}^2}2y \\ =y^2{e}^{x{y}^2}x\frac{\partial }{\partial y}{y}^2+e^{x{y}^2}2y \\ =y^2{e}^{x{y}^2}2xy+2y{e}^{x{y}^2} \\ =2x{y}^3{e}^{x{y}^2}+2y{e}^{x{y}^2} \end{gathered}$$

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

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

$=2x{y}^3{e}^{x{y}^2}+2y{e}^{x{y}^2}$

Since $f_{xy}(x,y)=2x{y}^3{e}^{x{y}^2}+2y{e}^{x{y}^2}=f_{yx}(x,y)$ so the function exists.

Integrate ( 2 ) with respect to $x$

$$\begin{gathered} f(x,y)=\int y^2{e}^{x{y}^2}dx+q\left(y\right) \\ =y^2\int e^{x^2}dx+q\left(y\right) \\ =y^2·\frac{e^{x{y}^2}}{y^2}+q\left(y\right) \\ =e^{x{y}^2}+q\left(y\right)\cdots \cdots \end{gathered}$$

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

Step 3: Calculation

Next, identify $q\left(y\right)$ to partially differentiate $\left(4\right)$ with regard to $y$

$\frac{\partial }{\partial y}f(x,y)=\frac{\partial }{\partial y}{e}^{x{y}^2}+\frac{\partial }{\partial y}q\left(y\right)$ $f_y(x,y)=e^{x^2}\frac{\partial }{\partial y}x{y}^2+q^{\text{'}}\left(y\right)$ $=e^{x{y}^2}·2xy+q^{\text{'}}\left(y\right)$$=2xy{e}^{x{y}^2}+q^{\text{'}}\left(y\right)$ $2xy{e}^{x{y}^2}=2xy{e}^{x{y}^2}+q^{\text{'}}\left(y\right)\left[\right.$ From $\left.\left(3\right);f_y(x,y)=2xy{e}^{x{y}^2}\right]$ $2xy{e}^{x{y}^2}-2xy{e}^{x{y}^2}=q^{\text{'}}\left(y\right)$ $0=q^{\text{'}}\left(y\right)$ $q^{\text{'}}\left(y\right)=0\dots \dots \left(5\right)$ Integrate $\left(5\right)$ with respect to

 $y$ $\int q^{\text{'}}\left(y\right)dy=\int 0dy+C$

$q\left(y\right)=0+C$

where $C$ is the constant of integration.

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

$f(x,y)=e^{x{y}^2}+C$

Therefore, the required function is

$f(x,y)=e^{x^2}+C\text{. }$