Q 56. - Calculus

Question

Find a function of two variables with the given gradient. 

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

Step-by-Step Solution

Answer

$f(x,y)=\sin x\cos y+x^2+C$

Part (a) Step 1: Given information

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

Step 2: Calculation

Consider the gradient 

$\nabla f(x,y)=(2x+\cos x\cos y)\mathrm{i}-\sin x\sin y\mathrm{j}\dots \dots \text{. (1) }$

The goal is to deduce the function from the gradient.

Rewrite ( 1) as

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

Equate both sides

$f_x(x,y)=2x+\cos x\cos y\dots \dots \text{ (2) }$

And

$f_y(x,y)=-\sin x\sin y \dots \dots .\left(3\right)$

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

$$\begin{gathered} f_{xy}(x,y)=2\frac{\partial }{\partial y}x+\frac{\partial }{\partial y}\left(\cos x\cos y\right) \\ =2·0+\cos x\frac{\partial }{\partial y}\cos y \\ =\cos x(-\sin y) \\ =-\cos x\sin y \end{gathered}$$

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

$$\begin{gathered} f_{yx}(x,y)=\frac{\partial }{\partial x}(-\sin x\sin y) \\ =-\sin y\frac{\partial }{\partial x}\sin x \\ =-\sin y\cos x \end{gathered}$$

Since $f_{xy}(x,y)=-\sin y\cos x=f_{yx}(x,y)$so the function exists.

Step 3: Calculation

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

$$\begin{gathered} f(x,y)=\int (-\sin x\sin y)dy+q\left(x\right) \\ =-\sin x\int \sin ydy+q\left(x\right) \\ =-\sin x(-\cos y)+q\left(x\right) \\ =\sin x\cos y+q\left(x\right)\dots \dots \end{gathered}$$

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

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

$\frac{\partial }{\partial x}f(x,y)=\frac{\partial }{\partial x}\sin x\cos y+\frac{\partial }{\partial x}q\left(x\right)$ $$\begin{gathered} f_x(x,y)=\cos y\frac{\partial }{\partial x}\sin x+q^{\text{'}}\left(x\right) \\ =\cos y\cos x+q^{\text{'}}\left(x\right) \\ 2x+\cos x\cos y=\cos y\cos x+q^{\text{'}}\left(x\right)\left[\text{ From }\left(2\right);f_x(x,y)=2x+\cos x\cos y\text{ ]}\right. \end{gathered}$$ $2x=q^{\text{'}}\left(x\right)$ $$\begin{gathered} q^{\text{'}}\left(x\right)=2x\cdots \cdots \left(5\right) \\ \text{ Integrate }\left(5\right)\text{ with respect to }y. \\ \int q^{\text{'}}\left(x\right)dx=\int 2xdx+C \\ q\left(x\right)=2\int xdx+C \\ =2·\frac{x^2}{2}+C \\ =x^2+C \end{gathered}$$

where $C$ is the constant of integration.

Put $q\left(x\right)=x^2+C$ in $( 4)$

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

As a result, the necessary function is $f(x,y)=\sin x\cos y+x^2+C\text{. }$