Chapter 4, Problem 4.36E

Which of the following functions are exact differentials?

(a) $dF=\frac{1}{x}dx+\frac{1}{y}dy$

(b) $dF=\frac{1}{\text{y}}dx+\frac{1}{x}dy$

(c) $dF=2x^2{y}^{2}dx+3x^3{y}^{3}dy$

(d) $dF=2x^2{y}^{3}dx+2x^3{y}^{2}dy$

(e) $dF=x^{n}dx+y^{n}dy,n=$ any integer

(f) $dF=\left(x^3\cdot \cos y\right)dx+\left(x^3\cdot \sin y\right)dy$

Interpretation Introduction

(a)

Interpretation:

Whether the function, $dF=\frac{1}{x}dx+\frac{1}{y}dy$ is an exact differential or not is to be stated.

Concept introduction:

A function is said to be an exact differential if it satisfies the relation given below.

${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y={\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$

In the above equation, $F$ is the function having variables $x$ and $y$.

Answer to Problem 4.36E

The given function, $dF=\frac{1}{x}dx+\frac{1}{y}dy$ is an exact differential.

Explanation of Solution

The given function is written below.

$dF=\frac{1}{x}dx+\frac{1}{y}dy$

Using the identity of exact differential, the function can be differentiated as shown below.

$\begin{array}{rll}{\left(\frac{\partial F}{\partial y}\right)}_x& =& -\frac{1}{y^2}\\ {\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y& =& 0\\ {\left(\frac{\partial F}{\partial x}\right)}_y& =& -\frac{1}{x^2}\\ {\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x& =& 0\end{array}$

Since, ${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y={\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$ for the given function. Therefore, it is an exact differential.

Conclusion

The given function, $dF=\frac{1}{x}dx+\frac{1}{y}dy$ is an exact differential.

Interpretation Introduction

(b)

Interpretation:

Whether the function, $dF=\frac{1}{\text{y}}dx+\frac{1}{x}dy$ is an exact differential or not is to be stated.

Concept introduction:

A function is said to be an exact differential if it satisfies the relation given below.

${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y={\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$

In the above equation, $F$ is the function having variables $x$ and $y$.

Answer to Problem 4.36E

The given function, $dF=\frac{1}{\text{y}}dx+\frac{1}{x}dy$ is not an exact differential.

Explanation of Solution

The given function is written below.

$dF=\frac{1}{\text{y}}dx+\frac{1}{x}dy$

Using the identity of exact differential, the function can be differentiated as shown below.

$\begin{array}{rll}{\left(\frac{\partial F}{\partial y}\right)}_x& =& \frac{1}{x}\\ {\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y& =& -\frac{1}{x^2}\\ {\left(\frac{\partial F}{\partial x}\right)}_y& =& \frac{1}{y}\\ {\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x& =& -\frac{1}{y^2}\end{array}$

Since, ${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y\ne {\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$ for the given function. Therefore, it is not an exact differential.

Conclusion

The given function, $dF=\frac{1}{\text{y}}dx+\frac{1}{x}dy$ is not an exact differential.

Interpretation Introduction

(c)

Interpretation:

Whether the function, $dF=2x^2{y}^{2}dx+3x^3{y}^{3}dy$ is an exact differential or not is to be stated.

Concept introduction:

A function is said to be an exact differential if it satisfies the relation given below.

${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y={\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$

In the above equation, $F$ is the function having variables $x$ and $y$.

Answer to Problem 4.36E

The given function, $dF=2x^2{y}^{2}dx+3x^3{y}^{3}dy$ is not an exact differential.

Explanation of Solution

The given function is written below.

$dF=2x^2{y}^{2}dx+3x^3{y}^{3}dy$

Using the identity of exact differential, the function can be differentiated as shown below.

$\begin{array}{rll}{\left(\frac{\partial F}{\partial y}\right)}_x& =& 9x^3{y}^2\\ {\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y& =& 27x^2{y}^2\\ {\left(\frac{\partial F}{\partial x}\right)}_y& =& 4x{y}^2\\ {\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x& =& 8xy\end{array}$

Since, ${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y\ne {\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$ for the given function. Therefore, it is not an exact differential.

Conclusion

The given function, $dF=2x^2{y}^{2}dx+3x^3{y}^{3}dy$ is not an exact differential.

Interpretation Introduction

(d)

Interpretation:

Whether the function, $dF=2x^2{y}^{3}dx+2x^3{y}^{2}dy$ is an exact differential or not is to be stated.

Concept introduction:

A function is said to be an exact differential if it satisfies the relation given below.

${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y={\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$

In the above equation, $F$ is the function having variables $x$ and $y$.

Answer to Problem 4.36E

The given function, $dF=2x^2{y}^{3}dx+2x^3{y}^{2}dy$ is not an exact differential.

Explanation of Solution

The given function is written below.

$dF=2x^2{y}^{3}dx+2x^3{y}^{2}dy$

Using the identity of exact differential, the function can be differentiated as shown below.

$\begin{array}{rll}{\left(\frac{\partial F}{\partial y}\right)}_x& =& 4x^{3}y\\ {\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y& =& 12x^{2}y\\ {\left(\frac{\partial F}{\partial x}\right)}_y& =& 4x{y}^3\\ {\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x& =& 12x{y}^2\end{array}$

Since, ${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y\ne {\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$ for the given function. Therefore, it is not an exact differential.

Conclusion

The given function, $dF=2x^2{y}^{3}dx+2x^3{y}^{2}dy$ is not an exact differential.

Interpretation Introduction

(e)

Interpretation:

Whether the function, $dF=x^{n}dx+y^{n}dy$ is an exact differential or not is to be stated.

Concept introduction:

A function is said to be an exact differential if it satisfies the relation given below.

${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y={\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$

In the above equation, $F$ is the function having variables $x$ and $y$.

Answer to Problem 4.36E

The given function, $dF=x^{n}dx+y^{n}dy$ is an exact differential.

Explanation of Solution

The given function is written below.

$dF=x^{n}dx+y^{n}dy$

Using the identity of exact differential, the function can be differentiated as shown below.

$\begin{array}{rll}{\left(\frac{\partial F}{\partial y}\right)}_x& =& n{y}^{n-1}\\ {\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y& =& 0\\ {\left(\frac{\partial F}{\partial x}\right)}_y& =& n{x}^{n-1}\\ {\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x& =& 0\end{array}$

Since, ${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y={\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$ for the given function. Therefore, it is an exact differential.

Conclusion

The given function, $dF=x^{n}dx+y^{n}dy$ is an exact differential.

Interpretation Introduction

(f)

Interpretation:

Whether the function, $dF=\left(x^3\cdot \cos y\right)dx+\left(x^3\cdot \sin y\right)dy$ is an exact differential or not is to be stated.

Concept introduction:

A function is said to be an exact differential if it satisfies the relation given below.

${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y={\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$

In the above equation, $F$ is the function having variables $x$ and $y$.

Answer to Problem 4.36E

The given function, $dF=\left(x^3\cdot \cos y\right)dx+\left(x^3\cdot \sin y\right)dy$ is not an exact differential.

Explanation of Solution

The given function is written below.

$dF=\left(x^3\cdot \cos y\right)dx+\left(x^3\cdot \sin y\right)dy$

Using the identity of exact differential, the function can be differentiated as shown below.

$\begin{array}{rll}{\left(\frac{\partial F}{\partial y}\right)}_x& =& x^3\cdot \cos y\\ {\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y& =& 3x^2\cdot \cos y\\ {\left(\frac{\partial F}{\partial x}\right)}_y& =& 3x^2\cdot \cos y\\ {\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x& =& -3x^2\cdot \sin y\end{array}$

Since, ${\left(\frac{\partial }{\partial x}{\left(\frac{\partial F}{\partial y}\right)}_x\right)}_y\ne {\left(\frac{\partial }{\partial y}{\left(\frac{\partial F}{\partial x}\right)}_y\right)}_x$ for the given function. Therefore, it is not an exact differential.

Conclusion

The given function, $dF=\left(x^3\cdot \cos y\right)dx+\left(x^3\cdot \sin y\right)dy$ is not an exact differential.