Question

Observe the following statements :

I: If $Z=\log \left(\frac{x^2+y^2}{xy}\right)$ then$\frac{{\partial }^{2}z}{\partial y^2}-\frac{{\partial }^{2}z}{\partial x^2}=0$

II: If $F=x^{2}yz-2x{z}^3+x{z}^2$ then $\frac{{\partial }^{2}F}{\partial x\partial y}$ at $(1,0,-3)$ is 6.

Which of the above statement is correct?

• A. only I
• B. only II
• C. both I and II
• D. Neither I nor II

Answer 1

The correct option is D Neither I nor II

$I$ $Z=\log \left(\frac{x^2+y^2}{xy}\right)=\log (\frac{x}{y}+\frac{y}{x})\frac{\partial Z}{\partial x}=\frac{xy}{x^2+y^2}(\frac{1}{y}-\frac{y}{x^2})=\frac{xy(x^2-y^2)}{(x^2+y^2)\left(x^{2}y\right)}=\frac{(x^2-y^2)}{(x^2+y^2)}\frac{1}{x}\frac{{\partial }^{2}Z}{{\partial x}^2}=\frac{x(x^2+y^2)\left(2x\right)-(x^2-y^2)(3x^2+y^2)}{x^2{(x^2+y^2)}^2}=\frac{-x^4+y^4+4x^2{y}^2}{x^2{(x^2+y^2)}^2}Similarly,\frac{\partial Z}{\partial y}=\frac{(y^2-x^2)}{(y^2+x^2)}\frac{1}{y}\frac{{\partial }^{2}Z}{{\partial y}^2}=\frac{-y^4+x^4+4x^2{y}^2}{y^2{(x^2+y^2)}^2}$

Just interchanging $x$ with $y$ in the results of $\frac{\partial Z}{\partial x}$ for that of $\frac{\partial Z}{\partial y}$

$\therefore \frac{-{\partial }^{2}Z}{{\partial x}^2}+\frac{{\partial }^{2}Z}{{\partial y}^2}=\frac{1}{{(x^2+y^2)}^2}\{-(-x^2+\frac{y^4}{x^2}+4y^2)+\{(-y^2)+\frac{x^4}{y^2}+4x^2\}\}=\frac{1}{{(x^2+y^2)}^2}\{5x^2-5y^2+(-\frac{y^4}{x^2}+\frac{x^4}{y^2})\}\Rightarrow \frac{{\partial }^{2}Z}{{\partial y}^2}-\frac{{\partial }^{2}Z}{{\partial x}^2}\ne 0$

Hence, $I$ is incorrect.

Now, $\frac{\partial F}{\partial x}=2xyz-2z^3+z^2\Rightarrow \frac{\partial F}{\partial x\cdot \partial y}=2xz$

$2xz$ at $(1,0,-3)=-6\ne 6$

$\Rightarrow II$ is incorrect.

Hence, neither $I$ nor $II$ is correct.