Exercise 8.4-Additional Problems - Chapter 8 Differentials and Partial Derivatives 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
If $u=\log (\tan x+\tan y+\tan z)$, prove that $\Sigma \sin 2 x \frac{\partial u}{\partial y}=2$
Solution:
$
\begin{gathered}
\frac{\partial u}{\partial y}=\frac{\sec ^2 x}{\tan x+\tan y+\tan z} \\
\sin 2 x \frac{\partial u}{\partial x}=\frac{2 \sin x \cos x \cdot \sec ^2 x}{\tan x+\tan y+\tan z}=\frac{2 \tan x}{\tan x+\tan y+\tan z} \\
\text { similarly, } \sin 2 y \frac{\partial u}{\partial y}=\frac{2 \tan y}{\tan x+\tan y+\tan z} \\
\sin 2 z \frac{\partial u}{\partial z}=\frac{2 \tan z}{\tan x+\tan y+\tan z} \\
\text { L.H.S. }=\Sigma \sin 2 x \frac{\partial u}{\partial y}=\frac{2(\tan x+\tan y+\tan z)}{\tan x+\tan y+\tan z}=2=\text { R.H.S. }
\end{gathered}
$

Question 2.
If $U=(x-y)(y-z)(z-x)$ then show that $U_x+U_y+U_z=0$
Solution:
$
\begin{aligned}
& U_x=(y-z)\{(x-y)(-1)+(z-x) \cdot 1\} \\
& =(y-z)[(z-x)-(x-y)] \\
& \text { Similarly } U_y=(z-x)[(x-y)-(y-z)] \\
& z=(x-y)[(y-z)-(z-x] \\
& U_x+U_y+U_z=(y-z)[(z-x)-(z-x)]+(x-y)[-(y-z)+(y-z)]+(z-x)[(x-y)-(x-y)] \\
& \therefore U_x+U_y+U_z=0 .
\end{aligned}
$
Hence proved.

Question 3.
If $u=x^2+3 x y+y^2$ Verify $\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial^2 u}{\partial y \partial x}$.
Solution:
$
\begin{aligned}
& \frac{\partial u}{\partial x}=2 x+3 y(1)+0=2 x+3 y \\
& \frac{\partial u}{\partial y}=0+3 x(1)+2 y=3 x+2 y \\
& \frac{\partial^2 u}{\partial x \partial y}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right)=\frac{\partial}{\partial x}(3 x+2 y)=3 \\
& \frac{\partial^2 u}{\partial y \partial x}=\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right)=\frac{\partial}{\partial y}(2 x+3 y)=3 \\
&
\end{aligned}
$

Question 4.
If $u=\frac{x}{y^2}-\frac{y}{x^2}$, show that $\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial^2 u}{\partial y \partial x}$
Solution:
$
\begin{aligned}
u & =\frac{x}{y^2}-\frac{y}{x^2} \\
\frac{\partial u}{\partial y} & =\frac{1}{y^2}(1)-y\left[\frac{-2}{x^3}\right]=\frac{1}{y^2}+\frac{2 y}{x^3} \\
\frac{\partial u}{\partial y} & =x\left[\frac{-2}{y^3}\right]-\frac{1}{x^2}(1)=\frac{-2 x}{y^3}-\frac{1}{x^2} \\
\frac{\partial^2 u}{\partial x \partial y} & =\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right)=\frac{\partial}{\partial x}\left(\frac{-2 x}{y^3}-\frac{1}{x^2}\right) \\
& =\frac{-2}{y^3}(1)-\left(\frac{-2}{x^3}\right)=\frac{-2}{y^3}+\frac{2}{x^3} \\
\frac{\partial^2 u}{\partial y \partial x} & =\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\right)=\frac{\partial}{\partial y}\left(\frac{1}{y^2}+\frac{2 y}{x^3}\right)=\frac{-2}{y^3}+\frac{2}{x^3} \\
\text { From (1) and (2) } & \frac{\partial^2 u}{\partial x \partial y}=\frac{\partial^2 u}{\partial y \partial x}
\end{aligned}
$