Exercise 8.7-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.
Verify Euler's theorem for $f(x, y)=\frac{1}{\sqrt{x^2+y^2}}$
Solution:

$
f(t x, t y)=\frac{1}{\sqrt{t^2 x^2+t^2 y^2}}=\frac{1}{t} f(x, y)=t^{-1} f(x, y)
$
$\therefore f$ is a homogeneous function of degree $-1$ and by Euler's theorem.
$
x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=-f
$
Verification:
$
\begin{aligned}
f_x & =-\frac{1}{2} \frac{2 x}{\left(x^2+y^2\right)^{\frac{3}{2}}}=\frac{-x}{\left(x^2+y^2\right)^{\frac{3}{2}}} \\
\text { Similarly, } f_y & =\frac{-y}{\left(x^2+y^2\right)^{\frac{3}{2}}} \\
x f_x+y f_y & =-\frac{x^2+y^2}{\left(x^2+y^2\right)^{\frac{3}{2}}}=\frac{-1}{\sqrt{x^2+y^2}}=-f
\end{aligned}
$

Question 2.
Using Euler's theorem, prove that $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=\frac{1}{2} \tan u$ if $u=\sin ^{-1}\left(\frac{x-y}{\sqrt{x}+\sqrt{y}}\right)$
Solution:
R.H.S. is not a homogeneous and hence
define $f=\sin u=\frac{x-y}{\sqrt{x}+\sqrt{y}} \Rightarrow f$ is homogeneous of degree $\frac{1}{2}$.
$\therefore$ By Euler's theorem, $x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=\frac{1}{2} f$
i.e., $x \cdot \frac{\partial}{\partial x}(\sin u)+y \frac{\partial}{\partial y}(\sin u)=\frac{1}{2} \sin u$
$x \frac{\partial u}{\partial x} \cos u+y \frac{\partial u}{\partial y} \cos u=\frac{1}{2} \sin u$
$x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=\frac{1}{2} \tan u$