Differentiation Question 97
Question: If $ u^{2}={{(x-a)}^{2}}+{{(y-b)}^{2}}+{{(z-c)}^{2}} $ , then $ \sum \frac{{{\partial }^{2}}u}{\partial x^{2}}= $
[Tamilnadu (Engg.) 2002]
Options:
A) $ \frac{2}{u} $
B) $ \frac{3}{u} $
C) 0
D) $ \frac{1}{u} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 2u\frac{\partial u}{\partial x}=2(x-a) $
Therefore $ u\frac{\partial u}{\partial x}=x-a $
Therefore $ u.\frac{{{\partial }^{2}}u}{\partial x^{2}}+{{( \frac{\partial u}{\partial x} )}^{2}}=1 $
Therefore $ u.\frac{{{\partial }^{2}}u}{\partial x^{2}}=1-{{( \frac{x-a}{u} )}^{2}} $
Therefore $ \frac{{{\partial }^{2}}u}{\partial x^{2}}=\frac{1}{u}-\frac{{{(x-a)}^{2}}}{u^{3}} $
Similarly, $ \frac{{{\partial }^{2}}u}{\partial y^{2}}=\frac{1}{u}-\frac{{{(y-b)}^{2}}}{u^{3}} $ , $ \frac{{{\partial }^{2}}u}{\partial z^{2}}=\frac{1}{u}-\frac{{{(z-b)}^{2}}}{u^{3}} $
$ \therefore $ $ \sum \frac{{{\partial }^{2}}u}{\partial x^{2}}=\frac{3}{u}-\frac{1}{u^{3}}[{{(x-a)}^{2}}+{{(y-b)}^{2}}+{{(z-c)}^{2}}] $
= $ \frac{3}{u}-\frac{1}{u^{3}}.(u^{2})=\frac{3}{u}-\frac{1}{u}=\frac{2}{u} $ .
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