Question

If $u=\sin \left\{\frac{\pi (x^2+y^2+zx{)}^{\frac{1}{2}}}{2(x^2+xy+2yz+z^2{)}^{\frac{1}{3}}}\right\}$, then the value of $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}$ for $x=0;y=1;z=2$ is

• A. $\frac{\pi }{2^{\frac{1}{12}}}$
• B. $\frac{\pi }{12\sqrt{2}}$
• C. $\frac{\pi }{12}$
• D. $\frac{\pi }{\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{\pi }{12\sqrt{2}}$

Given $u=\sin \left(\frac{\pi }{2}\frac{(x^2+y^2+2x{)}^{\frac{1}{2}}}{(x^2+xy+2yz+z^2{)}^{\frac{1}{3}}}\right)$

let $v=\frac{(x^2+y^2+2x{)}^{1/2}}{(x^2+xy+2yz+z^2{)}^{\frac{1}{3}}}$

$(x^2+xy+2yz+z^2{)}^{\frac{1}{3}}v=(x^2+y^2+2x{)}^{\frac{1}{2}}$

$(x^2+xy+2yz+z^2{)}^2{v}^6=(x^2+y^2+2x{)}^3$

Partially differentiate on both side w.r.t $x,y,z$

$2(x^2+xy+2yz+z^2)v^6(2x+y)+6(2x^2+xy+2yz+z^2{)}^2{v}^5\frac{\partial v}{\partial x}=3(x^2+y^2+xz{)}^2(2x+z)\to \left(1\right)$

$2(x^2+xy+2yz+z^2)v^6(x+2z)+6(2x^2+xy+2yz+z^2{)}^2{v}^5\frac{\partial v}{\partial y}=3(x^2+y^2+xz{)}^2\left(2y\right)\to \left(2\right)$

$2(x^2+xy+2yz+z^2)v^6(2y+2z)+6(2x^2+xy+2yz+z^2{)}^2{v}^5\frac{\partial v}{\partial z}=3(x^2+y^2+xz{)}^2\left(x\right)\to \left(3\right)$

$\left(1\right)\times x+\left(2\right)\times y+\left(3\right)\times z\Rightarrow$

$2(x^2+xy+2yz+z^2)v^{6}2(x^2+xy+xy+2zy+2yz+2z^2)+6(2x^2+xy+2yz+z^2{)}^2{v}^5(x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y}+z\frac{\partial v}{\partial z})$

$=3(x^2+y^2+xz{)}^2(2x^2+xz+2y^2+zx)$

$4(x^2+xy+2yz+z^2{)}^2{v}^6+6(x^2+xy+2yz+z^2{)}^2{v}^5(x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y}+z\frac{\partial v}{\partial z})=6(x^2+y^2+xz{)}^3$

$6(x^2+xy+2yz+z^2{)}^2{v}^5(x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y}+z\frac{\partial v}{\partial z})=2(x^2+xy+2yz+z^2{)}^2{v}^6$

$\therefore x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y}+z\frac{\partial v}{\partial z}=\frac{v}{3}$

$u=sin\left(\frac{\pi }{2}v\right)$

$\frac{\partial u}{\partial x}=\frac{\pi }{2}cos\left(\frac{\pi }{2}v\right)\frac{\partial v}{\partial x}$

$\frac{\partial u}{\partial y}=\frac{\pi }{2}cos\left(\frac{\pi }{2}v\right)\frac{\partial v}{\partial y}$

$\frac{\partial u}{\partial z}=\frac{\pi }{2}cos\left(\frac{\pi }{2}v\right)\frac{\partial v}{\partial z}$

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=\frac{\pi }{2}cos\left(\frac{\pi }{2}v\right)[x\frac{\partial v}{\partial x}+y\frac{\partial v}{\partial y}+z\frac{\partial v}{\partial z}]$

$=\frac{\pi }{6}vcos\left(\frac{\pi }{2}v\right)$

For$x=0,y=1,z=2$

$v=\frac{(x^2+y^2+zx{)}^{\frac{1}{2}}}{(x^2+xy+2yz+z^2{)}^{\frac{1}{3}}}$

$=\frac{(0+1+0{)}^{\frac{1}{2}}}{(0+0+4+4{)}^{\frac{1}{3}}}=\frac{1}{2}$

For$x=0$,$y=1,z=2$

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+z\frac{\partial u}{\partial z}=\frac{\pi }{12\sqrt{2}}$