Question

1. $u=\sqrt{{\sin }^{2}x+{\sin }^{2}y+{\sin }^{2}z}$ then $\frac{\partial u}{\partial z}$ at $(0,0,\frac{\pi }{4})=A$

2. $u=(x-y)(y-z)(z-x)$ then $u_x+u_y+u_z=B$

3. $u=\log (x^3+y^3+z^3-3xyz)$ then $2(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z})$ at $(1,-1,3)=C$

The ascending order in magnitude of $A,B,C$ is?

• A. $A,B,C$
• B. $B,C,A$
• C. $C,A,B$
• D. $B,A,C$

Answer 1

The correct option is D $B,A,C$

$1)\frac{\partial u}{\partial z}{|}_{0,0,\frac{\pi }{4}}=\frac{2\sin z\cos z}{2\sqrt{{\sin }^{2}x+{\sin }^{2}y+{\sin }^{2}z}}$

$=\frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{2}}=A$

$2)\frac{\partial u}{\partial x}=(y-z\left)\right(z-x)+0-z(x-y\left)\right(y-z)$

$\frac{\partial u}{\partial y}=-(y-z)(z-x)+(x-y)(z-x)$

$\frac{\partial u}{\partial z}=-(x-y)(z-x)+(y-z)(x-y)$

$\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=0=B$

$3)\frac{\partial u}{\partial x}=\frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}$

$\frac{\partial u}{\partial y}=\frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}$

$\frac{\partial u}{\partial z}=\frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}$

$2(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}){|}_{1-1,3}=2\left(\frac{3(x^2+y^2+z^2)-3(xy+yz+zx)}{x^3+y^3+z^3-3xyz}\right)=2=C$

Thus, the ascending order in magnitude is $B,A,C$.