Question

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

then $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=?$

• A. $\frac{z^2-c^2}{z}$
• B. $\frac{z^2+c^2}{z}$
• C. $\frac{c^2}{z}$
• D. $\frac{z^2}{2}$

Answer 1

The correct option is A $\frac{z^2-c^2}{z}$

$\frac{2x}{a^2}+\frac{2z}{c^2}\frac{dz}{dx}=0\Rightarrow \frac{dz}{dx}=\frac{-x}{a^2}\frac{c^2}{z}$

$\frac{2y}{b^2}+\frac{2z}{c^2}\frac{dz}{dy}=0\Rightarrow \frac{dz}{dy}=\frac{-y{c}^2}{b^{2}z}$

$x\frac{dz}{dx}+y\frac{dz}{dy}=-(\frac{x^2}{a^2}+\frac{y^2}{b^2})\frac{c^2}{z}$
$=-(1-\frac{z^2}{c^2})\frac{c^2}{z}$

$=\frac{(z^2-c^2)}{z}$