Question

If $a^2{x}^2+b^2{y}^2=c^2{z}^2$, then$\frac{1}{a^2}\frac{{\partial }^{2}z}{\partial x^2}+\frac{1}{b^2}\frac{{\partial }^{2}z}{\partial y^2}$ is equal to

• A. $\frac{1}{z^2}$
• B. $\frac{1}{z}$
• C. $\frac{1}{c^{2}z}$
• D. $1$

Answer 1

The correct option is C $\frac{1}{c^{2}z}$

The given information is:

$a^2{x}^2+b^2{y}^2=c^2{z}^2$

$\Rightarrow z=\left[\right(\frac{ax}{c}{)}^2+(\frac{by}{c}{)}^2{]}^{\frac{1}{2}}$

Now finding the respective derivatives,

$\frac{dz}{dx}=\frac{\frac{2ax}{c^2}}{2\left[\right(\frac{ax}{c}{)}^2+(\frac{by}{c}{)}^2{]}^{\frac{1}{2}}}$

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

Now again differentiating with respected to x we get,

$\frac{d^{2}z}{d{x}^2}=\frac{a^2}{c^2}\left(\frac{z-\frac{a^2{x}^2}{c^{2}z}}{z^2}\right)$

$\Rightarrow \frac{1}{a^2}\frac{d^{2}z}{d{x}^2}=\frac{1}{c^2}\left(\frac{z-\frac{a^2{x}^2}{c^{2}z}}{z^2}\right)$

$\Rightarrow \frac{1}{a^2}\frac{d^{2}z}{d{x}^2}=\frac{c^2{z}^2-a^2{x}^2}{c^4{z}^3}$ .....(I)

Now again differentiating with respected to y we get,

$\frac{d^{2}z}{d{y}^2}=\frac{b^2}{c^2}\left(\frac{z-\frac{b^2{y}^2}{c^{2}z}}{z^2}\right)$

$\Rightarrow \frac{1}{b^2}\frac{d^{2}z}{d{y}^2}=\frac{1}{c^2}\left(\frac{z-\frac{b^2{y}^2}{c^{2}z}}{z^2}\right)$

$\Rightarrow \frac{1}{b^2}\frac{d^{2}z}{d{y}^2}=\frac{c^2{z}^2-b^2{y}^2}{c^4{z}^3}$ .....(II)

So on adding equations (I) and (II) we get,

$\Rightarrow \frac{1}{a^2}\frac{d^{2}z}{d{x}^2}+\frac{1}{b^2}\frac{d^{2}z}{d{y}^2}=\frac{2c^2{z}^2-a^2{x}^2-b^2{y}^2}{c^4{z}^3}$

$\Rightarrow \frac{1}{a^2}\frac{d^{2}z}{d{x}^2}+\frac{1}{b^2}\frac{d^{2}z}{d{y}^2}=\frac{2c^2{z}^2-c^2{z}^2}{c^4{z}^3}$

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