If fx-y-z - x2-y2-z2 12 then - 2f- x2- 2f- y2- 2f- z2 is equal to

F(x, y, z) = ( x^2+y^2+z^2 )^ 12 ∂ f∂ x= 12 ( x^2+y^2+z^2 )^ 3/2(2x) ∂ f∂ x= x ( x^2+y^2+z^2 )^ 3/2 ∂^2f ∂ x^2= x ( 32 ) ( x^2+y^2+z^2 )^ 5/2× (2x) ( x^2+ ...

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Engineering Mathematics

Homogeneous Linear Differential Equations (General Form of LDE)

If fx,y,z = x...

Question

If f(x,y,z) = ${(x^{2} + y^{2} + z^{2})}^{\frac{1}{2}}$ then $\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2} f}{\partial z^{2}}$ is equal to

zero

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-3

{(x^{2} + y^{2} + z^{2})}^{\frac{5}{2}}

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Solution

The correct option is A zero
f(x, y, z) = ${(x^{2} + y^{2} + z^{2})}^{\frac{- 1}{2}}$

$\frac{\partial f}{\partial x} = - \frac{1}{2} {(x^{2} + y^{2} + z^{2})}^{- 3 / 2} (2 x)$

$\frac{\partial f}{\partial x} = - x {(x^{2} + y^{2} + z^{2})}^{- 3 / 2}$

$\frac{\partial^{2} f}{\partial x^{2}} = - x (\frac{- 3}{2}) {(x^{2} + y^{2} + z^{2})}^{- 5 / 2} \times$

(2x)- ${(x^{2} + y^{2} + z^{2})}^{- 3 / 2}$

$\frac{\partial^{2} f}{\partial x^{2}} = 3 x^{2} {(x^{2} + y^{2} + z^{2})}^{- 5 / 2} - {(x^{2} + y^{2} + z^{2})}^{- 3 / 2}$

Similarly

Z = f(x,y)

$\frac{\partial^{2} f}{\partial y^{2}} = 3 y^{2} (x^{2} + y^{2} + z^{2})^{- 5 / 2} - (x^{2} + y^{2} + z^{2})^{- 3 / 2}$

$\frac{\partial^{2} f}{\partial z^{2}} = 3 z^{2} (x^{2} + y^{2} + z^{2})^{- 5 / 2} - (x^{2} + y^{2} + z^{2})^{- 3 / 2}$

$\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2} f}{\partial z^{2}} = 3 (x^{2} + y^{2} + z^{2})$ $(x^{2} + y^{2} + z^{2})^{- 5 / 2} - 3 (x^{2} + y^{2} + z^{2})^{- 3 / 2} = 0$

Method II : Let $r^{2} = x^{2} + y^{2} = z^{2}$

Differentiating (1) partially w.r.t 'x'

$\frac{\partial}{\partial} (r^{2}) = \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (y^{2}) + \frac{\partial}{\partial x} (z^{2})$

$2 r \frac{\partial r}{\partial} = 2 x + 0 + 0$

i.e, $\frac{\partial r}{\partial x} = \frac{x}{r}$

Similarly, $\frac{\partial r}{\partial y} = \frac{y}{r} a n d \frac{\partial r}{\partial z} = \frac{z}{r}$

Now $f = (x^{2} + y^{2} + z^{2})^{- 1 / 2} = (r^{2})^{- 1 / 2} = \frac{1}{r}$

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (\frac{1}{r}) = - \frac{1}{r^{2}} \frac{\partial r}{\partial x}$

$= - \frac{1}{r^{2}} (\frac{x}{r}) = - \frac{x}{r^{3}}$

$\frac{\partial^{2} f}{\partial x^{2}} = \frac{\partial}{\partial x} (\frac{\partial f}{\partial x}) = \frac{\partial}{\partial x} [- \frac{x}{r^{3}}]$

$= - ⎡ ⎢ ⎢ ⎢ ⎣ \frac{r^{3} \frac{\partial}{\partial x} (x) - x \frac{\partial}{\partial x} (r^{3})}{r^{6}} ⎤ ⎥ ⎥ ⎥ ⎦$

$= - ⎡ ⎢ ⎣ \frac{r^{3} - x .3 r^{2} \frac{\partial r}{\partial x}}{r^{6}} ⎤ ⎥ ⎦$

$= - ⎡ ⎢ ⎢ ⎣ \frac{r^{3} - 3 r^{2} . x (\frac{x}{r})}{r^{6}} ⎤ ⎥ ⎥ ⎦ = - [\frac{1}{r^{3}} - \frac{3 x^{2}}{r^{5}}]$

Similarly $\frac{\partial^{2} f}{\partial y^{2}} = - [\frac{1}{r^{3}} - \frac{3 y^{2}}{r^{5}}]$

and $\frac{\partial^{2} f}{\partial z^{2}} = - [\frac{1}{r^{3}} - \frac{3 z^{2}}{r^{5}}]$

Now $\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2} f}{\partial z^{2}} = - \frac{3}{r^{3}} + \frac{3 (x^{2} + y^{2} + z^{2})}{r^{5}}$

$= - \frac{3}{r^{3}} + \frac{3}{r^{3}} = 0$

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If f(x,y,z) = (x2+y2+z2)12 then ∂2f∂x2+∂2f∂y2+∂2f∂z2 is equal to

If f(x,y,z) = ${(x^{2} + y^{2} + z^{2})}^{\frac{1}{2}}$ then $\frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2} f}{\partial z^{2}}$ is equal to