$$r=\sqrt{x^2+y^2+z^2}$$
$$u=\frac{1}{r}f(r-ct)$$
where c is constant
show that
$$\frac{\partial^2u }{\partial t^2}=c^2\left ( \frac{\partial^2 u}{\partial x^2} +\frac{\partial^2u }{\partial y^2}+\frac{\partial^2 u}{\partial z^2}\right )$$

Question

$$r=\sqrt{x^2+y^2+z^2}$$
$$u=\frac{1}{r}f(r-ct)$$
where c is constant
show that
$$\frac{\partial^2u }{\partial t^2}=c^2\left ( \frac{\partial^2 u}{\partial x^2} +\frac{\partial^2u }{\partial y^2}+\frac{\partial^2 u}{\partial z^2}\right )$$

ujjwal · Answer

@ganeshie8 @mathslover @myininaya

ujjwal · Answer

@Destinymasha

mathslover · Answer

This seems complicated.. No idea :/ Sorry.

ujjwal · Answer

@mathslover , Thank you anyways! :)

hartnn · Answer

did u try ?

hartnn · Answer

$\Large \dfrac{\partial u}{\partial x} = ... ?$

hartnn · Answer

chain rule will be very useful here

hartnn · Answer

oh, and 
$\Large \dfrac{\partial u}{\partial t} = $
would be more simpler here
because 'r' will be treated constant

ujjwal · Answer

r is not a constant.

hartnn · Answer

when partially differentiating w.r.t 't', every other variable is treated as constant.

ujjwal · Answer

yep!

ujjwal · Answer

wasn't that difficult upon actually trying! 
Actually i was just too lazy to continue just coz it was a bit lengthy.
Anyways thanks guys! ;)

hartnn · Answer

welcome ^_^