# How to find out $\frac{\partial^{2} y}{\partial t^{2}}$?

1. Nov 9, 2012

### iVenky

Let

y=F(z) where z=x-vt

Now I can find out the value of

$\frac{\partial y}{\partial t}$

but I don't know how to find out the

$\frac{\partial ^{2} y}{\partial t^{2}}$

I know that this is equal to

$\frac{d^{2}F}{dz^{2}} (-v)^2$

I tried to solve this in this way- just tell me if this is correct

$\frac{ \partial y}{ \partial t}= \frac{dF}{ dz} \frac{ \partial z}{ \partial t}=\frac{dF}{dz} (-v) \\ \frac{ \partial^{2} y}{ \partial t^{2}}= \frac{\partial }{\partial t} \frac{dF}{dz} (-v)= \frac{d}{dz}\frac{\partial z}{\partial t} \frac{dF}{dz} (-v) = \frac{d^{2}F}{dz^{2}} (-v)^2$

Just tell me if the above method is correct

Thanks a lot

Last edited: Nov 9, 2012
2. Nov 9, 2012

### Simon Bridge

Let me just tidy that up a bit, and, in the process, make sure I understand you ;)

You have $y(x,t)=F(z):z=x-vt$ and you want $$\frac{ \partial^{2} y}{ \partial t^{2}}$$... so you tried:
... that looks right to me (provided v is a constant with t). It will probably help with your thinking if you put: $$G(z)=-v\frac{dF}{dz} \Rightarrow \frac{ \partial G}{ \partial t}= \frac{dG}{ dz} \frac{ \partial z}{ \partial t}=-v\frac{dG}{dz}$$... then substitute back.