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How to find out [itex]\frac{\partial^{2} y}{\partial t^{2}} [/itex]?

  1. Nov 9, 2012 #1
    Let

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

    Now I can find out the value of

    [itex] \frac{\partial y}{\partial t} [/itex]

    but I don't know how to find out the

    [itex] \frac{\partial ^{2} y}{\partial t^{2}} [/itex]

    I know that this is equal to

    [itex] \frac{d^{2}F}{dz^{2}} (-v)^2 [/itex]


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

    [itex] \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
    [/itex]

    Just tell me if the above method is correct

    Thanks a lot
     
    Last edited: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2

    Simon Bridge

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    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.
     
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