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Partial derivatives etc

  1. May 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose z=ψ(2x-3y), Show that the second partial derivative of z with respect to x, is equal to the second partial derivative with respect to y multiplied by a scalar k.

    2. Relevant equations



    3. The attempt at a solution

    I thought this was too simple to be correct, and apparently it was. I just use the chain rule and found the first partial derivative with respect to x as 2ψ and with respect to y as -3ψ, hence both the second partial derivatives will be 0, and hence the scalar k is just 0. But apparently this is incorrect, apparently the scalar k is 4/9. I'm not exactly sure why, but our maths teacher told us to redefine z in terms of u, and then make u depend on x and y. Any help would be appreciated. Cheers.
     
  2. jcsd
  3. May 11, 2014 #2
    Is it because we can't assume that psi is not dependent on x and hence we have to redefine a function u which is only dependent on x and y, and use the chain rule to solve that zx=zu*ux?
     
  4. May 11, 2014 #3
    Your first partial involves ψ. So, when you take the second partial with respect to x, you need to take the partial of the ψ in the first partial with respect to x again. Same for y.

    Chet
     
  5. May 11, 2014 #4
    But don't you treat all the other variables as constant when you are doing partial derivatives?
     
  6. May 11, 2014 #5

    haruspex

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    You treat all other independent variables as constant. x and y are independent, but ψ depends on x and y.
     
  7. May 11, 2014 #6
    How do you know that, they're just a bunch of undefined variables. Are we just supposed to make the assumption?
     
  8. May 11, 2014 #7
    And I commented above I thought this may be the case, but it seems stupid that they've left the variables undefined not showing what they're dependent on, and just making you assume, without knowing if psi is dependent on x and y.
     
  9. May 11, 2014 #8
    Nope I can't figure it out. The answer says find psi', but it has evaluated psi' for both the x partial and y partial, saying they're the same thing, but on is psi with respect to x and one is psi with respect to y goddamit!
     
  10. May 11, 2014 #9
    Like i've sampled the answer by chucking psi=x+y and evaluating it and it doesn't work, stupid question is pissing me off now.
     
  11. May 11, 2014 #10
    Try thinking about it by making the substitution η=2x-3y, so that z = ψ(η). So, by the chain rule,

    [tex]\frac{∂ψ}{∂x}=\frac{dψ}{dη}\frac{∂η}{∂x}=ψ'(2)[/tex]

    Similarly,

    [tex]\frac{∂^2ψ}{∂x^2}=2\frac{d^2ψ}{dη^2}\frac{∂η}{∂x}=4ψ''[/tex]

    Hope this helps.

    Chet
     
  12. May 11, 2014 #11
    Yes that definitely does help, but it seems really weird haha. Thanks.
     
  13. May 11, 2014 #12
    Wait wait wait, but seeing as psi and eta are both dependent on x, doesn't this mean that z partial with x would require the product rule and it still doesn't work.
     
  14. May 11, 2014 #13
    Also how can you assume that psi is only dependent on eta, what if psi isn't a function that is a multiple of eta, then psi is a function of x and y as well.
     
  15. May 11, 2014 #14
    All we've been taught is the chain rule, and if i just draw z as a function of variables psi and eta, then eta is a function of x and y, and psi is a function of eta, but it can't just be a function of eta, because if it's not a scalar multiple of eta then you have to take on the difference in terms of x and y, so psi is of eta and x or y, and eta is a function of x and y so you end up with a stupid tree which doesn't even give me the right answer.
     
  16. May 11, 2014 #15
    Here's the file, question 6. The answer just makes no sense to me.
     

    Attached Files:

  17. May 11, 2014 #16
    If you put psi=x, and you solve for zxx and zyy, it is clearly not a scalar multiple of their suggested k out. Stupid question!
     
  18. May 11, 2014 #17

    haruspex

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    No, ψ is a function specifying how the variable z depends on η. z is not "a function of ψ".
    Yes it is. For any given value of η, the function ψ applied to that value tells you the value of z.
    sorry, you've lost me there.
    ψ is a function, not a function of a specific variable. z is a dependent variable, a function of η.
    E.g. you could have another dependent variable w = ψ(y). Same function, different variables.
     
  19. May 11, 2014 #18
    So is it just analogous to:

    y=f(x)?

    Not z=psi*(2x-3y)=psi2x-psi3y?
     
  20. May 11, 2014 #19
    If so why the heck didn't they just write z=f(2x-3y)!
     
  21. May 11, 2014 #20
    Suppose, for example, ψ(η)=η2+2η+1=(2x-3y)2+2(2x-3y)+1

    Then [tex]\frac{∂ψ}{∂x}=\frac{dψ}{dη}\frac{∂η}{∂x}[/tex]

    [tex]\frac{dψ}{dη}=ψ'=2η+2=2(2x-3y)+2[/tex]
    [tex]\frac{∂η}{∂x}=2[/tex]
    So,
    [tex]\frac{∂ψ}{∂x}=4η+4=4(2x-3y)+4[/tex]

    See what you get when you evaluate it by taking the partial derivative of (2x-3y)2+2(2x-3y)+1 with respect to x.

    Chet
     
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