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Multi Calculus- Partial Derivatives

  1. Mar 19, 2010 #1
    1. The problem statement, all variables and given/known data

    I am translating the question from another language so it might not be word to word with the original question.

    assume x(s,t) and y(s,t) determined by these two functions:

    sin(xt) +x+s=1
    eyt+y(s+1)=1

    The following function is defined H(x,y)=x2+y2-xy
    such that
    K(s,t)=H(x(s,t),y(s,t))

    find the partial derivative Kt(0,0) (the derivative of K in terms of t)

    2. Relevant equations



    3. The attempt at a solution

    first I wanted to solve the first two equation system in terms of x and y, but it didn't work.
    tried also plugging them in to H, no success.

    Would appreciate any help.
    Roni
     
  2. jcsd
  3. Mar 19, 2010 #2

    gabbagabbahey

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    Start by applying the chain rule to [itex]H[/itex]...
     
  4. Mar 19, 2010 #3
    what do you mean by the chain rule ?

    should I first replace the x,y with s,t ?
    I don't really see the lead ....
    would appreciate if you could give any more hints but I'm going to try it also
     
  5. Mar 19, 2010 #4
    I tried the chain rule, but looks like I'm doing something wrong.

    Kt=Kx*Xt+Ky*Yt

    I am assuming K(x,y)=H(x,y), a correct assumption ?
    I am getting 0 and that's not the answer.
    Kx=2x-y
    Ky=2y-x

    no matter what I multiply these by, I must get a zero but the answer is not zero.
    So I guess I am missing something
     
  6. Mar 19, 2010 #5

    gabbagabbahey

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    Well, sort of. [itex]K(s,t)[/itex] is explicitly a function of [itex]s[/itex] and [itex]t[/itex] only. You know that [itex]x[/itex] and [itex] y[/itex] are also some function of [itex]s[/itex] and [itex]t[/itex] and that

    [tex]K(s,t)=H(x(s,t),y(s,t))=[x(s,t)]^2+[y(s,t)]^2+x(s,t)y(s,t)[/itex]

    So, it is more correct to say

    [tex]K_t(s,t)=H_x x_t+H_y y_t[/itex]

    Well, [itex]H_x=2x-y[/itex] and [itex]H_y=2y-x[/itex]....but what about [itex]x_t[/itex] and [itex]y_t[/itex]....what are you getting for those?
     
  7. Mar 19, 2010 #6
    Xt=-x*cos(xt)
    Yt=(-yeyt)/(t*eyt+s+1)

    But it doesnt matter what I'm getting for these because I multiply each one of them by a zero.
     
  8. Mar 19, 2010 #7
    ohh hold on, the s and t are zeros , so it doesnt mean x and y are zeros
    wow

    Edit:
    got the answer,

    thanks for the help
     
  9. Mar 19, 2010 #8

    gabbagabbahey

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    Your expression for [itex]x_t[/itex] doesn't look quite right.

    I'm not sure why you think this. [itex]H_x=2x-y\neq 0[/itex] and [itex]H_y=2y-x\neq 0[/itex]
     
  10. Mar 19, 2010 #9
    oh, it didn't influence my answer so I didn't see the mistake

    I think it's gotta be
    -x*cos(xt)/(t*cos(xt)+1)

    Thank you.
     
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