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Loser66

  • one year ago

Given a function on the xy plane. f(x,y) such that f(2,2) = 20 and \(\dfrac{\partial f}{\partial x}=3x^2+3y\) , \(\dfrac{\partial f}{\partial y}= 3x+2y\). Then f(0,0)=?? Please, help

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  1. Loser66
    • one year ago
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    My attempt: \(f(0,0) -f(2,2) = \triangle f = 3x^2+3y+3x+2y\) At (2,2) , \(\triangle f = 28\), hence f(0,0) = 48, But it is very wrong. hahaha.... I don't know how to fix

  2. ganeshie8
    • one year ago
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    use approximation formula ? looks im getting \(\Delta f \approx f_x\Delta x+f_y\Delta y = (18)*(-2) + (10)*(-2) = -56\)

  3. Loser66
    • one year ago
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    The options are 1) 0 2) -1 3) -2 4) -3 5) -4

  4. ganeshie8
    • one year ago
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    \(f(0,0) =f(2,2)+\Delta f \approx 20 - 56 = -36\)

  5. Loser66
    • one year ago
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    Another thought: \(\dfrac{\partial f}{\partial x}=3x^2 +3y\\f(x,y)= x^3+3xy+h(y)\\\dfrac{\partial f}{\partial y}= 3x+h'(y) = 3x+2y\\h'(y) = 2y\\h(y) =y^2+C\) Hence \(f(x,y) = x^3+3xy +y^2+C \\f(2,2) = 2^3+3*2*2+2^2+C=20\\C= -4\) therefore \(f(x,y) = x^3+3xy+y^2-4\\f(0,0) =-4\)

  6. Astrophysics
    • one year ago
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    Interesting, so that's like finding the potential function haha.

  7. Loser66
    • one year ago
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    Yes, because from approximating method, I don't know how to check whether my logic is right or wrong.

  8. Astrophysics
    • one year ago
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    Yeah, that method looked weird to me

  9. Astrophysics
    • one year ago
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    -4 looks good but I'm not sure...@ganeshie8

  10. Astrophysics
    • one year ago
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    @ganeshie8

  11. Loser66
    • one year ago
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    Thank you, let's wait for him.

  12. ganeshie8
    • one year ago
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    -36 is correct using approximation formula -4 is correct by solving the function depends on what you want

  13. Loser66
    • one year ago
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    hey, no way!!! -4 is ffffffffffffffar different from -36

  14. ganeshie8
    • one year ago
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    Thats the reason it is called "approximation" formula

  15. Loser66
    • one year ago
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    but the error is so nonsense. If it is -4 and -4.5, I am ok.

  16. ganeshie8
    • one year ago
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    You just got lucky with the present problem because there exists a potential function, this is not the case with most functions and you will end up using approximation formula

  17. Loser66
    • one year ago
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    I know!! that is why I posted the question here. hehehe...

  18. Loser66
    • one year ago
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    Unfortunately, when using app. and the answer is -36, how can I check the correct choice? I don't have -36 options.

  19. ganeshie8
    • one year ago
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    how did u get the idea of finding f(x,y) directly instead of using the approximation formula ?

  20. Loser66
    • one year ago
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    I don't get what you mean by "how". I try all the ways out. I saw partial derivative and I linked it to DE. hehehe...

  21. ganeshie8
    • one year ago
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    Hmm try this : Given a function on the xy plane. f(x,y) such that f(2,2) = 20 and \(\dfrac{\partial f}{\partial x}=3x^2+3y\) , \(\dfrac{\partial f}{\partial y}= x+2y\). Then f(0,0)=??

  22. ganeshie8
    • one year ago
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    If you think of that as vector field, notice that the curl is nonzero. So potential function cannot exist. Let me know if you ever get successful finding \(f(x,y)\) :P

  23. Loser66
    • one year ago
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    as above, f(x,y) = x^3 + 3xy + h(y) f_y = 3x + h'(y) = x+2y h'(y) = -2x+2y h(y) -2xy +y^2 +C f(x,y) =x^3+3xy-2xy+y^2 +C = x^3 +xy +y^2 +C f(2,2)= 8+4+4 +C=20 C= 4 f(x,y) = x^3+xy +y^2 +4 f(0,0) =4

  24. ganeshie8
    • one year ago
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    f(x,y) = x^3+xy +y^2 +4 f_x = 3x^2 + y this is wrong

  25. ganeshie8
    • one year ago
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    It turns out that we cannot find \(f(x,y)\) because there is no such function whose partials with respect to \(x\) and \(y\) are \(3x^2+3y\) and \(x+2y\) respectively

  26. Loser66
    • one year ago
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    Yes, I see it. Thanks for pointing it out.

  27. Loser66
    • one year ago
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    How to solve my original problem and down the answer to one of the options?? Please

  28. ganeshie8
    • one year ago
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    My point is that question should make it clear. question seems to be asking the exact value but your initial attempt focused on finding the approximation.

  29. ganeshie8
    • one year ago
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    you've got f(0,0) = -4 done with the original prblem right ?

  30. Loser66
    • one year ago
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    YOu said that it is invalid. I am just lucky. I don't want it. I want the firm logic.

  31. ganeshie8
    • one year ago
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    Never said it is invalid.

  32. Loser66
    • one year ago
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    above : "You just got lucky with the present problem because there exists a potential function, this is not the case with most functions and you will end up using approximation formula"

  33. ganeshie8
    • one year ago
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    \(f(x,y) = x^3+3xy+y^2-4\) does satisfy all the given conditions, so we're good!

  34. ganeshie8
    • one year ago
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    what i meant was, you cannot solve the function always. recall the definition of exact differential equations or conservative vector fields

  35. Loser66
    • one year ago
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    Ok, got you. Thank you for being patient to me. :)

  36. ganeshie8
    • one year ago
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    It seems we can find the potential function if the mixed partials are equal : \[f_{xy} = f_{yx}\] Eventhough most functions that we encounter directly satisfy this, you can see it is easy to cook up the partials that violate above condition and thus lack solution(s).

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