## 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

1. Loser66

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

use approximation formula ? looks im getting $$\Delta f \approx f_x\Delta x+f_y\Delta y = (18)*(-2) + (10)*(-2) = -56$$

3. Loser66

The options are 1) 0 2) -1 3) -2 4) -3 5) -4

4. ganeshie8

$$f(0,0) =f(2,2)+\Delta f \approx 20 - 56 = -36$$

5. Loser66

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

Interesting, so that's like finding the potential function haha.

7. Loser66

Yes, because from approximating method, I don't know how to check whether my logic is right or wrong.

8. Astrophysics

Yeah, that method looked weird to me

9. Astrophysics

-4 looks good but I'm not sure...@ganeshie8

10. Astrophysics

@ganeshie8

11. Loser66

Thank you, let's wait for him.

12. ganeshie8

-36 is correct using approximation formula -4 is correct by solving the function depends on what you want

13. Loser66

hey, no way!!! -4 is ffffffffffffffar different from -36

14. ganeshie8

Thats the reason it is called "approximation" formula

15. Loser66

but the error is so nonsense. If it is -4 and -4.5, I am ok.

16. ganeshie8

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

I know!! that is why I posted the question here. hehehe...

18. Loser66

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

19. ganeshie8

how did u get the idea of finding f(x,y) directly instead of using the approximation formula ?

20. Loser66

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

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

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

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

f(x,y) = x^3+xy +y^2 +4 f_x = 3x^2 + y this is wrong

25. ganeshie8

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

Yes, I see it. Thanks for pointing it out.

27. Loser66

How to solve my original problem and down the answer to one of the options?? Please

28. ganeshie8

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

you've got f(0,0) = -4 done with the original prblem right ?

30. Loser66

YOu said that it is invalid. I am just lucky. I don't want it. I want the firm logic.

31. ganeshie8

Never said it is invalid.

32. Loser66

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

$$f(x,y) = x^3+3xy+y^2-4$$ does satisfy all the given conditions, so we're good!

34. ganeshie8

what i meant was, you cannot solve the function always. recall the definition of exact differential equations or conservative vector fields

35. Loser66

Ok, got you. Thank you for being patient to me. :)

36. ganeshie8

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).