Loser66
  • Loser66
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Loser66
  • 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
ganeshie8
  • ganeshie8
use approximation formula ? looks im getting \(\Delta f \approx f_x\Delta x+f_y\Delta y = (18)*(-2) + (10)*(-2) = -56\)
Loser66
  • Loser66
The options are 1) 0 2) -1 3) -2 4) -3 5) -4

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More answers

ganeshie8
  • ganeshie8
\(f(0,0) =f(2,2)+\Delta f \approx 20 - 56 = -36\)
Loser66
  • 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 \(f(x,y) = x^3+3xy+y^2-4\\f(0,0) =-4\)
Astrophysics
  • Astrophysics
Interesting, so that's like finding the potential function haha.
Loser66
  • Loser66
Yes, because from approximating method, I don't know how to check whether my logic is right or wrong.
Astrophysics
  • Astrophysics
Yeah, that method looked weird to me
Astrophysics
  • Astrophysics
-4 looks good but I'm not sure...@ganeshie8
Astrophysics
  • Astrophysics
@ganeshie8
Loser66
  • Loser66
Thank you, let's wait for him.
ganeshie8
  • ganeshie8
-36 is correct using approximation formula -4 is correct by solving the function depends on what you want
Loser66
  • Loser66
hey, no way!!! -4 is ffffffffffffffar different from -36
ganeshie8
  • ganeshie8
Thats the reason it is called "approximation" formula
Loser66
  • Loser66
but the error is so nonsense. If it is -4 and -4.5, I am ok.
ganeshie8
  • 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
Loser66
  • Loser66
I know!! that is why I posted the question here. hehehe...
Loser66
  • Loser66
Unfortunately, when using app. and the answer is -36, how can I check the correct choice? I don't have -36 options.
ganeshie8
  • ganeshie8
how did u get the idea of finding f(x,y) directly instead of using the approximation formula ?
Loser66
  • 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...
ganeshie8
  • 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)=??
ganeshie8
  • 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
Loser66
  • 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
ganeshie8
  • ganeshie8
f(x,y) = x^3+xy +y^2 +4 f_x = 3x^2 + y this is wrong
ganeshie8
  • 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
Loser66
  • Loser66
Yes, I see it. Thanks for pointing it out.
Loser66
  • Loser66
How to solve my original problem and down the answer to one of the options?? Please
ganeshie8
  • 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.
ganeshie8
  • ganeshie8
you've got f(0,0) = -4 done with the original prblem right ?
Loser66
  • Loser66
YOu said that it is invalid. I am just lucky. I don't want it. I want the firm logic.
ganeshie8
  • ganeshie8
Never said it is invalid.
Loser66
  • 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"
ganeshie8
  • ganeshie8
\(f(x,y) = x^3+3xy+y^2-4\) does satisfy all the given conditions, so we're good!
ganeshie8
  • ganeshie8
what i meant was, you cannot solve the function always. recall the definition of exact differential equations or conservative vector fields
Loser66
  • Loser66
Ok, got you. Thank you for being patient to me. :)
ganeshie8
  • 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).

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