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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
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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Loser66
 one year ago
Best ResponseYou've already chosen the best response.1My 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
 one year ago
Best ResponseYou've already chosen the best response.2use approximation formula ? looks im getting \(\Delta f \approx f_x\Delta x+f_y\Delta y = (18)*(2) + (10)*(2) = 56\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1The options are 1) 0 2) 1 3) 2 4) 3 5) 4

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(f(0,0) =f(2,2)+\Delta f \approx 20  56 = 36\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Another 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 \(f(x,y) = x^3+3xy+y^24\\f(0,0) =4\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Interesting, so that's like finding the potential function haha.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Yes, because from approximating method, I don't know how to check whether my logic is right or wrong.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, that method looked weird to me

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.04 looks good but I'm not sure...@ganeshie8

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Thank you, let's wait for him.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.236 is correct using approximation formula 4 is correct by solving the function depends on what you want

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1hey, no way!!! 4 is ffffffffffffffar different from 36

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Thats the reason it is called "approximation" formula

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1but the error is so nonsense. If it is 4 and 4.5, I am ok.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2You 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
 one year ago
Best ResponseYou've already chosen the best response.1I know!! that is why I posted the question here. hehehe...

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Unfortunately, when using app. and the answer is 36, how can I check the correct choice? I don't have 36 options.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2how did u get the idea of finding f(x,y) directly instead of using the approximation formula ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.2Hmm 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
 one year ago
Best ResponseYou've already chosen the best response.2If 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
 one year ago
Best ResponseYou've already chosen the best response.1as 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+3xy2xy+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
 one year ago
Best ResponseYou've already chosen the best response.2f(x,y) = x^3+xy +y^2 +4 f_x = 3x^2 + y this is wrong

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2It 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
 one year ago
Best ResponseYou've already chosen the best response.1Yes, I see it. Thanks for pointing it out.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1How to solve my original problem and down the answer to one of the options?? Please

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2My 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
 one year ago
Best ResponseYou've already chosen the best response.2you've got f(0,0) = 4 done with the original prblem right ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1YOu said that it is invalid. I am just lucky. I don't want it. I want the firm logic.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Never said it is invalid.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1above : "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
 one year ago
Best ResponseYou've already chosen the best response.2\(f(x,y) = x^3+3xy+y^24\) does satisfy all the given conditions, so we're good!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2what i meant was, you cannot solve the function always. recall the definition of exact differential equations or conservative vector fields

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Ok, got you. Thank you for being patient to me. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2It 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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