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

- katieb

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- 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

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

- Loser66

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

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## More answers

- ganeshie8

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

- 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

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

- Loser66

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

- Astrophysics

Yeah, that method looked weird to me

- Astrophysics

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

- Astrophysics

- Loser66

Thank you, let's wait for him.

- ganeshie8

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

- Loser66

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

- ganeshie8

Thats the reason it is called "approximation" formula

- Loser66

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

- 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

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

- Loser66

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

- ganeshie8

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

- 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

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

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

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

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

- 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

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

- Loser66

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

- 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

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

- Loser66

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

- ganeshie8

Never said it is invalid.

- 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

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

- ganeshie8

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

- Loser66

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

- 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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