Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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 \(f(x,y) = x^3+3xy+y^2-4\\f(0,0) =-4\)

Answer 6

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

Answer 7

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

Answer 8

Yeah, that method looked weird to me

Answer 9

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

Answer 10

@ganeshie8

Answer 11

Thank you, let's wait for him.

Answer 12

-36 is correct using approximation formula
-4 is correct by solving the function

depends on what you want

Answer 13

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

Answer 14

Thats the reason it is called "approximation" formula

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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...

Answer 21

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)=??

Answer 22

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

Answer 23

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

Answer 24

f(x,y) = x^3+xy +y^2 +4

f_x = 3x^2 + y

this is wrong

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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.

Answer 29

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

Answer 30

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

Answer 31

Never said it is invalid.

Answer 32

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"

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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).