How does one do this? | Mathematics | Page 1

noble rampart Apr 7, 2025, 4:55 PM

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I thought you take the find the integral for both.. and add them

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unique rapidsBOT Apr 7, 2025, 4:55 PM

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shell shale Apr 7, 2025, 5:02 PM

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For the first one you gotta integrate correctly in the first place

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Integral of x+6 wrt x doesn't get you x^2+6x

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And it's not exactly adding the two integrals you should do

noble rampart Apr 7, 2025, 5:05 PM

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i meann i thought you had to work backwards

shell shale Apr 7, 2025, 5:05 PM

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For the second one, ok you integrate 4x+4y wrt x, you can get 2x^2+4xy**+f(y) for some function f**

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4x+5y wrt y, you get 4xy+5/2y^2**+g(x) for some function g**

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Then if you can find f and g such that 2x^2+4xy+f(y) = 4xy+5/2y^2+g(x) then gg you got yourself a potential function

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It's really matching the terms between the two integrals, not just plain adding

noble rampart Apr 7, 2025, 5:17 PM

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OH so cause the 4xys cancel, the potential is 2x^2+ 5/2Y^2

shell shale Apr 7, 2025, 5:27 PM

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No they don't cancel

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You want the 2 sides to be equal for all x,y

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So if you pick f(y)=5/2y^2, g(x)=2x^2 you manage to match the two expressionz

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This expression you get in the end is your potential function

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So 2x^2 + 4xy + 5/2y^2

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You can check that the two partial derivatives are what you need

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@noble rampart

#How does one do this?