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Partial derivatives and small changes

Hi,
I’m confused how the answer to part ii of this question is 0.01, when I work through it I get a function of m1, m2 and r?

Here is the question and my working
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Reply 1
Each of the two masses are multiplied by 1.01, so it must be the same for r for the force to be the same. So thats the fractional change (1.01-1).
Reply 2
Original post by mqb2766
Each of the two masses are multiplied by 1.01, so it must be the same for r for the force to be the same. So thats the fractional change (1.01-1).


Do you not need to do any working on the equation for that? Like how do you know r changes proportionally?
Reply 3
Original post by mqb2766
Each of the two masses are multiplied by 1.01, so it must be the same for r for the force to be the same. So thats the fractional change (1.01-1).


Is it that you just use the equation normally actually? I don’t know why but I thought you would use the differential expression because that is part i
(edited 10 months ago)
Reply 4
Original post by grhas98
Do you not need to do any working on the equation for that? Like how do you know r changes proportionally?


It should be almost obvious? (1.01)^2 on both the num and denom to get the same force. You could put some working in, but ...

WHat diff expression did you get in i)?
(edited 10 months ago)
Reply 5
Original post by mqb2766
It should be almost obvious? (1.01)^2 on both the num and denom to get the same force. You could put some working in, but ...

WHat diff expression did you get in i)?


Well yea actually looking at F is is pretty obviously now I just thought to use the diff expression
dF = -Gm1m2/r^3 dr + Gm2/r^2 dm1 + Gm1/r^2 dm2 because of pary i!
Reply 6
Also for part b could you tell me if there is something wrong with my working, the answer is apparently -2root2c/9 but I get -4c/9 and I’ve looked over the working many times but can’t find an issue unless I am misinterpreting r^

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Reply 7
Original post by grhas98
Well yea actually looking at F is is pretty obviously now I just thought to use the diff expression
dF = -Gm1m2/r^3 dr + Gm2/r^2 dm1 + Gm1/r^2 dm2 because of pary i!


If you did use that and dF = 0 then
2 dr/r = dm1 / m1 + dm2 / m2 = 0.01 + 0.01
dr/r is the fractional change and similarly for the masses.
Reply 8
Original post by mqb2766
If you did use that and dF = 0 then
2 dr/r = dm1 / m1 + dm2 / m2 = 0.01 + 0.01
dr/r is the fractional change and similarly for the masses.


Ah I see, yes I was saying that dm1 = 0.01 but of course it is related to the current value of m1 thank you!
Reply 9
For b, youd use the unit vector for the direction in the dot product which gives a 1/sqrt(2)
(edited 10 months ago)

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