posted inside:)

Question

posted inside:)

anonymous · Answer

r=_______

?? thanks!!

phi · Answer

what have you tried ?

anonymous · Answer

i think you need to find the derivative right? 

but i'm unsure because there's the r^3 ' at the denominator on the second term and that's throwing me off :/

phi · Answer

Yes, finding a critical point of a function means  find the derivative and set equal to zero.

I would write the equation as
$$ f(x) = -A r^{-2} + B r^{-3} $$

anonymous · Answer

ahh okay but what about that ' symbol after r^3 ? :/

anonymous · Answer

does that affect the function? :/

phi · Answer

that is a comma, used to separate the equation from the rest of the sentence.  Commas are often a helpful grammatical device :)

anonymous · Answer

ohh okay haha :P

that makes sense now lol :P

so the derivative would be $$f'(x)=\frac{ 2ar-3b }{ r ^{4} }$$ ?

anonymous · Answer

the comma totally threw me off... haha i thought it was asking for some odd derivative rule that i hadn't heard about before lol :P

oops!!

phi · Answer

It helps if you ever read (or better, write) papers with equations.  Then you notice the punctuation marks.

I assume you can finish this ?

anonymous · Answer

ahh okay i'll have to google papers with equations haha :P

umm like this?

2ar-3b
------    =   0
  r^4

multiply both sides by r^4

you get 2ar-3b = 0
                +3b    +3b
--------------------
           2ar=3b
divide by 2a ?

r= 3b
    ----
      2a 

?? is that right? :/

anonymous · Answer

is that the value that minimizes? or is that the max? :/

phi · Answer

that is what I got

anonymous · Answer

okay:)

so is that the answer? when r>0, 3b/2a is the value that minimizes the force between the atoms? :O

phi · Answer

to test if it's a max or min I often try a few numbers.  Let A=B = 1 (keep it simple!)
then try r below 3/2, at 3/2 and above 3/2
it should be clear if we have a max or a min.

anonymous · Answer

okay, how do i do that? :/ i'm a bit confused by how to test :/

phi · Answer

$$ f(x) = -1/r^2 + 1/r^3 $$

the min (?) occurs at r= 3A/2B= 3/2
we get -4/9 + 8/27 = -4/27

anonymous · Answer

okay:)

so that's for the min:)

so the min=3/2

or would it still be 3A/2B for my problem?

phi · Answer

on the other hand, at r=1  (which is below r=3/2 = 1.5)  we get -1 + 1 =0 (bigger!)
and at r= 2 we get -¼ + ⅛ = -⅛ = -0.125 >  -4/27 (-0.148...)

phi · Answer

so r=3/2 is a minimum

anonymous · Answer

ahh okay:)
awesome!! thank you!!! :D