October 17th, 2016, 04:52 AM  #1 
Newbie Joined: Oct 2016 From: United Kingdom Posts: 13 Thanks: 0  Stationary Points
The potential energy of two molecules separated by a distance rr is given by U=U0((a/r)12−2(a/r)6)U=U0((a/r)12−2(a/r)6) where U0U0 and aa are positive constants. The equilibrium separation of the two molecules occurs when the potential energy is a minimum; find the equilibrium separation and give the value of the potential energy at this separation if U0=6.62×10−21JU0=6.62×10−21J. i dont know how to start this question: so any help is greatly appreciated! 
October 17th, 2016, 06:21 AM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555  Quote:
$U=U_0\bigg[\left(\dfrac{a}{r}\right)^{12}2\left(\dfrac{a}{r}\right)^6 \bigg]$ In any case, start by finding where $\dfrac{dU}{dr}=0$ to locate possible extrema.  
October 17th, 2016, 06:47 AM  #3 
Newbie Joined: Oct 2016 From: United Kingdom Posts: 13 Thanks: 0 
So where would i go from there?

October 17th, 2016, 07:07 AM  #4 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions  
October 17th, 2016, 07:08 AM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555 
once you find the value(s) of $r$ that make $\dfrac{dU}{dr}=0$, you can use the first derivative test or the second derivative test to classify the extrema.

October 17th, 2016, 07:18 AM  #6 
Newbie Joined: Oct 2016 From: United Kingdom Posts: 13 Thanks: 0 
How would i find r (really sorry im not very good at maths)

October 17th, 2016, 07:25 AM  #7 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Hmmm... Have a go at the following question Let: $\displaystyle y = 3x^{12} + 5x^6$ Can you: a) find $\displaystyle \frac{dy}{dx}$; then b) find the xvalue where $\displaystyle \frac{dy}{dx} = 0$? If you find this question tricky, we can help you with this 
October 17th, 2016, 07:27 AM  #8 
Math Team Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555  
October 17th, 2016, 07:39 AM  #9  
Newbie Joined: Oct 2016 From: United Kingdom Posts: 13 Thanks: 0  Quote:
how would i do the next part?  
October 17th, 2016, 07:41 AM  #10 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions  

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