My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Reply
 
LinkBack Thread Tools Display Modes
October 17th, 2016, 05:52 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!
Tom2000 is offline  
 
October 17th, 2016, 07:21 AM   #2
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,780
Thanks: 1431

Quote:
Originally Posted by Tom2000 View Post
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.
Wading through your double entries ... is this your function?

$U=U_0\bigg[\left(\dfrac{a}{r}\right)^{12}-2\left(\dfrac{a}{r}\right)^6 \bigg]$

In any case, start by finding where $\dfrac{dU}{dr}=0$ to locate possible extrema.
skeeter is offline  
October 17th, 2016, 07:47 AM   #3
Newbie
 
Joined: Oct 2016
From: United Kingdom

Posts: 13
Thanks: 0

So where would i go from there?
Tom2000 is offline  
October 17th, 2016, 08:07 AM   #4
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 2,132
Thanks: 717

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
Originally Posted by Tom2000 View Post
So where would i go from there?
Find the radius at the max/min and resubstitute it into the formula.
Benit13 is offline  
October 17th, 2016, 08:08 AM   #5
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,780
Thanks: 1431

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.
skeeter is offline  
October 17th, 2016, 08:18 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)
Tom2000 is offline  
October 17th, 2016, 08:25 AM   #7
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 2,132
Thanks: 717

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
Originally Posted by Tom2000 View Post
How would i find r (really sorry im not very good at maths)
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 x-value where $\displaystyle \frac{dy}{dx} = 0$?

If you find this question tricky, we can help you with this
Benit13 is offline  
October 17th, 2016, 08:27 AM   #8
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,780
Thanks: 1431

Quote:
Originally Posted by Tom2000 View Post
How would i find r (really sorry im not very good at maths)
what did you get for $\dfrac{dU}{dr}$ ?
skeeter is offline  
October 17th, 2016, 08:39 AM   #9
Newbie
 
Joined: Oct 2016
From: United Kingdom

Posts: 13
Thanks: 0

Quote:
Originally Posted by Benit13 View Post
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 x-value where $\displaystyle \frac{dy}{dx} = 0$?

If you find this question tricky, we can help you with this
so dy/dx = 36x^11+30x^5

how would i do the next part?
Tom2000 is offline  
October 17th, 2016, 08:41 AM   #10
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 2,132
Thanks: 717

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Quote:
Originally Posted by Tom2000 View Post
so dy/dx = 36x^11+30x^5

how would i do the next part?

Good start!

The maximum or minimum of that function occurs when the gradient (dy/dx) is zero:

dy/dx = 0

so try solving the equation

36x^11+30x^5 = 0

for x.
Benit13 is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
points, stationary



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Stationary Points ajrt3 Calculus 0 September 29th, 2014 01:10 PM
stationary points iulia.d Calculus 0 March 20th, 2014 01:48 AM
Stationary Points Agata78 Calculus 8 January 14th, 2013 03:47 AM
stationary points sahil112 Calculus 5 November 21st, 2012 12:51 PM
Stationary points .. wulfgarpro Algebra 5 May 30th, 2010 03:14 AM





Copyright © 2018 My Math Forum. All rights reserved.