Partial Derivation

  • Thread starter Bman900
  • Start date
  • #1
12
0
Now I solved something similar to this problem yesterday (https://www.physicsforums.com/showthread.php?t=447168) thanks to the help of p21bass but this one is really out there and I have no idea where to begin.

Homework Statement


secondproblem.jpg



Homework Equations





The Attempt at a Solution


I don't know where to even start as this is my first time ever seeing this problem. Where should I start?
 

Answers and Replies

  • #2
9
0
So is the question just asking you to find Fx, Fy, Fz?
If that is the case then its a fairly easy problem.
If you have not met partial differentiation before I wont go through what it is and how it comes about but i'll just tell you how to do it:
To differentiate du/dx (with curly d's to represent partial derivatives) you take all the variables which are not x and treat them as constants so for example d/dx (x^2yz)=2xyz
As with the minus signs you just need to find the derivatives and then multiply it by -1 .
Hopefully that helps a little.
 
  • #3
12
0
So is the question just asking you to find Fx, Fy, Fz?
If that is the case then its a fairly easy problem.
If you have not met partial differentiation before I wont go through what it is and how it comes about but i'll just tell you how to do it:
To differentiate du/dx (with curly d's to represent partial derivatives) you take all the variables which are not x and treat them as constants so for example d/dx (x^2yz)=2xyz
As with the minus signs you just need to find the derivatives and then multiply it by -1 .
Hopefully that helps a little.
so like this?

secondproblemqustion.jpg


But since am treating yz as constants wouldn't it be 0 if I take the derivative or am just taking the derivative of x and then multiplying it by yz?
 
  • #4
12
0
Ok so I read up on partial derivatives and came up with this:


secondproblemcopy.jpg



Am I right?
 
  • #5
44
0
Not quite. When you partially differentiate, you're treating the other variables as constant, but you still might be multiplying by the variable you're differentiating with respect to. For instance:

[tex]\frac{\partial }{\partial x} ( xyz ) = yz[/tex]

As you know

[tex]\frac{d}{dx} ( \alpha x ) = \alpha[/tex]

Remember: when you differentiate a constant on its own, you get 0, but a constant multiplying the variable you're differentiating with respect to is not zero!
 
  • #6
12
0
I really do appreciate the help here! Now is this any better?

secondproblemcopy-2.jpg
 
  • #7
44
0
Looks great, nice work!
 

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