Derivative of a function

  • I
  • Thread starter 1Keenan
  • Start date
  • #1
1Keenan
100
3
Hello,

I would need some help in calculating the derivative of the function T_el in the attached image.
I want to calculate d T_el /d yd, where yd is the variable and it appears in the term I called A_elSide. Its expression is again in the image.
Numbers you see are not important.


Just to explain what I am trying to do:
I want to calculate energy resolution of a magnetic spectrometer (which means distance between two energy point) and it can be calculated as (dT/dy)*spotsize, basically the derivative of the energy with respect to the position on the detector and multiplied by the "nominal" beam spot size.

This means I need to express the energy as a function of the beam position.
Beam position at the detector plane is yd=R*sin(theta)+(d*tg(theta)

First term (R*sin(theta)) is the y coordinate at the magnet output, second term ((d*tg(theta) ) is the additional displacement in the drift d between magnet and detector.
Kinetic energy is in the bending radius R, so doing some math I got an expression for the kinetic enexrgy, which is T_el in the picture


Expression for T_el has an element A_elSide, cotaining the position on the detector (yd), which is the variable with respect I should do the derivative.

An now I need help, I did it already 4 times and I got 4 different results...



Any help?
 

Attachments

  • Immagine.png
    Immagine.png
    17.7 KB · Views: 97

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
15,358
4,334
I got 4 different results...
I don't see any of them. Telepathic capabilities severely limited.
And please, ##\LaTeX##, not pictures. Do you want us to do it for you ?
 
  • #3
1Keenan
100
3
no, I don't want you to do it for me, off course!
I didn't posted them because they are wrong.

My strategy is to derive it as a sqrt((a+f(x))^2) but I guess it doesn't work properly...
 
  • #4
BvU
Science Advisor
Homework Helper
15,358
4,334
To me it seems the tough part is inverting
yd=R*sin(theta)+(d*tg(theta) )
so I wonder if ##\theta## is small enough to approximate linearly.
(by the way, the thing is called tan, not tg )

My strategy is to derive it as a sqrt((a+f(x))^2)
So what is $${d\over dz} \, \sqrt {a+z^2} \quad ? $$
I didn't posted them because they are wrong.
Post what you think is the best one and solicit comments
 
  • #5
1Keenan
100
3
Tan or tg is the same...
I'll redo the calculation tomorrow, I think I made a mess with all the other constant terms
 
  • #6
1Keenan
100
3
ok, I did calculation and the most reasonable result is attached here.

There is a problem, it doesn't work at all...
any help?
 

Attachments

  • Immagine.png
    Immagine.png
    6.7 KB · Views: 83
  • #7
BvU
Science Advisor
Homework Helper
15,358
4,334
I tried to help in post #4. Did it not help ?
 
  • #8
1Keenan
100
3
Yes, you did.
I forgot to tell you that theta is not small to approximate sin(theta) = theta.

I think I'm messing up with the constants. Or do you see any huge mistake in the derivative?
 

Suggested for: Derivative of a function

  • Last Post
Replies
14
Views
796
Replies
10
Views
988
Replies
1
Views
710
Replies
5
Views
2K
Replies
6
Views
630
  • Last Post
Replies
15
Views
916
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
28
Views
1K
Replies
15
Views
1K
Top