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Partial Differentiation

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Usage

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If you have a function of two variables such that

then the following equation is true in general

this allows you to calculate the rate of change of given the rate of change of and . For example, in the case of a cone

this means that the following is true

Ref Desk Q's

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electrons

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I asked a similar question this yesterday, though this one seems much more complicated to me. I will go through my calucaltions so far, so if there are any errors, you can help me rectify them.

If you have n electrons distributed evenly around a circle radius r, distance x from another electron, what is the force of repulsion experienced by the electron not on the circle from the circle of electrons. Electrostatic repulsion is a inverse square law, and the coefficent of electrostatic force can be considered as k.

  • r = radius (OA, OC)
  • x = CY
  • y = OB
  • a = AY
  • α = AOY
  • β = BAY

using cosine rule; so the force, being an inverse square is; now, seeing as always in the circle the components in y-direction will cancel, we can consider only those in the x-direction in order to caluclate a resultant force we need to consider only the x-directions components, to find this consider the triangle ABY, the angle β and distance OB are neccesary. using sine rule.
therefore using sine rule,

so the x direction component of F, or Fsinβ is;

So to consider all the electrons; then where N number of electrons. and k varies from 1 to N. Therfore a sum of all the forces would be.

Now this is it, I really need some help simplifying this massive mess, and like yesterday I really need the summation series out of the equation if that's possible. Thank you.

data handling

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sense proj

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(1*1.1*10^(-6))/(pi*((0.193*10^(-3))/2)^2)

Silicon proj

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known

known

therfore

for pure

for doped

Maths Problem

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Ok well I encountered this problem (I added the y to simplify some calculations a bit, but otherwise its exactly the same information i got). I tried to solve but I wasnt sure wether I had done it correctly. So I thought I'd post here for someone to maybe review it and find the faults or confirm wether its correct. So heres what I went about doing. I numbered the steps for reference in your replies.

  1. Well firstly isnce the triangles in the corners have x on both sides, we can assume they are isoceles right angled triangles, therefore the angles are 90, 45, and 45, thus meaning that all the angles in any of the shapes in this case are either 45 or 90.
  2. (pythagorus)
  3. The white triangle in the centre of the right hand edge is (assuming the two sides identical in length are variable z)
    1. (pythagorus)
  4. the L shape can now be divided into 3 segments, one square in the centre which is , and 2 identical rectangles which are y by z. And as such the grey area consists of
  5. As such the entire grey area works out to equal this when the left part of the shape is included
  6. Therefore, the entire size of the shape should equal
  7. The two triangles with their sides against the x by 2006 shaded areas two identiacal sides can be calculated (assuming the two sides identical in length are variable a)
  8. Substituting in the other formulae, to give in terms of x leaves
  9. So the formula for the entire square equals
  10. So now we have 2 formula for the entire shape, so we can substitute them together

Unfortunately this is where I got stuck. So if anyone knows where to go from here... The help would be appreciated. Philc TECI 21:04, 10 September 2006 (UTC)

Space

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Russian Launch Vehicles Template

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