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Right triangle, describe dx as dθ

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data
    This is part of a find the electric field problem. I've narrowed down the part I find confusing.

    Consider a right triangle, with legs of length x and R. Angle θ is opposite x. Leg x is a segment of a ray starting where lines x and R intersect. There is a differential length dx along line x (it is at a vertex of the right triangle). Find dx in terms of R and θ.

    2. Relevant equations
    standard trig functions, pythagorean theorem?

    3. The attempt at a solution
    The answer according to the solutions manual is dx = (R/cos^2 θ) dθ. Obviously, I cannot understand at all where that came from.

    I did get R tan ( θ + dθ) = x + dx but that isn't very helpful either
  2. jcsd
  3. Oct 2, 2009 #2


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    Since x = R tan(θ), then dx/dθ = R sec2θ.

    So what does this give for dx?
    Last edited: Oct 2, 2009
  4. Oct 2, 2009 #3
    man, this makes me feel dumb

    thanks for pointing out that dx = (1 / cos2θ) dθ. i would have never guessed that trick by myself

    but where does the R come into that equation?
  5. Oct 2, 2009 #4


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    Just a careless typo. I fixed it with an edit. dx/dθ = R sec2θ
  6. Oct 2, 2009 #5
    this stupid part is i didnt realize it was a typo either.. but thank you for this trick. i never thought like that before
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