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Why multiply by delta theta?

  1. Oct 21, 2011 #1

    I'm reading through one of my books and it's explaining how a vector is eqaul to multiplying sin[itex]\phi[/itex] and [itex]\Delta\vartheta[/itex]. the way it's written in the text is as followed,

    |[itex]\Delta[/itex]i| [itex]\approx[/itex] (sin[itex]\phi[/itex])[itex]\Delta[/itex][itex]\vartheta[/itex]

    I have never understood how things like this work. Could someone please explain to me why this is true and how it works? I included a picture of the figure it uses. Thank you.

    Attached Files:

  2. jcsd
  3. Oct 22, 2011 #2


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    You are talking about spherical coordinates, right? Also you are using "physics" notation which switches [itex]\theta[/itex] and [itex]\phi[/itex] from "mathematics" notation. In your notation, [itex]\phi[/itex] is the "longitude" and [itex]\theta[/itex] is the "co-latitude".

    In these coordinates, for a fixed [itex]\theta[/tex] and r, sweeping [itex]\phi[/itex]through 0 to [itex]2\pi[/itex], the point [itex](r, \phi, \theta)[/itex] sweeps through a circle, but not a circle of radius r. If we draw the vertical axis, the line from (0, 0, 0) to the point [itex](r, \phi, \theta)[/itex], and the line from that point perpendicular to the vertical axis, we get a right triangle with hypotenuse of length r and base angle of [itex]\phi[/itex]. If we call the opposite side to that angle "x" then [itex]sin(\phi)= x/r[/itex] so [itex]x= r sin(\phi)[/itex]. That will be the radius of the circle swept out.
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