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Finding a new way to calculate one side of a triangle

  1. Sep 22, 2013 #1

    nearc

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    hopefully this is the right place for this question, the first part is a trig/geometry question but it is really a integration question:

    i'm trying to find another way to compute d1 without using law of cosines because i don't know how to integrate (cos a)^.5, if someone knows how to do that please let me know.

    in the figure provided i'm attempting to find d1 as a function of angle a, r0 and r1 are constants and angles t, p and lengths d0, x can be computed trivially.

    i can use law of sines to find d1 if i can make b, m or c a function of a
     

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    Last edited: Sep 22, 2013
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  3. Sep 22, 2013 #2

    Simon Bridge

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    If r1=r0+x, then d1 is related to r0 and r1 by pythagoras.
     
  4. Sep 22, 2013 #3

    SteamKing

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    This diagram is a handy way to visualize the relationship between the trig functions and the angle using a unit circle: (credit: wikipedia)

    500px-Circle-trig6.svg.png
     
  5. Sep 22, 2013 #4

    nearc

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    yes, r1=r0+x, but there are no right angles in the figure
     
  6. Sep 22, 2013 #5

    nearc

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    thanks i'll reflect on this, but i suspect that this would only be useful in the case where i have right angles
     
  7. Sep 22, 2013 #6

    Simon Bridge

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    Ah - so d1 is not a tangent ... it looks a lot like it is supposed to be one in the figure.

    This is where I suspect you need to look for another way to set up the integral - or learn to use elliptic integrals.
     
  8. Sep 22, 2013 #7

    nearc

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    can you give me some good lead for elliptic integrals? thansk
     
  9. Sep 23, 2013 #8

    Simon Bridge

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  10. Oct 3, 2013 #9

    nearc

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    thanks for your help, but i've looked through some basic elliptical integrals and i'm not finding anything like (cos x)^.5 the basic form is 1-(sin x )^2

    also what i'm working with is not an ellipse it is a circle. the difference is that instead of a constant radius from the center tracing out the circle it is circle traced from a point off center and instead of a constant radius length the this segment length changes
     
  11. Oct 3, 2013 #10

    Simon Bridge

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    "elliptic integral" not "elliptical".
    It is the name for a class of functions - they have application beyond elliptic curves.

    How about:
    http://integrals.wolfram.com/index.jsp?expr=sqrt(cos(x))&random=false

    Um - note: if x is very small compared with r0, then cos(m)~1 and sin(m)~m and tan(m)~m.

    In the diagram in post #3 - your triangles are AOD and ADE.
    The exsec side is you side x.
    AO=OD=r0
    AD=d0
    ... like that.
     
    Last edited: Oct 3, 2013
  12. Dec 31, 2013 #11

    nearc

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    thanks to simon i've figured out the simple elliptic however i now have a more difficult version:

    $$\int_0^{2 \pi } \sqrt{\left(1-\frac{60}{61} \cos (x-179)\right) \left(1-\frac{60}{61} \cos (x-t)\right)} \, dx $$

    using mathematica i can find the solution to

    $$\int \sqrt{1-\frac{60}{61} \cos (x-t)} \, dx $$

    but mathematica was not able to solve the harder version, any ideas?

    also how do get the latex to show correctly?

    thanks mark44
     
    Last edited: Dec 31, 2013
  13. Dec 31, 2013 #12

    Mark44

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    Add $$ before and after your LaTeX. Here's what your stuff looks like with this addition.
    $$\int_0^{2 \pi } \sqrt{\left(1-\frac{60}{61} \cos (x-179)\right) \left(1-\frac{60}{61} \cos (x-t)\right)} \, dx $$

    $$\int \sqrt{1-\frac{60}{61} \cos (x-t)} \, dx
    $$
     
  14. Dec 31, 2013 #13

    Simon Bridge

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    $$\left(1-\frac{60}{61} \cos (x-179)\right) \left(1-\frac{60}{61} \cos (x-t)\right)$$ ... hmmm? expand the brackets and try trig identities?
    A quick path can be to expand the trig function by the Euler relations since manipulating exponentials is usually easier.
     
  15. Dec 31, 2013 #14

    nearc

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    thanks simon, are you referring to

    $$e^{\text{ix}}=x \cos + i x \sin $$
     
  16. Dec 31, 2013 #15

    Simon Bridge

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  17. Jan 1, 2014 #16

    nearc

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    yep thats it, mathematica likes to switch things?
     
  18. Jan 1, 2014 #17

    Simon Bridge

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    Dunno - never used mathematica.
     
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