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Finding the normal of a curve

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Screenshot2012-02-25at20521AM.png

    If t = pi/2, then that would equal 2, but they don't say that t = pi/2 so how do they get 2?

    Maybe since it's circular motion t = pi, that would work too, but I want to be sure before i move on.
     
    Last edited: Feb 25, 2012
  2. jcsd
  3. Feb 25, 2012 #2

    cepheid

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    It's a simple trignometric identity.

    sin2θ + cos2θ = 1
     
  4. Feb 25, 2012 #3
    Note the identity sin(kx)^2 + cos(kx)^2 = 1
    |v| becomes sqrt(4(1)) = 2
     
  5. Feb 25, 2012 #4
    What about the 2t? Unless the 2t gets subsumed in the theta and whatever theta is cos^2 and sin^2 add up to 1, is that right?
     
  6. Feb 25, 2012 #5
    Yes.

    If you let u = 2t, then we'd have
    sin(u)^2 + cos(u)^2

    Which equals 1, and we don't have a "u" to plug back into.
     
  7. Feb 25, 2012 #6

    cepheid

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    Theta is just a symbol for "the argument (or angle) that you pass into the trigonometric function." In this case, the argument to the function is 2t. In other words, θ = 2t here. The identity holds regardless of what theta is. I could have written the identity as:

    sin2x + cos2x = 1

    OR

    sin2ϕ+ cos2ϕ = 1

    OR

    sin2:smile:+ cos2:smile: = 1

    It doesn't matter. All that's changing here is the SYMBOL that I'm using to represent the argument. It doesn't change the meaning of the identity. All of these equations mean the same thing. Expressed in words, the identity is:

    "The square of the sine of an angle, plus the square of the cosine of that same angle, together must equal 1."
     
  8. Feb 25, 2012 #7

    Mark44

    Staff: Mentor

    sin2(whatever) + cos2(whatever) [itex]\equiv[/itex] 1.

    "Whatever" can be anything, as long as it is the same in both terms.

    This is one of the first trig identities presented in a trigonometry course. There aren't many trig identities that you need to memorize, but this is definitely one of them.
     
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