Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving a nasty set of equations

  1. Feb 16, 2009 #1
    Suppose I have the following set of equations
    [tex]x=3\pi r-r-2\sin\theta[/tex]
    [tex]y=r-2\cos\theta[/tex]
    [tex]\tan\theta=y/x[/tex]
    with r as a constant

    How would I go about finding out if an analytical solution exists? There must be a solution, as I distilled the equations from a fairly straightforward geometrical problem that I conjured up.
    I just can't seem to find it.

    The geometrical problem is basically finding the length of the inner tangent of two circles, between the points where the tangent and the circles touch plus parts of the circles. It's length would be [tex]\sqrt{x^2+y^2}[/tex] and the parts of the circle have a length [tex]2\theta r[/tex]. I hope this makes sense.
     
    Last edited: Feb 16, 2009
  2. jcsd
  3. Feb 16, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Mr.Miyag! :smile:

    (have a theta: θ :wink:)

    Rewrite sinθ and cosθ in terms of tanθ,

    then convert them to an expression in y/x, and solve. :wink:
     
  4. Feb 16, 2009 #3
    Hi:)

    I still don't see how I can solve for all three variables.
    [tex]\sin\theta=\tan\theta\cos\theta=\cos\theta\frac{y}{x}[/tex]
    [tex]\cos\theta=\tan\theta\frac{\cos^2\theta}{\sin\theta}=\tan\theta\cot\theta\cos\theta=\cot\theta\cos\theta\frac{y}{x}[/tex]
    then
    [tex]x=(3\pi-1)r-2\cos\theta\frac{y}{x}[/tex]
    [tex]y=r- 2\cot\theta\cos\theta\frac{y}{x}[/tex]
    I don't see a substitution that simplifies things.
     
  5. Feb 16, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Mr.Miyagi! :smile:

    (what happened to that θ i gave you? :redface:)
    You need to learn your trigonometric identities …

    in particular, sec2θ = tan2θ + 1 :wink:
     
  6. Feb 16, 2009 #5
    (What's up with the theta?:tongue:)

    Ah, I overlooked the identity. I can express x as a function of y and vice versa, with the trig identity now. But it doesn't seem to help me in finding θ as a function of r.(I shouldn't have said r is a constant. It is just a variable).
    I feel I'm overlooking something really obvious, but I can't seem to figure it out.
     
  7. Feb 16, 2009 #6
    I don't quite understand the geometry you are trying to explain. How are these circles defined again? Are they concentric.

    Also, if your x,y,r,and theta are both the coordinates, you could use any of the equations that do {x,y}-->{r,theta} and visa versa.

    For example,

    x = r Cos[theta]
    y = r Sin[theta]
     
  8. Feb 16, 2009 #7
    I apologize for the poor choice of variable names. I've made a drawing to explain what I was talking about. It's in the attachment.
    The line GH would be our x, DH would be our y, the radius of the circles is our r and the angles denoted apha are our theta.
    It is also known that the line BC has length [tex](3\pi-1)r[/tex]
    Now I'd like to know the length of the line DG and the angle of alpha.
    knotpart.jpg
     
    Last edited: Feb 16, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Solving a nasty set of equations
  1. Solving equation (Replies: 4)

  2. Solve an equation (Replies: 8)

  3. The nasty easy equations (Replies: 10)

Loading...