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Planar curves

  1. Mar 3, 2005 #1
    Recall that every point (x, y) in the plane is described by its radius-
    vector r = xi + yj. A planar curve [gamma] has the following geometric property: at every point on the curve the radius vector and the tangent intersect at a fixed angle [alpha].

    1. Derive a first order differential equation dy/dx = f(x, y, [alpha]) describing [gamma]
    Note that the right-hand side should depend on the parameter [alpha].

    For part 1, I drew a circle(not sure if it's suppose to be a circle) on the (x, y) plane. Then i drew a radial vector from the origin to any point. i let the radial vector make an angle [phi] with x-axis. extended the tangent all the way to the x axis and let the tangent make an angle theta with the x axis.

    then i got

    tan [phi]= y/x......(1)

    tan [theta]= dy/dx...(2)

    so [alpha]= [theta] - [phi]

    tan [alpha]= tan ([theta] - [phi]) = (tan [theta] - tan [phi])/(1+tan[theta]*tan[phi])......(3)

    dont know how to derive the differential from here.....and not sure about the rest of the problem...please help.....!!!!!

    2. Solve the equation from Part 1 for the case when [alpha] = pi/2 assuming that [gamma] passes through (1, 0). For which curve is the radius vector always perpendicular to the tangent?

    I think the answer is a circle but not sure how to solve it???/

    3. Assume that [alpha] = 3pi/4 and the curve passes through (1, 0). Apply Euler's method with step h = 0.5 to trace out the curve for this case.

    is this step function? if so how do i go about solving this???

    4. Solve the equation from Part 1 for [alpha]= 3pi/4 with (1, 0) as the initial condition. Superimpose the exact solution from this part with the numerical
    solution from Part 3. Do you know the name of this curve?

    I think in order to plot the curve, i have to switch to polar coordinates
    x = r cos [theta]; y = r sin [theta]:

  2. jcsd
  3. Mar 4, 2005 #2
    Answer to Question1: Put theta=arctan(dy/dx) & phi=arctan(y/x) in the formula you've got.
    But why not use polar coordinates? That'd solve all the problems.
    I'm,with great respect,
  4. Mar 4, 2005 #3


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    Sounds to me like you are making this a lot more complicated than necessary.

    If the radius vector is r(t) then the tangent vector is r'(t) Since than angle between any two vectors, u, v, is given by cos(θ)= u.v/(|u||v|), the condition that r and r' have constant angle, α, between them is that r'(t).r(t)/(|r||r'|)= cos(α). Since r= <x, y>, r'= <x', y'>, that becomes
    [tex]\frac{xx'+ yy'}{\sqrt{x^2+ y^2}\sqrt{x'^2+ y'^2}}= cos(\alpha)[/tex]
    [tex]xy'+ yy'= cos(\alpha)\sqrt{x^2+ y^2}\sqrt{x'^2+ y'^2}[/tex]

    "Dividing" both sides by x' (It's really using the chain rule.)
    [tex]x+ y\frac{dy}{dx}= cos(\alpha)\sqrt{x^2+ y^2}\sqrt{1+(\frac{dy}{dx})^2}[/tex]

    That's your differential equation.

    Of course, if [itex]\alpha= \frac{\pi}{2}[/itex] the cos(&alpha;)= 0 and your differential equation is just
    [tex] x+ y\frac{dy}{dx}= 0[/tex].
    That's a simple separable differential equation (and, yes, the solution is a circle.)
    Last edited: Mar 4, 2005
  5. Mar 4, 2005 #4


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    This is okay. If you write it out in x and y:

    [tex]\frac{dy}{dx}=\tan(\alpha + \phi)=\frac{\tan(\alpha)+\tan(\phi)}{1-\tan(\alpha)\tan(\phi)}=\frac{y/x+\tan(\alpha)}{1-\tan(\alpha)y/x}[/tex]

    There's your differential equation. But definately switch to polar coordinates.
  6. Mar 5, 2005 #5

    I understand how to plug in and get to this point, but how would i seperate the variables in this case? I tried plugging in a lot of different things but keep coming up with an unsolvable differential.

    i get tan(a) when i do it and when i plug in pi/2 i get undefined. i'm sure i'm doing somethign wrong but im not sure

  7. Mar 6, 2005 #6
    Does anyone know how to convert the above equation in to polar coordinates ^^^^
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