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Laws of area and conservation of momentum

  1. Mar 3, 2013 #1
    i am having problem getting the dθ in the laws of area.

    what the textbook states is,
    dθ=

    xdy-ydx
    x2+y2

    I cant get the equation..
    the only thing i could think of is trigo functions, but i got stuck...
    any help?
     
  2. jcsd
  3. Mar 3, 2013 #2
    almo,

    What are you trying to do? What are the "laws of area". What equation are you trying to find? How does the conservation of momentum fit into this?

    Ratch
     
  4. Mar 4, 2013 #3
    "Laws of Area" is vague, but the formula for dtheta is straightforward.

    [itex] \theta = \arctan (y/x)+C[/itex]

    Here C is an additive constant that depends on 1. Which region of the plane you are in, and 2. How you assign angles to points in the plane. For example, if you define theta to be between 0 and 2pi, then C=0 for (x,y) in the first quadrant, C=pi in the second and third quadrants, and C=2pi in the fourth quadrant. If you define theta to lie between -pi and +pi, then C works differently. But all that is irrelevant, because even though theta as a function of x and y has some ambiguity, no matter what your convention is for assigning angles to points, dtheta is well defined by the formula:

    [itex] d\theta = \frac{\partial \theta}{\partial x} dx +\frac{\partial \theta}{\partial y}dy[/itex]

    There are ways to derive the formula for dtheta without relying on a formula for theta as a function of x and y. For example, no matter how you assign angles, you have:

    [itex] x=r\cos(\theta), \hspace{1cm} y=r\sin(\theta).[/itex]

    Use these formulas to calculate dx and dy as linear functions of dr and dtheta (with a basepoint understood to be fixed). You get:

    [itex] dx = (x/r)dr -yd\theta,\hspace{1cm} dy= (y/r)dr+xd\theta[/itex].

    Multiply the first equation by y, the second equation by x and subtract to obtain:
    [itex] ydx - xdy = -(y^2+x^2)d\theta =-r^2d\theta[/itex]
    Dividing out the - (r squared) gives you the formula for dtheta.
     
    Last edited: Mar 4, 2013
  5. Mar 4, 2013 #4
    Thanks vargo!!!.. I figure out to just differentiate theta, which gives me the equation. I am too use to differentiate something against something else, that I forget about some basic stuff.

    The parametric method seems quite confusing when multiplication of x and y is involved though, but its brilliant.
     
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