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Finding a functional connection.

  1. Dec 6, 2006 #1
    im given:
    u=arcsin(x)+arccos(y)
    v=xsqrt(1-y^2)+ysqrt(1-x^2)
    and i need to find the functional connection between u and v.

    i know that:
    v'_x/u'_x=dv/du
    and i have got:
    dv/du=sqrt(1-x^2)sqrt(1-y^2)-xy
    now i need to show the rhs as a function of v or u, obviously of v should be much easier, the problem is i dont know how to simplify it.

    i mean i tried adding and substracting terms, but with no success can someone help me on this?

    thanks.
     
  2. jcsd
  3. Dec 8, 2006 #2
    i fixed my mistake, the first equation is u=arcsin(x)+arcsin(y)
    can now someone help me on this?
    thanks.
     
  4. Dec 8, 2006 #3

    dextercioby

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    How about

    [tex] v=x\left(1-y^{2}\right)\frac{\partial u}{\partial y} +y\left(1-x^{2}\right)\frac{\partial u}{\partial x} [/tex]

    Is this what you were looking for?

    Daniel.
     
  5. Dec 8, 2006 #4
    i need to find a functional connection like u=u(v) or v=v(u), or in simple terms v is a function of u or otherwise.
     
  6. Dec 8, 2006 #5

    HallsofIvy

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    [itex]cos(arcsin(x))= \sqrt{1- sin^2(arcsin(x)}= \sqrt{1- x^2}[/itex]

    Since sin(a+ b)= sin(a)cos(b)+ cos(a)cos(b), sin(arcsin(x)+ arcsin(y))= sin(arcsin(x))cos(arcsin(y))+ cos(arcsin(x))sin(arcsin(y)).

    That is, sin(arcsin(x)+ arcsin(y))= [itex]x\sqrt{1- y^2}+ y\sqrt{1-x^2}[/itex].

    In other words, v= sin(u)!
     
  7. Dec 9, 2006 #6
    yes i also found it by using derivatives we know that:
    v'_x=sqrt(1-y^2)-xy/sqrt(1-x^2)
    and u'_x=1/sqrt(1-x^2)
    so dv/du=sqrt(1-x^2)sqrt(1-y^2)-xy
    (sqrt(1-x^2)sqrt(1-y^2)-xy)^2=1-v^2
     
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