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Math Help - Multivaraible Calc. vectors

  1. #1
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    Multivaraible Calc. vectors

    ok..so here is what I am given for this problem:

    The vector function r sends a point (u,v) in the plane to a point (vector) in R3.
    r(u,v)= (1/2sqrt(v)*cosh U, sqrt(V)*sinhU, V)
    or in other words:
    x= 1/2sqrt(V)*coshU
    y= sqrt(V)*sinhU
    z=V
    these points satisfy the equation: z= 4x^2-y^2, and it is a hyperbolic parabloid.
    Question:
    Fix u=u(sub 0) and consider the parametised curve r(sub v), defined as rv(v)=r(u(sub )), v)
    Write explicitly r(subv)(v):

    What does this mean? does it mean to replace all u with u(sub 0) in the original vector equation? Help..please someone explain this to me.
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  2. #2
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    That should be it
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  3. #3
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    Excellent. If you are not too busy, can you check if I've done the next part of the problem right:
    So the orginal vector function is as follows:
    r(u,v)= (1/2sqrtv*coshU, sqrtv*sinhU, V)
    then to write v explicitly I am given that r(sub v)(V)=r(u(sub 0),v)
    I did what you said and replaced all the u's in the original vector function with u knot and got:
    r(u(sub 0),v)= (1/2sqrtV*coshu(sub0), sqrtv*sinhu(sub0), V)
    Then I am asked to compute the tangent vector of r(subv)(V), so I first took the derivitive and I got:
    r(v) prime= ( 1/4sqrtv*sinhu(sub0), 1/2sqrtv*coshu(sub0), 1)
    Then to get the tangent vector I also calculated it magnitude and got:
    the magnitude of r(v)prime= sqrt( everything squared)= sqrt (1/16v*sinh^2u(sub0)+ 1/4v*cosh^2u(sub0)+1)
    I then factored out a 1/4 v from the first two terms and got:

    sqrt( 1/4v(1/4*sinh^2u(sub0)+cosh^2u(sub0)) +1)
    Since I can't simplify this anymore the tangent vector would be the derivitive I calculated earlier over this magnitude.

    Sorry I know this is confusing to type/read but I would appreciate any corrections.
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