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Tangent vectors in the coordinate basis

  1. Apr 29, 2016 #1
    1. The problem statement, all variables and given/known data

    In Euclidean three-space, let ##p## be the point with coordinates ##(x,y,z)=(1,0,-1)##. Consider the following curves that pass through ##p##:

    ##x^{i}(\lambda)=(\lambda , (\lambda -1)^{2}, -\lambda)##
    ##x^{i}(\mu)=(\text{cos}\ \mu , \text{sin}\ \mu , \mu - 1)##
    ##x^{i}(\sigma)=(\sigma^{2} , \sigma^{3}+\sigma^{2} , \sigma )##

    (a) Calculate the components of the tangent vectors to these curves at ##p## in the coordinate basis ##\{\partial_{x},\partial_{y},\partial_{z}\}##.

    (b) Let ##f=x^{2}+y^{2}-yz##. Calculate ##\frac{df}{d\lambda}, \frac{df}{d\mu}## and ##\frac{df}{d\sigma}##.

    2. Relevant equations

    3. The attempt at a solution

    My understanding of the concept of tangent vectors and coordinate basis is a little shaky.

    I understand that, for a function ##f=f(x^{i}(\lambda))##, the tangent vector is ##\frac{df}{d\lambda}=\frac{dx^{i}}{d\lambda}\frac{df}{dx^{i}}=\frac{dx^{i}}{d\lambda}\partial_{i}f##, where the set of ##\frac{dx^{i}}{d\lambda}## are the components of the tangent vector and the set of ##\partial_{i}## is called the coordinate basis.

    I wonder why ##\frac{df}{d\lambda}## is called the tangent vector. Is it because ##\frac{df}{d\lambda}## points along the tangent to the curve ##f(\lambda)##?
     
  2. jcsd
  3. Apr 29, 2016 #2
    In the first case, write:

    $$x^1(\lambda)=x=\lambda$$
    $$x^2(\lambda)=y=(\lambda-1)^2$$
    $$x^3(\lambda)=z=-\lambda$$
    In terms of ##\lambda## and ## d\lambda##, what are dx1, dx2, and dx3 along the curve? In terms of the unit vectors in the x, y, and z directions, what is the tangent vector to the curve? In this coordinate system, the coordinate basis vectors are equal to the unit vectors.
     
  4. Apr 29, 2016 #3
    In terms of ##\lambda## and ##d\lambda##, ##dx^{1}=d\lambda##, ##dx^{2}=2(\lambda -1)d\lambda## and ##dx^{3}=-d\lambda##.

    In terms of the unit vectors in the ##x##, ##y##, and ##z## directions, the tangent vector to the curve is ##(1, 2(\lambda - 1), -1)##.

    I don't really see how, in the ##xyz##-coordinate system, the coordinate basis vectors are equal to the unit vectors.
     
  5. Apr 29, 2016 #4
    At the point P, ##\lambda =1##. So....?
    In ##xyz##, system, a position vector from the origin to an arbitrary point is:
    ##\vec{s}=x\vec{i}_x+y\vec{i}_y+z\vec{i}_z=x^1\vec{i}_1+x^2\vec{i}_2+x^3\vec{i}_3##
    So, what are the partial derivatives of ##\vec{s}## with respect to each of the three coordinates?
     
  6. Apr 29, 2016 #5
    At the point ##p##, ##\lambda =1## so that the tangent vector to the curve at ##p## is ##(1,0,-1)##.

    ##\partial_{1}\vec{s} = \vec{i}_{1}##, ##\partial_{2}\vec{s} = \vec{i}_{2}## and ##\partial_{3}\vec{s} = \vec{i}_{3}##.

    How might this help?
     
  7. Apr 29, 2016 #6
    The partial derivatives of the position vector with respect to the coordinates are the definition of the coordinate basis vectors. So, the tangent vector is expressible as: $$\vec{t}=+1\vec{i}_1-1\vec{i}_3$$. At least this is the kind of notation I was taught when I was learning this stuff.
     
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