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Tangent vectors in the coordinate basis

  • #1
1,344
32

Homework Statement



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}##.

Homework Equations



The Attempt at a Solution


[/B]
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)##?
 

Answers and Replies

  • #2
19,962
4,109
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.
 
  • #3
1,344
32
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.
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.
 
  • #4
19,962
4,109
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.
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)##.
At the point P, ##\lambda =1##. So....?
I don't really see how, in the ##xyz##-coordinate system, the coordinate basis vectors are equal to the unit vectors.
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?
 
  • #5
1,344
32
At the point P, ##\lambda =1##. So....?
At the point ##p##, ##\lambda =1## so that the tangent vector to the curve at ##p## is ##(1,0,-1)##.

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?
##\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?
 
  • #6
19,962
4,109
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?
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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