Tangent vectors in the coordinate basis

In summary: Your notation may differ.In summary, the conversation discussed the concept of tangent vectors and coordinate basis in Euclidean three-space. Three different curves passing through a given point were considered and their tangent vectors were calculated in terms of the coordinate basis. The conversation also explored the calculation of the derivative of a function in terms of these tangent vectors. It was also mentioned that in the xyz-coordinate system, the coordinate basis vectors are equal to the unit vectors.
  • #1
spaghetti3451
1,344
33

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)##?
 
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  • #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.
 
  • #3
Chestermiller said:
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
failexam said:
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.
failexam said:
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
Chestermiller said:
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)##.

Chestermiller said:
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
failexam said:
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.
 

1. What is a tangent vector in the coordinate basis?

A tangent vector in the coordinate basis is a vector that is tangent to a specific point on a surface or curve. It is defined by its direction and magnitude, and can be represented by a set of coordinates in a given coordinate system.

2. How is a tangent vector calculated in the coordinate basis?

To calculate a tangent vector in the coordinate basis, the partial derivatives of a function at a specific point must be taken with respect to each of the coordinate variables. These partial derivatives are then used to construct a vector that is tangent to the surface or curve at that point.

3. What is the relationship between tangent vectors and curves?

Tangent vectors in the coordinate basis are closely related to curves, as they represent the direction and rate of change of a curve at a specific point. The tangent vector at any given point on a curve is parallel to the curve at that point.

4. How are tangent vectors used in differential geometry?

Tangent vectors in the coordinate basis are essential in differential geometry, as they allow for the measurement and analysis of curves and surfaces. They are used to define the tangent space at a specific point on a manifold and are crucial in understanding the geometry and topology of the space.

5. Can the coordinate basis be used to represent all tangent vectors?

No, the coordinate basis cannot represent all tangent vectors. This is because the coordinate basis is dependent on the specific coordinate system chosen, and different coordinate systems may result in different tangent vectors. Therefore, it is important to use a more general basis, such as the standard basis, to represent all possible tangent vectors.

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