# Tangent vectors in the coordinate basis

## 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}##.

## 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)##?

Chestermiller
Mentor
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 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.

Chestermiller
Mentor
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?

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?

Chestermiller
Mentor
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.