# Tangent vectors in the coordinate basis

1. Apr 29, 2016

### spaghetti3451

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. Apr 29, 2016

### Staff: 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.

3. Apr 29, 2016

### spaghetti3451

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. Apr 29, 2016

### Staff: Mentor

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?

5. Apr 29, 2016

### spaghetti3451

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?

6. Apr 29, 2016

### Staff: Mentor

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.