# Tangent vector

(Redirected from Tangent vectors)

In mathematics, a tangent vector is a vector that is tangent to a curve or surface at a given point. Tangent vectors are described in the differential geometry of curves in the context of curves in Rn. More generally, tangent vectors are elements of a tangent space of a differentiable manifold. Tangent vectors can also be described in terms of germs. Formally, a tangent vector at the point ${\displaystyle x}$ is a linear derivation of the algebra defined by the set of germs at ${\displaystyle x}$.

## Motivation

Before proceeding to a general definition of the tangent vector, we discuss its use in calculus and its tensor properties.

### Calculus

Let ${\displaystyle \mathbf {r} (t)}$  be a parametric smooth curve. The tangent vector is given by ${\displaystyle \mathbf {r} ^{\prime }(t)}$ , where we have used a prime instead of the usual dot to indicate differentiation with respect to parameter t.[1] The unit tangent vector is given by

${\displaystyle \mathbf {T} (t)={\frac {\mathbf {r} ^{\prime }(t)}{|\mathbf {r} ^{\prime }(t)|}}\,.}$

#### Example

Given the curve

${\displaystyle \mathbf {r} (t)=\{(1+t^{2},e^{2t},\cos {t})|\ t\in \mathbb {R} \}}$

in ${\displaystyle \mathbb {R} ^{3}}$ , the unit tangent vector at ${\displaystyle t=0}$  is given by

${\displaystyle \mathbf {T} (0)={\frac {\mathbf {r} ^{\prime }(0)}{\|\mathbf {r} ^{\prime }(0)\|}}=\left.{\frac {(2t,2e^{2t},\ -\sin {t})}{\sqrt {4t^{2}+4e^{4t}+\sin ^{2}{t}}}}\right|_{t=0}=(0,1,0)\,.}$

### Contravariance

If ${\displaystyle \mathbf {r} (t)}$  is given parametrically in the n-dimensional coordinate system xi (here we have used superscripts as an index instead of the usual subscript) by ${\displaystyle \mathbf {r} (t)=(x^{1}(t),x^{2}(t),\ldots ,x^{n}(t))}$  or

${\displaystyle \mathbf {r} =x^{i}=x^{i}(t),\quad a\leq t\leq b\,,}$

then the tangent vector field ${\displaystyle \mathbf {T} =T^{i}}$  is given by

${\displaystyle T^{i}={\frac {dx^{i}}{dt}}\,.}$

Under a change of coordinates

${\displaystyle u^{i}=u^{i}(x^{1},x^{2},\ldots ,x^{n}),\quad 1\leq i\leq n}$

the tangent vector ${\displaystyle {\bar {\mathbf {T} }}={\bar {T}}^{i}}$  in the ui-coordinate system is given by

${\displaystyle {\bar {T}}^{i}={\frac {du^{i}}{dt}}={\frac {\partial u^{i}}{\partial x^{s}}}{\frac {dx^{s}}{dt}}=T^{s}{\frac {\partial u^{i}}{\partial x^{s}}}}$

where we have used the Einstein summation convention. Therefore, a tangent vector of a smooth curve will transform as a contravariant tensor of order one under a change of coordinates.[2]

## Definition

Let ${\displaystyle f:\mathbb {R} ^{n}\rightarrow \mathbb {R} }$  be a differentiable function and let ${\displaystyle \mathbf {v} }$  be a vector in ${\displaystyle \mathbb {R} ^{n}}$ . We define the directional derivative in the ${\displaystyle \mathbf {v} }$  direction at a point ${\displaystyle \mathbf {x} \in \mathbb {R} ^{n}}$  by

${\displaystyle D_{\mathbf {v} }f(\mathbf {x} )=\left.{\frac {d}{dt}}f(\mathbf {x} +t\mathbf {v} )\right|_{t=0}=\sum _{i=1}^{n}v_{i}{\frac {\partial f}{\partial x_{i}}}(\mathbf {x} )\,.}$

The tangent vector at the point ${\displaystyle \mathbf {x} }$  may then be defined[3] as

${\displaystyle \mathbf {v} (f(\mathbf {x} ))\equiv (D_{\mathbf {v} }(f))(\mathbf {x} )\,.}$

## Properties

Let ${\displaystyle f,g:\mathbb {R} ^{n}\to \mathbb {R} }$  be differentiable functions, let ${\displaystyle \mathbf {v} ,\mathbf {w} }$  be tangent vectors in ${\displaystyle \mathbb {R} ^{n}}$  at ${\displaystyle \mathbf {x} \in \mathbb {R} ^{n}}$ , and let ${\displaystyle a,b\in \mathbb {R} }$ . Then

1. ${\displaystyle (a\mathbf {v} +b\mathbf {w} )(f)=a\mathbf {v} (f)+b\mathbf {w} (f)}$
2. ${\displaystyle \mathbf {v} (af+bg)=a\mathbf {v} (f)+b\mathbf {v} (g)}$
3. ${\displaystyle \mathbf {v} (fg)=f(\mathbf {x} )\mathbf {v} (g)+g(\mathbf {x} )\mathbf {v} (f)\,.}$

## Tangent vector on manifolds

Let ${\displaystyle M}$  be a differentiable manifold and let ${\displaystyle A(M)}$  be the algebra of real-valued differentiable functions on ${\displaystyle M}$ . Then the tangent vector to ${\displaystyle M}$  at a point ${\displaystyle x}$  in the manifold is given by the derivation ${\displaystyle D_{v}:A(M)\rightarrow \mathbb {R} }$  which shall be linear — i.e., for any ${\displaystyle f,g\in A(M)}$  and ${\displaystyle a,b\in \mathbb {R} }$  we have

${\displaystyle D_{v}(af+bg)=aD_{v}(f)+bD_{v}(g)\,.}$

Note that the derivation will by definition have the Leibniz property

${\displaystyle D_{v}(f\cdot g)(x)=D_{v}(f)(x)\cdot g(x)+f(x)\cdot D_{v}(g)(x)\,.}$