In mathematics, a **tangent vector** it is one that is parallel (or tangent) to a curve or surface at a given point. In differential curve geometry, they are defined in terms of curves in R^{n} or more generally, in differential geometry of variables, as a member of the tangent space. The direction of this vector is the same as the slope of the tangent line.

**Definition**

Be ** r**(t) = x î + y ĵ + z k̂ a differentiable curve, its tangent vector is defined as:

** T**(t) =

**'(t) / |**

*r***'(t) |**

*r*Where ** r**'(t) ≠ 0. Therefore, to find the tangent vector

**(t) of a curve described by**

*T***(t), we must:**

*r*- Determine the derivative
'(t).*r* - Calculate the magnitude of the previous vector.
- Divide the vector we found in step 2 by the magnitude of step 3.

The direction of the tangent vector is the same as the slope of the tangent line.

**Examples**

- Find the vector tangent to the curve given by:

** r**(t) = 3cos (t) î + 3sin (t) ĵ

We know that the tangent vector is given by:

** T**(t) =

**'(t) / |**

*r***'(t) |**

*r*Thus:

** r**'(t) = - 3sin (t) î + 3cos (t) ĵ

Y

|** r**'(t) | = √ [9without

^{2}(t) + 9cos

^{2}(t)] = √9 = 3

So:

** T**(t) = [- 3sin (t) î + 3cos (t) ĵ] / 3 = - sin (t) î + cos (t) ĵ

- Find the vector tangent to the curve given by:

** r**(t) = t î + 1/9 t

^{3}ĵ

We know that the tangent vector is given by:

** T**(t) =

**'(t) / |**

*r***'(t) |**

*r*Thus:

** r**'(t) = î + 1/3 t

^{2}ĵ

Y

|** r**'(t) | = √ [(1)

^{2}+ (1/3 t

^{2})

^{2}] = √ (1 + 1/9 t

^{4})

So:

** T**(t) = (î + 1/3 t

^{2}ĵ) / √ (1 + 1/9 t

^{4})

When t = 3

** T**(t) = (î + 1/3 (3)

^{2}ĵ) / √ (1 + 1/9 (3)

^{4}) = (î + 9/3 ĵ) / √ (1 + 81) = 1 / √10 î + 3 / √10 ĵ

**Tangent vector and normal vector**

Defined as a normal vector ** n** one whose direction is perpendicular to a curve, surface, or any vector parallel (or tangent) to it. Therefore, at a given point P,

**(t) and**

*n***(t) are octagonal.**

*T*