# Tangent vector 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 Rn 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) = r'(t) / |r'(t) |

Where r'(t) ≠ 0. Therefore, to find the tangent vector T(t) of a curve described by r(t), we must:

1. Determine the derivative r'(t).
2. Calculate the magnitude of the previous vector.
3. 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

1. 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) = r'(t) / |r'(t) |

Thus:

r'(t) = - 3sin (t) î + 3cos (t) ĵ

Y

|r'(t) | = √ [9without2(t) + 9cos2(t)] = √9 = 3

So:

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

1. Find the vector tangent to the curve given by:

r(t) = t î + 1/9 t3 ĵ

We know that the tangent vector is given by:

T(t) = r'(t) / |r'(t) |

Thus:

r'(t) = î + 1/3 t2 ĵ

Y

|r'(t) | = √ [(1)2 + (1/3 t2)2] = √ (1 + 1/9 t4)

So:

T(t) = (î + 1/3 t2 ĵ) / √ (1 + 1/9 t4)

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, n(t) and T(t) are octagonal. 