Tangent vector

tangent vectorIn 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.

tangent vector