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Tangent Vector — Definition, Formula & Examples

A tangent vector is a vector that points in the direction a curve is heading at a specific point. You find it by taking the derivative of the position vector that traces out the curve.

Given a smooth parametric curve r(t)=⟨x(t), y(t), z(t)⟩\mathbf{r}(t) = \langle x(t),\, y(t),\, z(t) \rangle, the tangent vector at parameter value tt is r′(t)=⟨x′(t), y′(t), z′(t)⟩\mathbf{r}'(t) = \langle x'(t),\, y'(t),\, z'(t) \rangle. The unit tangent vector is T(t)=r′(t)∥r′(t)∥\mathbf{T}(t) = \dfrac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}, defined wherever r′(t)≠0\mathbf{r}'(t) \neq \mathbf{0}.

Key Formula

T(t)=r′(t)∥r′(t)∥\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}
Where:
  • r(t)\mathbf{r}(t) = Position vector tracing the curve as a function of parameter t
  • r′(t)\mathbf{r}'(t) = Derivative of the position vector; the (non-normalized) tangent vector
  • T(t)\mathbf{T}(t) = Unit tangent vector (magnitude 1)

How It Works

Start with a vector-valued function r(t)\mathbf{r}(t) that parametrizes your curve. Differentiate each component with respect to tt to get r′(t)\mathbf{r}'(t)—this is the tangent vector. Its magnitude equals the speed along the curve, and its direction is the instantaneous direction of travel. To get a pure direction with no speed information, normalize it by dividing by its magnitude to obtain the unit tangent vector T(t)\mathbf{T}(t).

Worked Example

Problem: Find the unit tangent vector to the helix r(t)=⟨3cos⁡t, 3sin⁡t, 4t⟩\mathbf{r}(t) = \langle 3\cos t,\, 3\sin t,\, 4t \rangle at t=0t = 0.
Differentiate: Take the derivative of each component with respect to t.
r′(t)=⟨−3sin⁡t, 3cos⁡t, 4⟩\mathbf{r}'(t) = \langle -3\sin t,\, 3\cos t,\, 4 \rangle
Evaluate at t = 0: Plug in t = 0 to get the tangent vector at that point.
r′(0)=⟨0, 3, 4⟩\mathbf{r}'(0) = \langle 0,\, 3,\, 4 \rangle
Normalize: Compute the magnitude and divide to get the unit tangent vector.
∥r′(0)∥=02+32+42=5,T(0)=⟨0, 35, 45⟩\|\mathbf{r}'(0)\| = \sqrt{0^2 + 3^2 + 4^2} = 5, \quad \mathbf{T}(0) = \left\langle 0,\, \frac{3}{5},\, \frac{4}{5} \right\rangle
Answer: The unit tangent vector at t=0t = 0 is T(0)=⟨0, 35, 45⟩\mathbf{T}(0) = \left\langle 0,\, \tfrac{3}{5},\, \tfrac{4}{5} \right\rangle.

Why It Matters

Tangent vectors are essential for computing curvature, writing equations of tangent lines to space curves, and setting up line integrals in multivariable calculus. In physics, the tangent vector to a particle's trajectory gives the velocity direction, which is foundational in mechanics and electromagnetic theory.

Common Mistakes

Mistake: Using r′(t)\mathbf{r}'(t) as the unit tangent vector without normalizing.
Correction: r′(t)\mathbf{r}'(t) has magnitude equal to the speed, not 1. You must divide by ∥r′(t)∥\|\mathbf{r}'(t)\| to obtain the unit tangent vector T(t)\mathbf{T}(t).