The formula this page uses
T(t) = r′(t) / |r′(t)|
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Unit Tangent Vector Calculator
Solve unit tangent vector problems with clear steps, notation, and a final check.
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Evaluate the governing formula locally in your browser. Define each known quantity once, then change values to test another case instantly.
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Unit Tangent Vector Calculator explained
The short version
- The unit tangent vector T(t) points along a curve in the direction of travel, with length 1.
- Differentiate the position vector to get velocity, then divide by its own speed.
- T tells you the direction only; the speed you divided out is stored separately in |r′(t)|.
What each part means
SymbolWhat it means
r(t) — Position vectorWhere the point is at time t, one component per dimension.
r′(t) — Velocity vectorDifferentiate each component separately. Points along the curve.
|r′(t)| — SpeedThe magnitude of the velocity. It must not be zero, or T is undefined there.
T(t) — Unit tangentDirection of travel, length 1, no units.
Show your work: a full example
- The curve, a helixr(t) = (3 cos t, 3 sin t, 4t)
- Differentiate each componentr′(t) = (−3 sin t, 3 cos t, 4)
- Square and add the components9 sin²t + 9 cos²t + 16
- Use sin²t + cos²t = 19(1) + 16 = 25
- Take the root|r′(t)| = √25 = 5, the same at every t, so the point moves at constant speed
- DivideT(t) = (−0.6 sin t, 0.6 cos t, 0.8)
- Evaluate at t = 0T(0) = (0, 0.6, 0.8), and 0.6² + 0.8² = 1 confirms the length
A second, different case
- A different case: the speed changes with tr(t) = (t, t²), a plain parabola
- Differentiater′(t) = (1, 2t)
- Speed|r′(t)| = √(1 + 4t²), which grows as t grows
- General unit tangentT(t) = (1, 2t) / √(1 + 4t²)
- At t = 0|r′| = 1, so T(0) = (1, 0), pointing straight along the x-axis
- At t = 1|r′| = √5 ≈ 2.236, so T(1) ≈ (0.4472, 0.8944)
- Check against the ordinary slope0.8944 ÷ 0.4472 = 2, and dy/dx = 2t = 2 at t = 1
Copy-ready example
r(t) = (3 cos t, 3 sin t, 4t)
The curve, a helix
Questions about the Unit Tangent Vector Calculator
Does T depend on how fast the curve is traced?
No, and that is the point. Reparametrising r(t) to run twice as fast doubles r′ and doubles |r′|, so the ratio is unchanged. T is a property of the path's shape, not of the schedule.
What happens where the speed is zero?
T is undefined there. On the curve (t², t³) at t = 0 the velocity is (0, 0), which is a cusp: the path stops and turns, so there is no single direction of travel to report.
How does T relate to the tangent line?
T gives the tangent line's direction. The line through r(1) = (1, 1) on the parabola with direction (0.4472, 0.8944) has slope 2, which is exactly the derivative dy/dx there.
Why is T unit length rather than just r′?
Because the Frenet frame that follows, T, N and B, needs three mutually perpendicular unit vectors. Keeping the speed inside T would make curvature and torsion depend on the parametrisation instead of the curve.