Unit Tangent and Unit Normal Vectors

The unit tangent vector T(t) = r′(t)/‖r′(t)‖ points in the direction a curve is moving, and the unit normal vector N(t) = T′(t)/‖T′(t)‖ points in the direction the curve is bending. Both have length 1, and N is always perpendicular to T.

Before this lesson: Vector Functions and Space Curves, Arc Length of a Space Curve and Quotient Rule

Direction of motion and direction of bending

The velocity r′(t) points along the curve, but its length is the speed, which can change. Dividing by the speed keeps only the direction. That direction is the unit tangent vector T(t).

As you move along a curve, T(t) turns. Its derivative T′(t) shows which way it is turning. Because T always has length 1, T′ is perpendicular to T, so it points sideways, toward the inside of the bend. Scaling T′ to length 1 gives the unit normal vector N(t).

Think of driving a car. T is the way the car points. N is the way you turn the steering wheel: toward the center of the curve you are driving around.

Unit tangent and unit normal vector formulas

\mathbf{T}(t) = \dfrac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \qquad \mathbf{N}(t) = \dfrac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|}

• T(t) needs r′(t) ≠ 0. N(t) needs T′(t) ≠ 0; on a straight line T never changes, so N is not defined.
• T · N = 0 at every point, because T · T = 1 for all t and differentiating gives 2T · T′ = 0.
• N points toward the concave side of the curve (the inside of the bend).
• To find N at one point, differentiate T first, then substitute the value of t. Substituting first turns T into a constant, whose derivative is 0.

Unit tangent and unit normal vector examples

Example 1: Unit tangent of a line

Find T(t) for the line r(t) = ⟨1 + 2t, 3t, 2 − 6t⟩. Does N(t) exist?

\mathbf{r}'(t) = \langle 2, 3, -6 \rangle

\|\mathbf{r}'(t)\| = \sqrt{4 + 9 + 36}

= \sqrt{49} = 7

\mathbf{T}(t) = \left\langle \dfrac{2}{7}, \; \dfrac{3}{7}, \; -\dfrac{6}{7} \right\rangle

\mathbf{T}'(t) = \langle 0, 0, 0 \rangle

Answer: T(t) = ⟨2/7, 3/7, −6/7⟩ for every t. A line does not bend, so T′ = 0 and N(t) is not defined.

Example 2: T and N for a helix

Find T(t) and N(t) for the helix r(t) = ⟨3 cos t, 3 sin t, 4t⟩.

\mathbf{r}'(t) = \langle -3\sin t, \; 3\cos t, \; 4 \rangle

\|\mathbf{r}'(t)\| = \sqrt{9\sin^2 t + 9\cos^2 t + 16}

= \sqrt{9 + 16} = 5

\mathbf{T}(t) = \left\langle -\dfrac{3}{5}\sin t, \; \dfrac{3}{5}\cos t, \; \dfrac{4}{5} \right\rangle

\mathbf{T}'(t) = \left\langle -\dfrac{3}{5}\cos t, \; -\dfrac{3}{5}\sin t, \; 0 \right\rangle

\|\mathbf{T}'(t)\| = \sqrt{\dfrac{9}{25}\cos^2 t + \dfrac{9}{25}\sin^2 t} = \dfrac{3}{5}

\mathbf{N}(t) = \langle -\cos t, \; -\sin t, \; 0 \rangle

Answer: T(t) = ⟨−(3/5) sin t, (3/5) cos t, 4/5⟩ and N(t) = ⟨−cos t, −sin t, 0⟩. N is horizontal and points from the curve straight toward the z-axis, the center of the bend.

Example 3: Unit tangent at a point

Find T(t) for r(t) = ⟨eᵗ, e⁻ᵗ, √2 t⟩, and then T(0).

\mathbf{r}'(t) = \langle e^t, \; -e^{-t}, \; \sqrt{2} \rangle

\|\mathbf{r}'(t)\|^2 = e^{2t} + e^{-2t} + 2

= (e^t + e^{-t})^2

\|\mathbf{r}'(t)\| = e^t + e^{-t}

\mathbf{T}(t) = \dfrac{\langle e^t, \; -e^{-t}, \; \sqrt{2} \rangle}{e^t + e^{-t}}

\mathbf{T}(0) = \dfrac{\langle 1, \; -1, \; \sqrt{2} \rangle}{2}

\|\mathbf{T}(0)\| = \dfrac{\sqrt{1 + 1 + 2}}{2} = 1

Answer: T(0) = ⟨1/2, −1/2, √2/2⟩.

Example 4: T and N at a point when the speed changes

Find T(1) and N(1) for r(t) = ⟨t², 2t, ln t⟩.

\mathbf{r}'(t) = \left\langle 2t, \; 2, \; \dfrac{1}{t} \right\rangle

\|\mathbf{r}'(t)\| = 2t + \dfrac{1}{t} = \dfrac{2t^2 + 1}{t}

\mathbf{T}(t) = \left\langle 2t, \; 2, \; \dfrac{1}{t} \right\rangle \cdot \dfrac{t}{2t^2 + 1}

= \dfrac{\langle 2t^2, \; 2t, \; 1 \rangle}{2t^2 + 1}

\mathbf{T}(1) = \dfrac{\langle 2, 2, 1 \rangle}{3}

\left( \dfrac{2t^2}{2t^2 + 1} \right)' = \dfrac{4t(2t^2 + 1) - 2t^2(4t)}{(2t^2 + 1)^2} = \dfrac{4t}{(2t^2 + 1)^2}

\left( \dfrac{2t}{2t^2 + 1} \right)' = \dfrac{2(2t^2 + 1) - 2t(4t)}{(2t^2 + 1)^2} = \dfrac{2 - 4t^2}{(2t^2 + 1)^2}

\left( \dfrac{1}{2t^2 + 1} \right)' = \dfrac{-4t}{(2t^2 + 1)^2}

\mathbf{T}'(1) = \dfrac{\langle 4, \; -2, \; -4 \rangle}{9}

\|\mathbf{T}'(1)\| = \dfrac{\sqrt{16 + 4 + 16}}{9} = \dfrac{6}{9} = \dfrac{2}{3}

\mathbf{N}(1) = \dfrac{\langle 4, -2, -4 \rangle}{9} \cdot \dfrac{3}{2} = \dfrac{\langle 2, -1, -2 \rangle}{3}

\mathbf{T}(1) \cdot \mathbf{N}(1) = \dfrac{4 - 2 - 2}{9} = 0

Answer: T(1) = ⟨2/3, 2/3, 1/3⟩ and N(1) = ⟨2/3, −1/3, −2/3⟩.

Example 5: Driving up a spiral ramp (word problem)

A car drives up a spiral parking-garage ramp along r(t) = ⟨12 cos t, 12 sin t, 5t⟩, in meters. Find T and N at t = π/2, and describe which way the car is heading and which way it is turning.

\mathbf{r}'(t) = \langle -12\sin t, \; 12\cos t, \; 5 \rangle

\|\mathbf{r}'(t)\| = \sqrt{144 + 25} = 13

\mathbf{T}(t) = \dfrac{\langle -12\sin t, \; 12\cos t, \; 5 \rangle}{13}

\mathbf{T}'(t) = \dfrac{\langle -12\cos t, \; -12\sin t, \; 0 \rangle}{13}

\|\mathbf{T}'(t)\| = \dfrac{12}{13}

\mathbf{N}(t) = \langle -\cos t, \; -\sin t, \; 0 \rangle

\mathbf{r}\left(\dfrac{\pi}{2}\right) = \left\langle 0, \; 12, \; \dfrac{5\pi}{2} \right\rangle

\mathbf{T}\left(\dfrac{\pi}{2}\right) = \left\langle -\dfrac{12}{13}, \; 0, \; \dfrac{5}{13} \right\rangle

\mathbf{N}\left(\dfrac{\pi}{2}\right) = \langle 0, \; -1, \; 0 \rangle

Answer: at (0, 12, 5π/2) the car heads in the direction ⟨−12/13, 0, 5/13⟩: mostly in the −x direction while climbing. N = ⟨0, −1, 0⟩ points from the car straight toward the ramp’s center post on the z-axis, so the car is turning left, toward the center.

Common mistakes

Related methods

The binormal vector

The binormal vector B = T × N is a third unit vector, perpendicular to both T and N. Together, T, N and B form a moving set of axes called the TNB (or Frenet) frame. For the helix in Example 2, the cross product gives B = ⟨(4/5) sin t, −(4/5) cos t, 3/5⟩.

A shortcut in the plane

For a curve in the xy-plane with T = ⟨a, b⟩, the normal is one of the two perpendicular unit vectors ⟨−b, a⟩ or ⟨b, −a⟩. Choose the one that points toward the inside of the bend, which for a graph y = f(x) is upward where it is concave up.

What comes next

How fast T turns, compared with distance traveled, is the curvature of the curve. Later in this unit, T and N also split acceleration into a part along the path and a part toward the center of the bend.

Practice problems

1. Find T(t) for the line r(t) = ⟨3 − t, 2 + 2t, 4 + 2t⟩.
2. Find T(t) for the helix r(t) = ⟨cos t, sin t, t⟩.
3. Find N(t) for the helix in Problem 2.
4. Find T(1) for r(t) = ⟨t, t², t³⟩.
5. Find T(t) and N(t) for r(t) = ⟨t, cos 2t, sin 2t⟩.
6. For the parabola r(t) = ⟨t, t²⟩, find T(1), then use the shortcut in the plane to find N(1).
7. Show that T(t) · N(t) = 0 for the helix in Problems 2 and 3.
8. A drone flies along r(t) = ⟨6t, 3t², t³⟩. Find T(1) and N(1).
9. Find the binormal vector B(t) = T(t) × N(t) for the helix in Problems 2 and 3.
10. A rider on a carousel moves along r(t) = ⟨4 cos 3t, 4 sin 3t, 2⟩ (meters, seconds). Find T and N at t = π/6, and show that N points from the rider to the center of the carousel at (0, 0, 2).

\mathbf{r}'(t) = \langle -1, 2, 2 \rangle

\|\mathbf{r}'(t)\| = \sqrt{1 + 4 + 4} = 3

\mathbf{T}(t) = \left\langle -\dfrac{1}{3}, \; \dfrac{2}{3}, \; \dfrac{2}{3} \right\rangle

\mathbf{r}'(t) = \langle -\sin t, \; \cos t, \; 1 \rangle

\|\mathbf{r}'(t)\| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}

\mathbf{T}(t) = \dfrac{\langle -\sin t, \; \cos t, \; 1 \rangle}{\sqrt{2}}

\mathbf{T}'(t) = \dfrac{\langle -\cos t, \; -\sin t, \; 0 \rangle}{\sqrt{2}}

\|\mathbf{T}'(t)\| = \dfrac{1}{\sqrt{2}}

\mathbf{N}(t) = \langle -\cos t, \; -\sin t, \; 0 \rangle

\mathbf{r}'(t) = \langle 1, \; 2t, \; 3t^2 \rangle

\mathbf{r}'(1) = \langle 1, 2, 3 \rangle

\|\mathbf{r}'(1)\| = \sqrt{1 + 4 + 9} = \sqrt{14}

\mathbf{T}(1) = \dfrac{\langle 1, 2, 3 \rangle}{\sqrt{14}}

\mathbf{r}'(t) = \langle 1, \; -2\sin 2t, \; 2\cos 2t \rangle

\|\mathbf{r}'(t)\| = \sqrt{1 + 4\sin^2 2t + 4\cos^2 2t} = \sqrt{5}

\mathbf{T}(t) = \dfrac{\langle 1, \; -2\sin 2t, \; 2\cos 2t \rangle}{\sqrt{5}}

\mathbf{T}'(t) = \dfrac{\langle 0, \; -4\cos 2t, \; -4\sin 2t \rangle}{\sqrt{5}}

\|\mathbf{T}'(t)\| = \dfrac{4}{\sqrt{5}}

\mathbf{N}(t) = \langle 0, \; -\cos 2t, \; -\sin 2t \rangle

\mathbf{r}'(1) = \langle 1, 2 \rangle

\mathbf{T}(1) = \dfrac{\langle 1, 2 \rangle}{\sqrt{5}}

\dfrac{\langle -2, 1 \rangle}{\sqrt{5}} \text{ or } \dfrac{\langle 2, -1 \rangle}{\sqrt{5}}

\mathbf{N}(1) = \dfrac{\langle -2, 1 \rangle}{\sqrt{5}}

\mathbf{T} \cdot \mathbf{N} = \dfrac{(-\sin t)(-\cos t) + (\cos t)(-\sin t) + (1)(0)}{\sqrt{2}}

= \dfrac{\sin t\cos t - \sin t\cos t + 0}{\sqrt{2}}

= 0

\mathbf{T}(t) = \dfrac{\langle 6, \; 6t, \; 3t^2 \rangle}{3t^2 + 6} = \dfrac{\langle 2, \; 2t, \; t^2 \rangle}{t^2 + 2}

\mathbf{T}(1) = \dfrac{\langle 2, 2, 1 \rangle}{3}

\mathbf{T}'(t) = \dfrac{\langle -4t, \; 4 - 2t^2, \; 4t \rangle}{(t^2 + 2)^2}

\mathbf{T}'(1) = \dfrac{\langle -4, \; 2, \; 4 \rangle}{9}

\|\mathbf{T}'(1)\| = \dfrac{\sqrt{16 + 4 + 16}}{9} = \dfrac{2}{3}

\mathbf{N}(1) = \dfrac{\langle -2, 1, 2 \rangle}{3}

\mathbf{T} \times \mathbf{N} = \dfrac{1}{\sqrt{2}} \langle -\sin t, \; \cos t, \; 1 \rangle \times \langle -\cos t, \; -\sin t, \; 0 \rangle

x: \; (\cos t)(0) - (1)(-\sin t) = \sin t

y: \; (1)(-\cos t) - (-\sin t)(0) = -\cos t

z: \; (-\sin t)(-\sin t) - (\cos t)(-\cos t) = 1

\mathbf{r}'(t) = \langle -12\sin 3t, \; 12\cos 3t, \; 0 \rangle

\mathbf{T}(t) = \langle -\sin 3t, \; \cos 3t, \; 0 \rangle

\mathbf{T}'(t) = \langle -3\cos 3t, \; -3\sin 3t, \; 0 \rangle

\mathbf{N}(t) = \langle -\cos 3t, \; -\sin 3t, \; 0 \rangle

\mathbf{T}\left(\dfrac{\pi}{6}\right) = \langle -1, 0, 0 \rangle, \; \mathbf{N}\left(\dfrac{\pi}{6}\right) = \langle 0, -1, 0 \rangle

\langle 0, 0, 2 \rangle - \langle 0, 4, 2 \rangle = \langle 0, -4, 0 \rangle

Unit tangent and unit normal vectors FAQ

How do you find the unit tangent vector?

Differentiate the position to get r′(t), find its length, and divide: T(t) = r′(t)/‖r′(t)‖. The result points the way the curve moves and has length 1.

How do you find the unit normal vector?

Differentiate the unit tangent vector and divide by its length: N(t) = T′(t)/‖T′(t)‖. It points toward the inside of the bend.

Why is the unit normal vector perpendicular to the unit tangent vector?

T always has length 1, so T · T = 1. Differentiating both sides gives 2T · T′ = 0, so T′, and therefore N, is perpendicular to T.

What is the binormal vector?

The binormal vector is B = T × N. It is a unit vector perpendicular to both T and N, and it completes the TNB frame.

Summary

• Unit tangent vector: T(t) = r′(t)/‖r′(t)‖, the direction of motion.
• Unit normal vector: N(t) = T′(t)/‖T′(t)‖, the direction of bending, toward the inside of the curve.
• T · N = 0, and both have length 1.
• Differentiate before you substitute a value of t.
• A straight line has a constant T and no N; the binormal is B = T × N.

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