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Arc Length and Curvature · Tutorial 190 of 999

Unit Tangent Vectors for Non-Unit-Speed Curves

Use the velocity vector and speed to find a curve’s unit tangent, and interpret what normalization changes.

Intermediate 9 min read

What You'll Learn

  • Define the unit tangent vector for a regular parametrized curve.
  • Calculate velocity, speed, and unit tangent from a vector-valued function.
  • Check that a proposed unit tangent has length one.
  • Distinguish a velocity vector’s direction from its magnitude.
  • Recognize when the unit tangent formula is undefined.
  • Understand how reversing parameter direction affects the unit tangent.

From Velocity to Unit Tangent

A parametrized curve comes with a velocity vector: the derivative of its position vector. That velocity carries two kinds of information at once. Its direction indicates the curve’s instantaneous direction of travel, while its magnitude tells how quickly the parameterized point is moving.

For some questions, we need the direction without the speed. We obtain it by dividing the velocity by its length. The resulting vector has length one and points along the curve in the direction of increasing parameter.

Definition: Let \(\mathbf{r}(t)\) be a differentiable space curve. At a parameter value \(t\) where \(\mathbf{r}'(t)\ne\mathbf{0}\), the curve is regular and its unit tangent vector is \[ \mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}. \] The denominator \(\|\mathbf{r}'(t)\|\) is the speed.

Regularity matters: if the velocity is zero, the formula would require division by zero, so it does not define a unit tangent there. A curve may be regular at some parameter values but not at others.

Why Normalizing the Velocity Gives the Tangent Direction

The derivative \(\mathbf{r}'(t)\) describes the limiting direction of nearby secants to the curve. To see this, take a small positive increment \(h\). The vector from \(\mathbf{r}(t)\) to \(\mathbf{r}(t+h)\), divided by its length, is a unit secant direction:

$$ \frac{\mathbf{r}(t+h)-\mathbf{r}(t)} {\|\mathbf{r}(t+h)-\mathbf{r}(t)\|}. $$

For \(h>0\), the numerator can be written as \(h\) times the difference quotient, and the same factor \(h\) cancels from the numerator and denominator. Thus this unit secant direction equals

$$ \frac{\dfrac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}} {\left\|\dfrac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}\right\|}. $$

As \(h\) approaches zero, the difference quotient approaches \(\mathbf{r}'(t)\). If \(\mathbf{r}'(t)\ne\mathbf{0}\), its length approaches the positive number \(\|\mathbf{r}'(t)\|\). Therefore, the unit secant direction approaches \(\mathbf{r}'(t)/\|\mathbf{r}'(t)\|\). This proves that the formula gives the limiting tangent direction for increasing parameter values.

Key distinction: The velocity is \(\mathbf{r}'(t)\), and its magnitude is the speed \(\|\mathbf{r}'(t)\|\). The unit tangent \(\mathbf{T}(t)\) retains the velocity’s direction but has magnitude exactly one.

A Reliable Calculation

For a curve given by \(\mathbf{r}(t)\), calculate its unit tangent in three steps: differentiate each component, find the length of the resulting velocity vector, and divide every velocity component by that length. Keep the speed in the denominator until all components have been divided by it.

1
Differentiate.
Find \(\mathbf{r}'(t)\), the velocity vector.
2
Find the speed.
Compute \(\|\mathbf{r}'(t)\|\) and check that it is nonzero at the parameter values under consideration.
3
Normalize.
Divide the entire velocity vector by its speed, then check that the result has length one.

Worked Example: A Helix with Constant Speed

Find the unit tangent to \(\mathbf{r}(t)=\langle 3\cos t,3\sin t,4t\rangle\). Also find the velocity, speed, and unit tangent at \(t=0\).

Differentiate each component:

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

The speed is the length of this vector. Using \(\sin^2 t+\cos^2 t=1\),

$$ \|\mathbf{r}'(t)\| =\sqrt{(-3\sin t)^2+(3\cos t)^2+4^2} =\sqrt{9\sin^2t+9\cos^2t+16} =\sqrt{9+16} =5. $$

Since the speed is positive for every \(t\), the curve is regular everywhere. Divide each velocity component by \(5\):

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

At \(t=0\), the position is \(\mathbf{r}(0)=\langle 3,0,0\rangle\), and the velocity is

$$ \mathbf{r}'(0)=\langle -3\sin 0,3\cos 0,4\rangle =\langle 0,3,4\rangle. $$

The speed is \(5\), so \(\mathbf{T}(0)=\langle 0,3/5,4/5\rangle\). Check its length:

$$ \|\mathbf{T}(0)\| =\sqrt{0^2+(3/5)^2+(4/5)^2} =\sqrt{9/25+16/25} =1. $$

The velocity has length \(5\), while the unit tangent has length \(1\). Both point in the same direction.

Variable Speed Does Not Change the Normalization Rule

The speed need not be constant. In that case, it remains inside the formula as a function of the parameter. The unit tangent can change direction along the curve even though its length stays fixed at one.

Worked Example: A Regular Curve with Variable Speed

Find the unit tangent to \(\mathbf{r}(t)=\langle t^2,2t,3t\rangle\), and evaluate it at \(t=2\).

Differentiate the components:

$$ \mathbf{r}'(t)=\langle 2t,2,3\rangle. $$

Its speed is

$$ \|\mathbf{r}'(t)\| =\sqrt{(2t)^2+2^2+3^2} =\sqrt{4t^2+4+9} =\sqrt{4t^2+13}. $$

Because \(4t^2+13\ge 13>0\), the speed never vanishes, so the curve is regular for every real \(t\). Therefore,

$$ \mathbf{T}(t) =\frac{\langle 2t,2,3\rangle}{\sqrt{4t^2+13}} =\left\langle \frac{2t}{\sqrt{4t^2+13}}, \frac{2}{\sqrt{4t^2+13}}, \frac{3}{\sqrt{4t^2+13}} \right\rangle. $$

At \(t=2\), the velocity and speed are

$$ \mathbf{r}'(2)=\langle 4,2,3\rangle, \qquad \|\mathbf{r}'(2)\|=\sqrt{4^2+2^2+3^2}=\sqrt{29}. $$

Thus

$$ \mathbf{T}(2)=\left\langle\frac{4}{\sqrt{29}},\frac{2}{\sqrt{29}},\frac{3}{\sqrt{29}}\right\rangle. $$

Check by computing its length:

$$ \|\mathbf{T}(2)\| =\sqrt{\frac{16}{29}+\frac{4}{29}+\frac{9}{29}} =\sqrt{\frac{29}{29}} =1. $$

The speed depends on \(t\), but division by that speed still removes the magnitude of the velocity at each parameter value.

Worked Example: Simplifying Before Evaluating

Find the unit tangent to \(\mathbf{r}(t)=\langle 2t,t^2,2t\rangle\), then evaluate it at \(t=1\).

The velocity is

$$ \mathbf{r}'(t)=\langle 2,2t,2\rangle. $$

Its speed is

$$ \|\mathbf{r}'(t)\| =\sqrt{2^2+(2t)^2+2^2} =\sqrt{4+4t^2+4} =\sqrt{4(t^2+2)} =2\sqrt{t^2+2}. $$

This speed is positive for all \(t\), since \(t^2+2>0\). Dividing gives

$$ \mathbf{T}(t) =\frac{\langle 2,2t,2\rangle}{2\sqrt{t^2+2}} =\left\langle\frac{1}{\sqrt{t^2+2}},\frac{t}{\sqrt{t^2+2}},\frac{1}{\sqrt{t^2+2}}\right\rangle. $$

At \(t=1\), the velocity is \(\langle 2,2,2\rangle\), its speed is \(2\sqrt{3}\), and

$$ \mathbf{T}(1) =\left\langle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right\rangle. $$

Its length checks as

$$ \|\mathbf{T}(1)\| =\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}} =1. $$

At this parameter value, the velocity points in the direction \(\langle 1,1,1\rangle\) and has length \(2\sqrt{3}\). The unit tangent points in that same direction but has length \(1\).

Orientation and Regularity

The unit tangent describes the direction of increasing parameter. If a curve is traversed in reverse, its velocity points the opposite way, and so does its unit tangent. For example, if a new parameter \(u\) is related to \(t\) by a differentiable function \(t=t(u)\), the chain rule gives the new velocity as \(\mathbf{r}'(t(u))\,dt/du\). When \(dt/du\ne 0\), normalizing this vector preserves the original unit tangent if \(dt/du>0\), and reverses it if \(dt/du<0\). The tangent line is the same, but its chosen direction depends on the orientation of the parameter.

Regularity is a local condition on the velocity: at the parameter value in question, the derivative must be nonzero. If it is zero, do not apply the unit tangent formula there by cancellation or by assuming the curve has an ordinary tangent direction. First recognize that this particular formula is undefined at that value.

Common Mistake: Treating Velocity as a Unit Tangent

A frequent error is to differentiate the curve and stop, calling the result the unit tangent. For the helix in the first example, this would give \(\langle -3\sin t,3\cos t,4\rangle\). At \(t=0\), that vector is \(\langle 0,3,4\rangle\), whose length is \(5\), not \(1\). It is the velocity, not the unit tangent.

The correction is to divide by the speed, which is \(5\) in this example:

$$ \frac{\langle 0,3,4\rangle}{5} =\left\langle 0,\frac{3}{5},\frac{4}{5}\right\rangle, \qquad \sqrt{0^2+\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2}=1. $$

A related error is to divide by the position vector’s length \(\|\mathbf{r}(t)\|\) instead of the velocity’s length \(\|\mathbf{r}'(t)\|\). The unit tangent is defined from velocity. The position vector points from the origin to the curve point; in general, it does not give the curve’s tangent direction.

Why the Unit Tangent Is Useful

Velocity combines direction and rate of motion, and both are useful in motion problems. But when the question concerns the curve’s direction at a point, speed is irrelevant. The unit tangent provides a consistent direction vector even when the parameter moves quickly in one part of the curve and slowly in another.

It is also an essential ingredient in later descriptions of how a curve’s direction changes. For now, the central calculation is simple: differentiate, compute the speed, and divide. The regularity check ensures the denominator is nonzero, and the length check confirms that the result is genuinely a unit vector.

Check Your Understanding

For each question, identify the velocity and speed separately from the unit tangent.

  1. For \(\mathbf{r}(t)=\langle 4\cos t,4\sin t,3t\rangle\), find \(\mathbf{r}'(t)\), the speed, and \(\mathbf{T}(t)\).
  2. For \(\mathbf{r}(t)=\langle t,2t^2,0\rangle\), find \(\mathbf{T}(t)\) and state whether the curve is regular for every real \(t\).
  3. Why is \(\mathbf{T}(t)\) undefined at any value where \(\mathbf{r}'(t)=\mathbf{0}\)?
  4. If a parameter change reverses the direction of traversal, what happens to the unit tangent’s direction?
  5. A proposed tangent vector has length \(7\). What does that tell you about whether it is a unit tangent?