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How sharply a curve bends, measured as $|d\mathbf{T}/ds|$ so that the answer belongs to the curve and not to the schedule tracing it, with the cross-product formula and the tangent, normal and binormal vectors the curve carries with it.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute the curvature of a space curve from the cross-product formula, read it as the reciprocal of the radius of the osculating circle, and build the unit tangent, principal normal and binormal at a point. You will also be able to say why curvature is measured against distance rather than time, and what that buys.
The arc length parametrisation from the last lesson, the unit tangent from lesson 8, and the fact that a vector of constant length is perpendicular to its own derivative. All three are used in the first paragraph below, and the third is what makes the normal vector exist at all.
The curvature $\kappa$ measures how sharply a curve bends, per unit of distance along it. The radius of curvature is $1/\kappa$, and the osculating circle is the circle of that radius that best matches the curve at a point. $\mathbf{T}$, $\mathbf{N}$ and $\mathbf{B}$ — the unit tangent, principal unit normal and binormal — are three mutually perpendicular unit vectors carried along by the moving point, together called the moving frame. The plane of $\mathbf{T}$ and $\mathbf{N}$ is the osculating plane: the plane the curve is momentarily lying in.
Let $\mathbf{T}$ be the unit tangent, $\mathbf{T} = \mathbf{r}'/|\mathbf{r}'|$. The curvature is
$$\kappa = \left| \frac{d\mathbf{T}}{ds} \right|,$$
the rate at which the direction of travel turns, measured against distance rather than against time. That choice is the whole design of the definition: a driver who slows down turns no more sharply, and dividing by $ds$ rather than $dt$ is what makes the answer a fact about the road.
The formula you will actually use. Since $ds/dt = |\mathbf{r}'|$,
$$\kappa = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^{3}}.$$
The second form is the practical one: it needs only two derivatives and one cross product, and it never asks for the arc length parametrisation — which is fortunate, because that parametrisation is almost never available.
The two benchmarks. A straight line has $\mathbf{T}$ constant, so $\kappa = 0$ everywhere. A circle of radius $a$ has $\kappa = 1/a$ everywhere. Every other curvature is read against those: $\kappa$ at a point says the curve is bending there exactly as a circle of radius $1/\kappa$ would.
The moving frame. $\mathbf{T}$ has constant length, so $\mathbf{T}'$ is perpendicular to it. Provided $\mathbf{T}' \ne \mathbf{0}$, define
$$\mathbf{N} = \frac{\mathbf{T}'}{|\mathbf{T}'|}, \qquad \mathbf{B} = \mathbf{T} \times \mathbf{N}.$$
$\mathbf{N}$ is a unit vector pointing into the bend; $\mathbf{B}$ is a unit vector perpendicular to both. The three are mutually perpendicular and right-handed, and they travel with the point — a coordinate system built by the curve for itself, owing nothing to the axes somebody chose.
Another way: picture
Drive along the curve and watch the steering wheel. Curvature is how far the wheel is turned, not how fast you are covering ground: a hairpin taken slowly has the same curvature as a hairpin taken quickly. The osculating circle is the circle you would trace if you froze the wheel where it is, and its radius is the radius of curvature.
Another way: steps
To find the curvature of a curve given as $\mathbf{r}(t)$:
Write $\mathbf{r}' = |\mathbf{r}'|\,\mathbf{T}$, speed times direction. Differentiating with the product rule,
$$\mathbf{r}'' = |\mathbf{r}'|'\,\mathbf{T} + |\mathbf{r}'|\,\mathbf{T}'.$$
Now cross $\mathbf{r}'$ with that. The first term dies, because it is a multiple of $\mathbf{T}$ and $\mathbf{r}'$ is too, and a vector crossed with a parallel vector is zero. What survives is
$$\mathbf{r}' \times \mathbf{r}'' = |\mathbf{r}'|^{2}\,(\mathbf{T} \times \mathbf{T}').$$
Since $\mathbf{T}$ and $\mathbf{T}'$ are perpendicular unit-ish vectors, the length of that cross product is $|\mathbf{T}'|$, and dividing by $|\mathbf{r}'|^{3}$ leaves $|\mathbf{T}'|/|\mathbf{r}'|$, which is $|d\mathbf{T}/ds|$.
The derivation is worth following once because of what it reveals: the component of acceleration along the motion contributes nothing to curvature. Only the sideways part bends the path. That observation is the whole content of the next lesson, arriving a lesson early.
Differentiating $\mathbf{T}$ with respect to $t$ and calling it $\kappa$. $|\mathbf{T}'(t)|$ is turning per unit time; curvature is turning per unit distance, so it needs a division by the speed.
Forgetting the cube. The denominator is $|\mathbf{r}'|^{3}$, not $|\mathbf{r}'|$. It is a cube precisely so that the two chain-rule factors cancel under a change of parameter.
Expecting $\mathbf{N}$ to exist everywhere. Where $\mathbf{T}' = \mathbf{0}$ — along any straight stretch — there is no direction of turning and $\mathbf{N}$ is undefined. The frame is defined where the curve actually bends.
Reading $\mathbf{N}$ as pointing outward. It points into the bend, towards the centre of the osculating circle, which is the same side the passengers are thrown away from.
In first-year calculus the second derivative was the measure of bending, and for a graph $y = f(x)$ near a point where $f'(x) = 0$ the two do agree: $\kappa = |f''|/(1 + f'^{2})^{3/2}$ reduces to $|f''|$. Everywhere else they differ, and the difference is not a refinement — it is a correction for the fact that $f''$ measures bending against the $x$ axis while curvature measures it against the curve itself.
The clearest case is a straight line through the origin at a steep angle, parametrised as $\mathbf{r}(t) = \langle t, 100t, 0 \rangle$. Its second derivative is zero and its curvature is zero, so they agree. Now reparametrise it as $\langle t^{3}, 100t^{3}, 0 \rangle$: the second derivative is no longer zero, yet the curve is still a straight line and its curvature is still zero. The second derivative reports on the schedule; curvature reports on the road.
For $\mathbf{r}(t) = \langle 3\cos t,\, 3\sin t,\, 4t \rangle$, $\mathbf{r}'(t) = \langle -3\sin t,\, 3\cos t,\, 4 \rangle$ and $\mathbf{r}''(t) = \langle -3\cos t,\, -3\sin t,\, 0 \rangle$.
Two derivatives, componentwise.
The cross product works out to $\langle 12\sin t,\, -12\cos t,\, 9 \rangle$, whose length is $\sqrt{144 + 81} = 15$; and $|\mathbf{r}'| = 5$, so $|\mathbf{r}'|^{3} = 125$.
The identity collapses the trigonometric terms again.
So $\kappa = 15/125 = 3/25$, the same at every point, and the radius of curvature is $25/3$ — larger than the helix's own radius of $3$, because climbing straightens the bend out.
A helix has constant curvature, like a circle.
For $\mathbf{r}(t) = \langle a\cos t,\, a\sin t,\, 0 \rangle$ with $a > 0$, the speed is $a$ and $\mathbf{T} = \langle -\sin t,\, \cos t,\, 0 \rangle$.
Divide the velocity by the speed.
$\mathbf{T}'(t) = \langle -\cos t,\, -\sin t,\, 0 \rangle$ already has length one, so $\mathbf{N} = \langle -\cos t,\, -\sin t,\, 0 \rangle$, which points from the moving point straight at the centre.
The normal points into the bend.
$\mathbf{B} = \mathbf{T} \times \mathbf{N} = \langle 0, 0, 1 \rangle$, constant. A constant binormal is exactly what it means for a curve to be flat: the osculating plane never tips, so the whole curve lies in one plane.
A constant binormal means a plane curve.
Take $\mathbf{r}(t) = \langle t,\, t^{2},\, 0 \rangle$, so $\mathbf{r}'(t) = \langle 1,\, 2t,\, 0 \rangle$ and $\mathbf{r}''(t) = \langle 0,\, 2,\, 0 \rangle$.
The vertex is at the parameter value zero.
The cross product is $\langle 0, 0, 2 \rangle$ for every $t$, of length $2$; at $t = 0$ the speed is $1$, so $\kappa = 2/1^{3} = 2$.
The radius of curvature at the vertex is therefore $1/2$. Away from the vertex the speed grows and the curvature falls like $2/(1 + 4t^{2})^{3/2}$: a parabola is sharpest exactly at its turning point and flattens for ever after, which is why the vertex is the only place its curvature is a round number.
Match each object to what it is, for a smooth curve traced at speed $6$.
| A unit vector along the direction of travel | A unit vector pointing the way the curve turns | A unit vector perpendicular to the plane the bend lies in | The rate the direction turns, per unit of distance | |
|---|---|---|---|---|
| The unit tangent $\mathbf{T}$ | ||||
| The unit normal $\mathbf{N}$ | ||||
| The binormal $\mathbf{B}$ | ||||
| The curvature $\kappa$ |
For the circle $\mathbf{r}(t) = \langle 3\cos t,\; 3\sin t,\; 0 \rangle$, fill in the three quantities below.
| Value | |
|---|---|
| The speed | |
| The curvature | |
| The radius of curvature |
For $\mathbf{r}(t) = \langle t,\; 7t^{2},\; 0 \rangle$, find the curvature at $t = 0$.
Answer:
A curve is retraced $4$ times as fast, so that $\mathbf{q}(t) = \mathbf{r}(4t)$. What happens to the curvature at each point of the curve?
Match each object to what it is, for a smooth curve traced at speed $8$.
| A unit vector along the direction of travel | A unit vector pointing the way the curve turns | A unit vector perpendicular to the plane the bend lies in | The rate the direction turns, per unit of distance | |
|---|---|---|---|---|
| The unit tangent $\mathbf{T}$ | ||||
| The unit normal $\mathbf{N}$ | ||||
| The binormal $\mathbf{B}$ | ||||
| The curvature $\kappa$ |
A cycle path is designed so that every bend has radius of curvature at least $8$ metres. What is the largest curvature the path may have anywhere?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For the circle $\mathbf{r}(t) = \langle 3\cos t,\; 3\sin t,\; 0 \rangle$, fill in the three quantities below.
| Value | |
|---|---|
| The speed | |
| The curvature | |
| The radius of curvature |
You can find a curve's curvature and its moving frame, and say why retracing the curve faster changes neither. Next: the acceleration split along that frame, into the part that changes speed and the part that turns.
10. Your turn: the curvature of a parabola at its vertex, step 3