Unit 2

Vector Functions & Motion in Space

A vector function feeds in one number — usually time — and hands back a point in space. As \(t\) runs, the point traces a curve: the flight of a projectile, the spiral of an electron in a magnetic field, the path you will integrate a field along in Unit 5. This unit does single-variable calculus on those curves, one component at a time, and then asks the geometric questions: how long is it, how sharply does it bend, and how does an object moving along it accelerate? The lab in Section 4 puts a moving frame on the curve so you can watch the answers.

1 · Space curves

\[ \mathbf{r}(t) = \langle f(t),\ g(t),\ h(t)\rangle = f(t)\,\mathbf{i} + g(t)\,\mathbf{j} + h(t)\,\mathbf{k} \] Three ordinary functions of one variable, stacked. The domain is wherever all three are defined. The graph is the set of tips of \(\mathbf{r}(t)\) — a curve in space, with a direction of travel as \(t\) increases.

Three curves you should recognise instantly:

The helix is the curve of this unit: an electron in a uniform magnetic field, a solenoid winding, a spring.
Curve	\(\mathbf{r}(t)\)	How to see it
Line	\(\mathbf{r}_0 + t\mathbf{v}\)	Unit 1. Constant velocity.
Circle of radius \(a\) in the \(xy\)-plane	\(\langle a\cos t,\ a\sin t,\ 0\rangle\)	\(x^2+y^2 = a^2\) always; counterclockwise seen from \(+z\).
Helix	\(\langle a\cos t,\ a\sin t,\ bt\rangle\)	The circle, climbing at constant rate \(b\). Lives on the cylinder \(x^2+y^2 = a^2\). Climbs \(2\pi b\) per turn (the pitch).

Sketching a space curve

Find a surface it lives on by eliminating \(t\) from two of the components. \(\langle \cos t, \sin t, t^2\rangle\) has \(x^2+y^2 = 1\), so it winds around that cylinder; the third component tells you how it moves along the surface. \(\langle t, t^2, t^3\rangle\) (the twisted cubic) lives on the parabolic cylinder \(y = x^2\) and also on \(z = x^3\). Two surfaces pin a curve down as their intersection.

2 · Derivatives and integrals

\[ \mathbf{r}'(t) = \lim_{h\to0}\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h} = \langle f'(t),\ g'(t),\ h'(t)\rangle \] Differentiate each component. That is the whole rule. Geometrically, \(\mathbf{r}'(t)\) is the tangent vector — it points along the curve in the direction of increasing \(t\), and its length is the speed.

\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \] The unit tangent: direction of travel only, length stripped out. Defined wherever \(\mathbf{r}'\ne\mathbf{0}\) — a curve with \(\mathbf{r}' = \mathbf{0}\) somewhere can have a cusp there.

Differentiation rules

Everything from Calculus I survives, with a dot or a cross in the right place:

\[ \frac{d}{dt}\big[\mathbf{u}\cdot\mathbf{v}\big] = \mathbf{u}'\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v}' \qquad \frac{d}{dt}\big[\mathbf{u}\times\mathbf{v}\big] = \mathbf{u}'\times\mathbf{v} + \mathbf{u}\times\mathbf{v}' \qquad \frac{d}{dt}\big[\mathbf{u}(f(t))\big] = f'(t)\,\mathbf{u}'(f(t)) \] In the cross-product rule the order must be kept — \(\mathbf{u}\times\mathbf{v}'\), not \(\mathbf{v}'\times\mathbf{u}\). Also \((f\mathbf{u})' = f'\mathbf{u} + f\mathbf{u}'\) for a scalar \(f\).

A vector of constant length is perpendicular to its derivative

If \(|\mathbf{u}(t)| = c\), then \(\mathbf{u}\cdot\mathbf{u} = c^2\). Differentiate: \(2\,\mathbf{u}\cdot\mathbf{u}' = 0\). So \(\mathbf{u}'\perp\mathbf{u}\). This tiny fact is why the normal vector in Section 4 is really normal, why uniform circular motion has centripetal acceleration, and why a magnetic field cannot change a particle's speed.

Integration is also componentwise, and the constant of integration is a vector:

\[ \int\mathbf{r}(t)\,dt = \left\langle \int f\,dt,\ \int g\,dt,\ \int h\,dt\right\rangle + \mathbf{C} \]

3 · Arc length

\[ L = \int_a^b |\mathbf{r}'(t)|\,dt = \int_a^b \sqrt{f'(t)^2 + g'(t)^2 + h'(t)^2}\;dt \] Speed integrated over time is distance travelled. This is the Calculus II arc-length formula with a third term under the root. As before, the integrand is usually a nasty radical; textbook problems are rigged so it simplifies.

The arc-length function \(s(t) = \int_a^t|\mathbf{r}'(u)|\,du\) measures distance along the curve from the starting point. If you can invert it, \(t = t(s)\), then \(\mathbf{r}(t(s))\) is the curve parametrised by arc length: one unit of \(s\) is one unit of distance, and the speed is identically 1. It is the geometrically honest parametrisation — every property of the curve's shape (as opposed to how fast it is traced) is defined in terms of \(s\).

The helix by arc length

For \(\mathbf{r}(t) = \langle \cos t, \sin t, t\rangle\), \(|\mathbf{r}'| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt2\), so \(s = \sqrt2\,t\) and \(\mathbf{r}(s) = \langle\cos(s/\sqrt2), \sin(s/\sqrt2), s/\sqrt2\rangle\). Constant speed makes this the one case where the inversion is trivial. Try it on \(\langle t, t^2, t^3\rangle\) and you will see why it is usually not.

4 · Curvature and the TNB frame

Curvature measures how fast the direction of travel turns, per unit of distance. Per distance, not per time — a car turning a corner slowly and one turning it fast follow the same curve.

\[ \kappa = \left|\frac{d\mathbf{T}}{ds}\right| = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} = \frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \] Three forms. The first is the definition. The second is the chain rule (\(dT/ds = (dT/dt)/(ds/dt)\)). The third is the one you use: no need to normalise anything first. For a straight line \(\mathbf{r}'' = \mathbf{0}\) and \(\kappa = 0\). For a circle of radius \(a\), \(\kappa = 1/a\) everywhere — small circles curve hard.

\[ \kappa(x) = \frac{|f''(x)|}{\big[1 + f'(x)^2\big]^{3/2}} \qquad\text{for the plane curve } y = f(x) \] Set \(\mathbf{r}(x) = \langle x, f(x), 0\rangle\) in the cross-product formula and this drops out. Note \(\kappa\) is largest where \(f''\) is large and the slope is small.

The frame

\[ \mathbf{T} = \frac{\mathbf{r}'}{|\mathbf{r}'|} \qquad \mathbf{N} = \frac{\mathbf{T}'}{|\mathbf{T}'|} \qquad \mathbf{B} = \mathbf{T}\times\mathbf{N} \] T points along the curve. N (the principal unit normal) points the way the curve is turning — toward the inside of the bend; it is perpendicular to \(\mathbf{T}\) by the constant-length fact above. B (the binormal) is perpendicular to both. Together they are a right-handed set of axes that rides along the curve.

The plane spanned by \(\mathbf{T}\) and \(\mathbf{N}\) is the osculating plane — the plane the curve is locally lying in.
The osculating circle lies in that plane, is tangent to the curve at the point, has radius \(\rho = 1/\kappa\) and centre \(\mathbf{r} + \rho\,\mathbf{N}\). It is the circle that best fits the curve there — a road engineer's “design radius”.
\(\mathbf{B}\) is normal to the osculating plane. Its rate of change measures torsion — how fast the curve twists out of its plane. A plane curve has zero torsion and constant \(\mathbf{B}\).

Interactive lab

Space curve & moving frame

A point rides along the curve; the three arrows are \(\mathbf{T}\), \(\mathbf{N}\) and \(\mathbf{B}\) at that point. The dashed circle is the osculating circle — watch it shrink where the curve bends hard and blow up where the curve is nearly straight. The graph on the right plots \(\kappa\) and speed against \(t\). Compare the helix (both constant) with the twisted cubic (neither constant) — then press play.

T N B velocity acceleration

Drag to orbit. The solid part of the curve has been traversed; the point moves in the direction of increasing \(t\).

κ(t) and speed |r′(t)| along the curve. The vertical line is where you are.

Parameter t0.00

Show

Speed1.0×

r(t)—

Speed |r′|—

Curvature κ—

Radius 1/κ—

a_T (tangential)—

a_N = κv²—

Curvature, live

5 · Velocity and acceleration

\[ \mathbf{v}(t) = \mathbf{r}'(t),\qquad \text{speed} = |\mathbf{v}(t)|,\qquad \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t) \] Given \(\mathbf{a}\) and initial conditions, integrate twice — componentwise — to recover \(\mathbf{r}\). Newton's second law \(\mathbf{F} = m\mathbf{a}\) is three scalar equations, one per component.

The deep result of this section is that acceleration splits into two parts, one along the curve and one toward its centre of curvature:

\[ \mathbf{a} = a_T\,\mathbf{T} + a_N\,\mathbf{N},\qquad a_T = \frac{d}{dt}|\mathbf{v}| = \frac{\mathbf{r}'\cdot\mathbf{r}''}{|\mathbf{r}'|},\qquad a_N = \kappa|\mathbf{v}|^2 = \frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|} \] There is no \(\mathbf{B}\) component — acceleration always lies in the osculating plane. \(a_T\) is how fast you are speeding up. \(a_N\) is the centripetal part: it grows with the square of speed and with the sharpness of the bend. Doubling your speed through a corner quadruples the sideways force. Check: \(a_T^2 + a_N^2 = |\mathbf{a}|^2\).

Constant speed does not mean zero acceleration

On any curved path, \(a_N = \kappa v^2\) is nonzero even when the speedometer does not move. Uniform circular motion has \(a_T = 0\) and \(a_N = v^2/r\), pointed at the centre. The exam question “the particle moves at constant speed; find its acceleration” is testing exactly this.

Projectile motion

With \(\mathbf{a} = \langle 0, -g, 0\rangle\) (or \(-g\mathbf{k}\) in 3D) and launch velocity \(v_0\) at angle \(\alpha\) from height \(h\):

\[ \mathbf{r}(t) = \left\langle (v_0\cos\alpha)\,t,\;\; h + (v_0\sin\alpha)\,t - \tfrac12 g t^2\right\rangle \] Integrate \(\mathbf{a}\) twice and use \(\mathbf{v}(0)\) and \(\mathbf{r}(0)\) to fix the two vector constants. Range on level ground is \(v_0^2\sin 2\alpha/g\), maximised at \(45^\circ\).

6 · Charged particles in fields

This is where the unit pays for itself. A charge \(q\) with velocity \(\mathbf{v}\) in a magnetic field \(\mathbf{B}\) feels \(\mathbf{F} = q\,\mathbf{v}\times\mathbf{B}\), so Newton's law reads

\[ m\,\mathbf{r}''(t) = q\,\mathbf{r}'(t)\times\mathbf{B} \]

Take \(\mathbf{B} = B\,\mathbf{k}\) (uniform, along \(z\)). Write out the cross product with \(\mathbf{v} = \langle v_x, v_y, v_z\rangle\): \(\mathbf{v}\times B\mathbf{k} = \langle v_y B,\ -v_x B,\ 0\rangle\). So

\[ v_x' = \omega v_y,\qquad v_y' = -\omega v_x,\qquad v_z' = 0,\qquad \omega = \frac{qB}{m} \] The \(z\)-velocity is untouched. The \(x\) and \(y\) velocities chase each other — that is the system for a rotation. Solution: \(v_x = v_\perp\cos\omega t,\ v_y = -v_\perp\sin\omega t\), where \(v_\perp\) is the initial speed perpendicular to \(\mathbf{B}\). Integrate once more and \(x, y\) trace a circle.

\[ \mathbf{r}(t) = \left\langle R\sin\omega t,\; R\cos\omega t,\; v_z t\right\rangle + \mathbf{r}_0, \qquad R = \frac{m v_\perp}{|q|B},\qquad \omega = \frac{|q|B}{m} \] A helix. Radius set by the perpendicular momentum, angular frequency (the cyclotron frequency) set by charge-to-mass ratio and field strength, pitch \(2\pi v_z/\omega\) set by the parallel velocity. If \(v_z = 0\) it is a circle. The sense of rotation flips with the sign of \(q\).

Speed is constant. \(\mathbf{F}\perp\mathbf{v}\), so \(\mathbf{F}\cdot\mathbf{v} = 0\) — the magnetic force does no work. Check it: \(\tfrac{d}{dt}|\mathbf{v}|^2 = 2\mathbf{v}\cdot\mathbf{v}' = \tfrac{2q}{m}\mathbf{v}\cdot(\mathbf{v}\times\mathbf{B}) = 0\).
The cyclotron frequency does not depend on speed. Faster particles make bigger circles in the same time. This is the principle behind the cyclotron, the mass spectrometer, and the way a magnetron in a microwave oven works.
Add an electric field \(\mathbf{E}\) and you get the full Lorentz force \(q(\mathbf{E} + \mathbf{v}\times\mathbf{B})\). With \(\mathbf{E}\perp\mathbf{B}\) the circle drifts sideways at velocity \(\mathbf{E}\times\mathbf{B}/B^2\) — the E-cross-B drift.

Set the lab above to Helix: that is this trajectory. Note that \(\mathbf{N}\) always points at the axis of the cylinder — straight at the centre of the circular motion — and that \(a_T = 0\), consistent with constant speed.

7 · Worked examples

1. Tangent line to a space curve

Find the tangent line to \(\mathbf{r}(t) = \langle t^2, \ln t, 2t\rangle\) at the point \((1, 0, 2)\).

The point corresponds to \(t = 1\) (check: \(1^2 = 1\), \(\ln 1 = 0\), \(2\cdot1 = 2\) ✓).

\[ \mathbf{r}'(t) = \langle 2t,\ 1/t,\ 2\rangle,\qquad \mathbf{r}'(1) = \langle 2, 1, 2\rangle \] \[ \text{Tangent line: } x = 1 + 2s,\quad y = s,\quad z = 2 + 2s \]

A different parameter letter for the line, so it is not confused with the curve's \(t\). The unit tangent there is \(\mathbf{T}(1) = \langle 2,1,2\rangle/3\) since \(|\langle 2,1,2\rangle| = 3\).

2. Arc length of a helix segment

Find the length of \(\mathbf{r}(t) = \langle 3\cos t, 3\sin t, 4t\rangle\) for \(0\le t\le 2\pi\).

\[ \mathbf{r}'(t) = \langle -3\sin t, 3\cos t, 4\rangle,\qquad |\mathbf{r}'| = \sqrt{9\sin^2 t + 9\cos^2 t + 16} = \sqrt{25} = 5 \] \[ L = \int_0^{2\pi} 5\,dt = 10\pi \approx 31.4 \]

Sanity check: one turn of the circle alone is \(2\pi\cdot 3 \approx 18.8\); the climb alone is \(4\cdot2\pi \approx 25.1\). By Pythagoras on the unrolled cylinder, \(\sqrt{18.8^2 + 25.1^2} = \sqrt{354 + 632} \approx 31.4\). ✓ A helix unrolls into a straight line — that is why its arc length is so clean.

3. Curvature of the twisted cubic at the origin

Find \(\kappa\) for \(\mathbf{r}(t) = \langle t, t^2, t^3\rangle\) at \(t = 0\) and at \(t = 1\).

\[ \mathbf{r}' = \langle 1, 2t, 3t^2\rangle,\qquad \mathbf{r}'' = \langle 0, 2, 6t\rangle \] \[ \mathbf{r}'\times\mathbf{r}'' = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&2t&3t^2\\0&2&6t\end{vmatrix} = \langle 12t^2 - 6t^2,\ -(6t - 0),\ 2\rangle = \langle 6t^2, -6t, 2\rangle \]

At \(t = 0\): \(\mathbf{r}' = \langle 1,0,0\rangle\), \(|\mathbf{r}'| = 1\); \(\mathbf{r}'\times\mathbf{r}'' = \langle 0,0,2\rangle\).

\[ \kappa(0) = \frac{2}{1^3} = 2 \]

At \(t = 1\): \(|\mathbf{r}'| = \sqrt{1+4+9} = \sqrt{14}\); \(|\langle 6,-6,2\rangle| = \sqrt{36+36+4} = \sqrt{76}\).

\[ \kappa(1) = \frac{\sqrt{76}}{14\sqrt{14}} \approx \frac{8.72}{52.4} \approx 0.166 \]

The curve bends hardest near the origin and straightens out as it shoots off — exactly what the lab graph shows for the twisted cubic. Radius of curvature at the origin is \(1/2\).

4. Tangential and normal components of acceleration

For \(\mathbf{r}(t) = \langle t, t^2, t^3\rangle\), find \(a_T\) and \(a_N\) at \(t = 1\).

From the previous example, at \(t = 1\): \(\mathbf{r}' = \langle 1,2,3\rangle\), \(\mathbf{r}'' = \langle 0,2,6\rangle\), \(|\mathbf{r}'| = \sqrt{14}\), \(|\mathbf{r}'\times\mathbf{r}''| = \sqrt{76}\).

\[ a_T = \frac{\mathbf{r}'\cdot\mathbf{r}''}{|\mathbf{r}'|} = \frac{0 + 4 + 18}{\sqrt{14}} = \frac{22}{\sqrt{14}} \approx 5.88 \] \[ a_N = \frac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|} = \frac{\sqrt{76}}{\sqrt{14}} \approx 2.33 \]

Check: \(|\mathbf{a}|^2 = 0 + 4 + 36 = 40\), and \(a_T^2 + a_N^2 = 484/14 + 76/14 = 560/14 = 40\). ✓ Most of the acceleration is going into speeding up, not turning — consistent with the small curvature.

5. Recovering position from acceleration

A particle has \(\mathbf{a}(t) = \langle 2, 6t, 0\rangle\), \(\mathbf{v}(0) = \langle 1, 0, 1\rangle\), \(\mathbf{r}(0) = \langle 0, 0, 0\rangle\). Find \(\mathbf{r}(t)\).

\[ \mathbf{v}(t) = \int\mathbf{a}\,dt = \langle 2t, 3t^2, 0\rangle + \mathbf{C}_1,\qquad \mathbf{v}(0) = \mathbf{C}_1 = \langle 1, 0, 1\rangle \] \[ \mathbf{v}(t) = \langle 2t + 1,\ 3t^2,\ 1\rangle \] \[ \mathbf{r}(t) = \langle t^2 + t,\ t^3,\ t\rangle + \mathbf{C}_2,\qquad \mathbf{C}_2 = \mathbf{r}(0) = \mathbf{0} \]

So \(\mathbf{r}(t) = \langle t^2 + t,\ t^3,\ t\rangle\). Two integrations, two vector constants, each fixed by an initial condition.

6. Electron in a magnetic field

An electron (\(m = 9.11\times10^{-31}\) kg, \(|q| = 1.60\times10^{-19}\) C) enters \(\mathbf{B} = 0.01\,\mathbf{k}\) T with \(\mathbf{v}_0 = \langle 2\times10^6, 0, 1\times10^6\rangle\) m/s. Describe its path.

\[ \omega = \frac{|q|B}{m} = \frac{(1.6\times10^{-19})(0.01)}{9.11\times10^{-31}} \approx 1.76\times10^{9}\ \text{rad/s} \] \[ R = \frac{m v_\perp}{|q|B} = \frac{v_\perp}{\omega} = \frac{2\times10^6}{1.76\times10^9} \approx 1.14\ \text{mm} \]

Period \(T = 2\pi/\omega \approx 3.57\) ns. In that time it climbs \(v_z T = (10^6)(3.57\times10^{-9}) \approx 3.6\) mm along \(\mathbf{B}\).

Path: a helix of radius 1.14 mm and pitch 3.6 mm around a line parallel to the \(z\)-axis, traced at constant speed \(\sqrt{4+1}\times10^6 \approx 2.24\times10^6\) m/s. The cyclotron frequency \(\omega/2\pi \approx 280\) MHz is independent of how fast the electron entered — only \(R\) and the pitch depend on the velocity.

Checklist

I differentiate and integrate vector functions componentwise, and keep the constant of integration as a vector.
I can find a tangent line from \(\mathbf{r}'(t_0)\), using a different parameter letter.
I compute arc length as \(\int|\mathbf{r}'|\,dt\) and can reparametrise a constant-speed curve by arc length.
I use \(\kappa = |\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^3\) and know \(\kappa = 1/a\) for a circle of radius \(a\).
I know \(\mathbf{N}\) points toward the inside of the bend and \(\mathbf{B} = \mathbf{T}\times\mathbf{N}\).
I can split \(\mathbf{a}\) into \(a_T\mathbf{T} + a_N\mathbf{N}\) and check \(a_T^2 + a_N^2 = |\mathbf{a}|^2\).
I know constant speed on a curve still means \(a_N = \kappa v^2 \ne 0\).
I can derive helical motion in a uniform \(\mathbf{B}\) from \(\mathbf{F} = q\mathbf{v}\times\mathbf{B}\) and know \(\omega = |q|B/m\) is independent of speed.