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 Open it in Interactive Laboratories →

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.