Difficulty: Intermediate | Prerequisites: Calculus I integration, trigonometric identities, vectors in 3D
This material covers how to describe paths through three-dimensional space using parametric equations, and how to extract geometric information (speed, length, the plane a curve sits in) from those descriptions. It builds directly on vector arithmetic.
Parametric equations describe paths in 3D by giving each coordinate as a function of a parameter t. To find how fast a particle moves, differentiate to get the velocity vector, then take its magnitude for speed. Arc length is the integral of speed over an interval. Two spheres' closest points lie along the line connecting their centres.
Parametric line segment
A line from point A to point B, written as r(t) = (1 − t)A + tB, where t ranges from 0 to 1. At t = 0 you are at A; at t = 1 you are at B.
In simple terms, this is a recipe that slides you from one point to another as t goes from 0 to 1.
Sphere
The set of all points at a fixed distance (the radius r) from a fixed point (the centre C). Equation: (x − c₁)² + (y − c₂)² + (z − c₃)² = r².
Think of it as: a ball's surface in 3D, defined entirely by its centre and radius.
Velocity vector
The derivative of a position vector: r′(t) = ⟨x′(t), y′(t), z′(t)⟩. It points in the direction of motion and its magnitude is the speed.
Speed
|r′(t)|, the magnitude of the velocity vector. A scalar quantity telling you how fast the particle is moving at time t, with no directional information.
Arc length
The total distance travelled along a curve from t = a to t = b. Computed as the integral of speed: L = ∫ from a to b of |r′(t)| dt.
Think of it as: if you unrolled the curve into a straight line, arc length is how long that line would be.
To parametrise the segment from point C to point D with t in [0, 1]: r(t) = (1 − t)C + tD, or equivalently r(t) = C + t(D − C).
Example: C = (−2, 1, 3) and D = (2, −7, 11). Then r(t) = ⟨−2 + 4t, 1 − 8t, 3 + 8t⟩.
Check: at t = 0, r(0) = (−2, 1, 3) = C. At t = 1, r(1) = (2, −7, 11) = D.
Given two spheres with centres C and D, the closest pair of points (one on each sphere) lies along the line segment CD.
First compute |CD|, the distance between centres. If |CD| > r₁ + r₂, the spheres do not overlap, and the gap between them is |CD| − r₁ − r₂.
The closest point P on sphere 1 sits at distance r₁ from C along the direction toward D. Using the unit vector from C to D: P = C + r₁ × (D − C)/|D − C|.
The closest point Q on sphere 2 sits at distance r₂ from D along the direction toward C: Q = D − r₂ × (D − C)/|D − C|.
Example: C = (−2, 1, 3) with r₁ = 3, D = (2, −7, 11) with r₂ = 6. The vector CD = ⟨4, −8, 8⟩, with |CD| = √(16 + 64 + 64) = 12. Unit vector = ⟨1/3, −2/3, 2/3⟩. So P = (−2 + 1, 1 − 2, 3 + 2) = (−1, −1, 5) and Q = (2 − 2, −7 + 4, 11 − 4) = (0, −3, 7).
Given r(t) = ⟨x(t), y(t), z(t)⟩, differentiate each component: r′(t) = ⟨x′(t), y′(t), z′(t)⟩.
Speed = |r′(t)| = √(x′(t)² + y′(t)² + z′(t)²).
Example: r(t) = ⟨√2(t − sin t), 2 − cos t, 3 − cos t⟩. Then r′(t) = ⟨√2(1 − cos t), sin t, sin t⟩.
Speed² = 2(1 − cos t)² + sin² t + sin² t = 2(1 − cos t)² + 2sin² t.
Expand: 2(1 − 2cos t + cos² t) + 2sin² t = 2 − 4cos t + 2cos² t + 2sin² t = 2 − 4cos t + 2 = 4(1 − cos t).
Using the identity 1 − cos t = 2sin²(t/2): speed² = 8sin²(t/2), so speed = 2√2 |sin(t/2)|.
Arc length from t = 0 to t = π: L = ∫ from 0 to π of 2√2 sin(t/2) dt (sin(t/2) ≥ 0 on this interval).
Substitution: let u = t/2, du = dt/2, so dt = 2du. When t = 0, u = 0; when t = π, u = π/2.
L = 2√2 × 2 ∫ from 0 to π/2 of sin(u) du = 4√2 [−cos(u)] from 0 to π/2 = 4√2 (0 − (−1)) = 4√2.
If you can find constants A, B, C, D such that Ax(t) + By(t) + Cz(t) = D for all t, then the curve lies in the plane Ax + By + Cz = D.
Example: for r(t) = ⟨√2(t − sin t), 2 − cos t, 3 − cos t⟩, notice that the y- and z-components satisfy z(t) − y(t) = (3 − cos t) − (2 − cos t) = 1 for all t.
So the curve lies in the plane z − y = 1, or equivalently −y + z = 1.
\mathbf{r}(t) = (1-t)\,C + t\,D \quad (t \in [0,1])(x - c_1)^2 + (y - c_2)^2 + (z - c_3)^2 = r^2|\mathbf{r}'(t)| = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}L = \int_a^b |\mathbf{r}'(t)|\, dt\sin^2 x = \frac{1 - \cos(2x)}{2}, \qquad 1 - \cos t = 2\sin^2\!\left(\tfrac{t}{2}\right)Students often confuse speed (a scalar, |r′(t)|) with velocity (a vector, r′(t)). The exam asks for speed, not velocity, when it writes |r′(t)|.
When computing speed², students frequently forget to apply trig identities to simplify the expression before taking the square root. The identity 1 − cos t = 2sin²(t/2) is essential for this exam.
Students sometimes write the parametrisation of a line segment with t going from 0 to the length of the segment. For the standard form r(t) = (1 − t)C + tD, the parameter t runs from 0 to 1.
When finding closest points on two spheres, students occasionally project from the wrong centre, or move in the wrong direction along the line. P is on sphere 1, so it is r₁ away from C toward D, not away from D.
⚠️ Parametric lines are tested as a building block: you parametrise a segment, then use it to find a point at a specific fraction of the way along.
⚠️ Arc length problems on this exam require trig simplification before integration. If your integrand looks messy, you probably need a half-angle identity.
⚠️ "The curve lies in a plane" questions ask you to find a linear relationship among x(t), y(t), z(t) that holds for all t. Look for components that differ by a constant.
True or false: the parametrisation r(t) = (1 − t)A + tB gives point A at t = 1. (False: A is at t = 0.)
Fill in the blank: speed is the ____ of the velocity vector. (Magnitude.)
True or false: arc length can be negative. (False: it is always non-negative.)
Fill in the blank: the closest points on two non-overlapping spheres lie along the ____ connecting their centres. (Line segment.)
Q: Parametrise the segment from C = (−2, 1, 3) to D = (2, −7, 11) with t in [0, 1].
A: r(t) = ⟨−2 + 4t, 1 − 8t, 3 + 8t⟩.
Q: Two spheres have centres C = (−2, 1, 3) and D = (2, −7, 11) with radii 3 and 6. Find the points P (on sphere 1) and Q (on sphere 2) that minimise the distance between the spheres.
A: |CD| = 12. Unit vector from C to D = ⟨1/3, −2/3, 2/3⟩. P = (−2, 1, 3) + 3⟨1/3, −2/3, 2/3⟩ = (−1, −1, 5). Q = (2, −7, 11) − 6⟨1/3, −2/3, 2/3⟩ = (0, −3, 7).
Q: For r(t) = ⟨√2(t − sin t), 2 − cos t, 3 − cos t⟩, find the speed |r′(t)|.
A: r′(t) = ⟨√2(1 − cos t), sin t, sin t⟩. Speed = 2√2 |sin(t/2)|.
Q: Find the arc length of the above curve from t = 0 to t = π.
A: L = ∫ from 0 to π of 2√2 sin(t/2) dt = 4√2.
Q: Show that the curve r(t) = ⟨√2(t − sin t), 2 − cos t, 3 − cos t⟩ lies in a plane and find the equation of that plane.
A: z(t) − y(t) = (3 − cos t) − (2 − cos t) = 1 for all t. The plane is z − y = 1, or equivalently −y + z = 1.
Parametric lines reappear when computing tangent lines to curves and surfaces. Arc length leads into arc-length parametrisation, which is used to define curvature. The technique of showing a curve lies in a plane connects to level sets and to finding equations of planes (covered in the next set of notes).
Related Terms / Search Tags: parametric line segment, parametric curve, velocity vector, speed, arc length, arc length integral, sphere equation, closest points on spheres, half-angle identity, trig substitution, curve in a plane, MATH 215, Calculus III, University of Michigan, parametrisation