Difficulty: Intermediate | Prerequisites: Calculus I and II (derivatives, integrals, parametric curves in 2D), basic vectors and dot/cross products (Chapter 12).
This unit lifts parametric curves out of the flat plane and into three dimensions. Where Calculus II had you tracing curves with x = f(t), y = g(t), you now add a third component and work with vector-valued functions r(t) = ⟨x(t), y(t), z(t)⟩. The payoff is that you can describe paths through space, from planetary orbits to roller-coaster tracks, and extract geometric information (tangent direction, intersection angles, the plane a curve lives on) directly from the parametrisation. You should already be comfortable differentiating and integrating component-wise, and with the dot product and cross product from Chapter 12.
A vector-valued function r(t) traces a curve in 3D. You differentiate it component by component to get a tangent vector, use tangent vectors to find angles of intersection between curves, and test whether a curve lies in a plane by checking that all its tangent vectors are perpendicular to a single normal. Parametrising circles on tilted planes requires building an orthonormal pair of vectors in that plane.
Vector-valued function
A function r(t) = ⟨f(t), g(t), h(t)⟩ whose output is a vector rather than a scalar. Each component is an ordinary real-valued function of the parameter t.
In simple terms, it is a rule that assigns a position vector in space to every value of t, so as t varies you trace out a curve.
Space curve
The set of points (x, y, z) traced out by a vector-valued function as t ranges over its domain.
Think of it as the path you would see if you attached a marker to a moving particle and watched the trail it left behind.
Parametrisation / parametric equations
A way of describing a curve by expressing each coordinate as a function of an independent variable (the parameter), typically t. The same geometric curve can have many different parametrisations.
In simple terms, instead of writing y = f(x), you let both x and y (and z) each depend on a third variable t.
Tangent vector
The derivative r'(t) = ⟨f'(t), g'(t), h'(t)⟩. At any point on the curve where r'(t) ≠ 0, it points in the instantaneous direction of travel.
Think of it as the velocity vector of a particle following the curve.
Unit tangent vector
T(t) = r'(t) / |r'(t)|. The tangent vector scaled to length 1, giving direction without magnitude.
In simple terms, it tells you which way the curve is heading, with the speed information stripped away.
Angle of intersection
When two curves meet at a point, the angle between them is defined as the angle between their tangent vectors at that point, found via the dot-product formula: cos θ = (r₁' · r₂') / (|r₁'| |r₂'|).
Normal vector to a plane
A vector n perpendicular to every vector lying in the plane. If a curve lies entirely in a plane, every tangent vector of the curve is perpendicular to n, so r'(t) · n = 0 for all t.
Orthonormal vectors
Two (or more) vectors that are mutually perpendicular and each of unit length. When parametrising a circle on a plane in 3D, you need two orthonormal vectors that both lie in that plane.
A vector-valued function maps a real number t to a position vector:
\mathbf{r}(t) = \langle f(t),\, g(t),\, h(t) \rangleThe domain is the intersection of the domains of f, g, and h. The image of r is a space curve.
Limits and continuity work component-wise: lim(t→a) r(t) = ⟨lim f(t), lim g(t), lim h(t)⟩, provided each component limit exists.
Differentiation is also component-wise: r'(t) = ⟨f'(t), g'(t), h'(t)⟩.
Integration follows the same pattern: ∫ r(t) dt = ⟨∫ f(t) dt, ∫ g(t) dt, ∫ h(t) dt⟩ + C.
The derivative r'(t) gives the tangent vector at the point r(t). Where r'(t) ≠ 0, the curve is called smooth at that point. The tangent line at t = a is:
\mathbf{L}(t) = \mathbf{r}(a) + t\,\mathbf{r}'(a)Lines in 3D
A line through point p in direction v is r(t) = p + tv. A line segment from p to q is r(t) = (1 - t)p + tq, with 0 ≤ t ≤ 1.
Circles on tilted planes
A circle of radius r centred at point c, lying in a plane with normal n, needs two orthonormal vectors u and v that are both perpendicular to n (and to each other). The parametrisation is then:
\mathbf{r}(t) = \mathbf{c} + r\cos t\;\mathbf{u} + r\sin t\;\mathbf{v}, \quad 0 \le t \le 2\piTo build u and v from n:
Pick any vector w not parallel to n.
Set u = (n x w) / |n x w|.
Set v = (n x u) / |n x u|.
Worked example: Parametrise the circle of radius 2 centred at (1, 2, 0) on the plane x + y + z = 3.
The normal to x + y + z = 3 is n = ⟨1, 1, 1⟩. Choose w = ⟨1, 0, 0⟩. Then n x w = ⟨0, 1, -1⟩, which after normalising gives u = (1/√2)⟨0, 1, -1⟩. Next, n x u ∝ ⟨-2, 1, 1⟩, which normalised gives v = (1/√6)⟨-2, 1, 1⟩. The parametrisation is:
\mathbf{r}(t) = \langle 1,2,0 \rangle + 2\cos t\,\frac{1}{\sqrt{2}}\langle 0,1,-1 \rangle + 2\sin t\,\frac{1}{\sqrt{6}}\langle -2,1,1 \rangleTwo circles intersecting at exactly two points
To build two circles C₁ and C₂ of the same radius that meet at exactly two prescribed points and nowhere else, place each circle's centre so that both points lie on the circle, and choose the planes so the circles are not coplanar. The line through the two intersection points is the "hinge" where the two planes meet. Each circle's centre lies on the perpendicular bisector of the segment joining the two intersection points, within its own plane.
Finding intersections of two space curves
Given r₁(t) and r₂(s), set r₁(t) = r₂(s) and solve the resulting system component by component. Use different parameter names (t and s) because the two curves may pass through the same point at different parameter values.
Important: two curves can pass through the same point without colliding (the particles are there at different times). Intersection means the geometric curves share a point, regardless of timing.
Worked example: r₁(t) = ⟨t, 2t², -1/t⟩ and r₂(s) = ⟨1 - s, 2 cos s, sin s - 1⟩.
From the first component: t = 1 - s, so s = 1 - t. Substituting into the second: 2t² = 2 cos(1 - t). Substituting into the third: -1/t = sin(1 - t) - 1. Testing t = 1 gives s = 0: second component 2(1) = 2 cos 0 = 2 ✓, third component -1 = sin 0 - 1 = -1 ✓. The intersection point is (1, 2, -1).
Computing the angle of intersection
At an intersection point, compute the tangent vectors of each curve (at the parameter values that give that point) and use the dot-product angle formula:
\cos\theta = \frac{\mathbf{r}_1'(t_0) \cdot \mathbf{r}_2'(s_0)}{|\mathbf{r}_1'(t_0)|\;|\mathbf{r}_2'(s_0)|}If the dot product is negative, the tangent vectors point in roughly opposite directions, and the acute angle of intersection is π - θ.
Continuing the worked example: r₁'(t) = ⟨1, 4t, 1/t²⟩ so r₁'(1) = ⟨1, 4, 1⟩. r₂'(s) = ⟨-1, -2 sin s, cos s⟩ so r₂'(0) = ⟨-1, 0, 1⟩. The dot product is -1 + 0 + 1 = 0, so the curves meet at a right angle (θ = π/2).
The Bernoulli spiral and the constant angle property
The curve r(t) = ⟨eᵗ cos 2t, eᵗ sin 2t⟩ has the property that the angle between the position vector r(t) and the tangent vector r'(t) is the same for every t. This angle φ satisfies:
\cos\varphi = \frac{\mathbf{r}(t) \cdot \mathbf{r}'(t)}{|\mathbf{r}(t)|\;|\mathbf{r}'(t)|}Computing the dot product and the magnitudes, the eᵗ factors and the trig cancel, leaving cos φ = 1/√5, so φ = arccos(1/√5) ≈ 1.11 radians. The constancy of this angle is what defines a logarithmic (Bernoulli) spiral.
The strategy
If a curve lies in a plane, all of its tangent vectors are parallel to that plane, which means they are all perpendicular to the plane's normal n. So the test is:
Compute r'(t).
Look for a constant (non-zero) vector n such that r'(t) · n = 0 for all t.
If such n exists, the curve lies in a plane with normal n. Find the equation by plugging any one point on the curve into n · ⟨x, y, z⟩ = d.
Another way to see it: if r'(t) and r''(t) are not parallel, they span the plane the curve lies in, and n = r'(t) x r''(t). If this cross product is constant (same direction for all t), the curve is planar.
Worked example: r(t) = ⟨t² - 1, 6 + 3t - t², 4 - 6t⟩.
r'(t) = ⟨2t, 3 - 2t, -6⟩. We need n = ⟨a, b, c⟩ with 2ta + (3 - 2t)b - 6c = 0 for all t.
Separating the constant and t-dependent parts: 3b - 6c = 0 and 2a - 2b = 0. So a = b and c = b/2. Choosing b = 2 gives n = ⟨2, 2, 1⟩.
Plugging in t = 0, the point (-1, 6, 4): 2(-1) + 2(6) + 1(4) = 14. The plane is 2x + 2y + z = 14.
\mathbf{r}'(t) = \langle f'(t),\, g'(t),\, h'(t) \rangleTangent line at t = a:
\mathbf{L}(t) = \mathbf{r}(a) + t\,\mathbf{r}'(a)Unit tangent vector:
\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}Angle between two tangent vectors:
\cos\theta = \frac{\mathbf{r}_1' \cdot \mathbf{r}_2'}{|\mathbf{r}_1'|\;|\mathbf{r}_2'|}Circle of radius r on a plane, centred at c with orthonormal in-plane vectors u, v:
\mathbf{r}(t) = \mathbf{c} + r\cos t\;\mathbf{u} + r\sin t\;\mathbf{v}Test for planarity: the curve lies in a plane if and only if there is a constant vector n ≠ 0 such that r'(t) · n = 0 for all t. Equivalently, r'(t) x r''(t) is constant in direction.
Vector-valued functions describe particle trajectories in physics (planetary orbits, charged particles in electromagnetic fields), and the tangent vector at any point is the velocity. In computer graphics and CAD, parametric curves in 3D are the basis for rendering smooth surfaces and animation paths. The technique of parametrising a circle on a tilted plane is used routinely in robotics and CNC machining when a tool must trace a circular path on an angled surface.
Students often use the same parameter for both curves when finding intersections (writing r₁(t) = r₂(t) instead of r₁(t) = r₂(s)). This forces the curves to be at the same point at the same parameter value, which is a collision, not a geometric intersection. Use different letters.
Students sometimes forget that the angle between two curves could be obtuse from the dot-product formula and report that directly, when the question asks for the acute angle of intersection. Take π - θ if the initial result exceeds π/2.
When showing a curve lies in a plane, students sometimes check only a few points rather than proving the condition r'(t) · n = 0 holds for all t. Checking finitely many values of t is not a proof.
When parametrising a circle on a plane, students sometimes pick two vectors in the plane that are not orthogonal. If u and v are not perpendicular, the result is an ellipse, not a circle.
⚠️ "Show the curve lies in a plane" is a classic exam question. The expected method is the dot-product test with r'(t) · n = 0, not graphing or checking sample points.
⚠️ Parametrising a circle on a given plane is frequently tested. You must produce orthonormal vectors in the plane. The Gram-Schmidt-lite procedure (cross product, normalise, cross again) is the standard approach.
⚠️ Finding the angle of intersection between two curves requires different parameters (t and s), finding the tangent vectors at the intersection, and applying the dot-product formula. All three steps are expected.
⚠️ The Bernoulli spiral's constant-angle property is a "show that" problem. You need to compute the dot product r(t) · r'(t) and the magnitudes, then show the ratio is independent of t.
True or false: r'(t) gives the position of the particle at time t. (False, it gives the tangent/velocity vector.)
Fill in the blank: to parametrise a circle of radius r on a plane with normal n, you need two ______ vectors that lie in the plane. (Orthonormal.)
True or false: if r₁(t) = r₂(t) has no solution, the two curves do not intersect. (False, you should use different parameters, r₁(t) = r₂(s).)
Fill in the blank: a space curve lies in a plane with normal n if and only if r'(t) · n = ______ for all t. (Zero.)
True or false: the angle between two intersecting curves is found using the position vectors at the intersection point. (False, use the tangent vectors.)
Q: Show that the curve r(t) = ⟨t, t², 1 - t⟩ lies in a plane, and find the equation of that plane.
A: r'(t) = ⟨1, 2t, -1⟩. We need n = ⟨a, b, c⟩ with a + 2tb - c = 0 for all t. The constant part gives a - c = 0 and the t-coefficient gives 2b = 0, so b = 0 and a = c. Taking a = c = 1 gives n = ⟨1, 0, 1⟩. At t = 0, the point is (0, 0, 1): 1(0) + 0(0) + 1(1) = 1. The plane is x + z = 1.
Q: Two curves meet at the point (2, 1, 3). At that point, the tangent vectors are ⟨1, 0, 1⟩ and ⟨0, 1, -1⟩. What is the angle of intersection?
A: Dot product = 0 + 0 + (-1) = -1. Magnitudes are both √2. cos θ = -1/2, so θ = 2π/3. The acute angle of intersection is π - 2π/3 = π/3, or 60°.
Q: Parametrise the circle of radius 3 centred at the origin, lying in the plane z = 0.
A: The normal is n = ⟨0, 0, 1⟩. Two orthonormal in-plane vectors are u = ⟨1, 0, 0⟩ and v = ⟨0, 1, 0⟩. So r(t) = ⟨3 cos t, 3 sin t, 0⟩, 0 ≤ t ≤ 2π.
Q: For r(t) = ⟨eᵗ cos 2t, eᵗ sin 2t⟩, compute |r(t)| and |r'(t)|.
A: |r(t)| = eᵗ√(cos²2t + sin²2t) = eᵗ. r'(t) = ⟨eᵗ cos 2t - 2eᵗ sin 2t, eᵗ sin 2t + 2eᵗ cos 2t⟩ = eᵗ⟨cos 2t - 2 sin 2t, sin 2t + 2 cos 2t⟩. |r'(t)| = eᵗ√(1 + 4) = √5 eᵗ.
This material connects directly to arc length and curvature (§13.3 and §13.4): the tangent vector's magnitude |r'(t)| is the speed, and integrating it gives arc length. Curvature measures how fast the unit tangent vector T changes direction, which builds on everything here.
The dot-product and cross-product techniques from Chapter 12 (vectors and geometry of space) are the engine behind every calculation in this unit. If those feel shaky, reviewing §12.3 and §12.4 will pay off immediately.
Parametric curves also return in Chapter 16 (line integrals), where you integrate a function along a curve described by exactly these parametrisations.
vector-valued function, space curve, parametric equations 3D, parametrisation, tangent vector, unit tangent vector, tangent line to space curve, angle of intersection curves, curve intersection 3D, circle on a plane, orthonormal vectors, normal vector, planarity test, Bernoulli spiral, logarithmic spiral, MATH 215, Stewart Multivariable Calculus §13.1, §13.2, Calculus III