Difficulty: Intermediate | Prerequisites: Calculus I and II (derivatives, integrals, trig identities), familiarity with vectors in 3D, dot product, cross product.
This topic introduces vector-valued functions, which assign a vector to each value of a parameter (usually t). Instead of graphing y = f(x) on a flat plane, you are now tracing curves through three-dimensional space. This is where Calculus III starts feeling genuinely different from what came before: motion, geometry, and calculus merge into one framework. You should already be comfortable differentiating and integrating single-variable functions, and you will need your trig identities (particularly the Pythagorean identity) close at hand.
A vector function r(t) = ⟨f(t), g(t), h(t)⟩ traces a curve in space as t varies. You can find where two surfaces intersect by parametrising their shared points. Derivatives of vector functions give tangent vectors, and integrating the magnitude of the derivative gives arc length.
Vector-valued function (vector function)
A function r(t) = ⟨f(t), g(t), h(t)⟩ whose input is a scalar t and whose output is a vector in ℝ² or ℝ³. Think of it as: a set of instructions that tells you where to be at every moment t.
Space curve
The set of all points (x, y, z) traced out by a vector function as t ranges over an interval. In simple terms, it is the path a point follows through 3D space.
Parameter / parametrisation
The independent variable t that drives the components of a vector function. In simple terms, t is the "dial" you turn to move along the curve. Different parametrisations can trace the same geometric curve at different speeds or directions.
Component functions
The individual scalar functions f(t), g(t), h(t) that make up each coordinate of r(t). Think of them as separate x-, y-, and z-instructions running in parallel.
Tangent vector
The derivative r′(t) = ⟨f′(t), g′(t), h′(t)⟩, which points in the direction the curve is heading at parameter value t. In simple terms, it is the velocity of a particle travelling along the curve.
Arc length
The total distance along a curve between two parameter values, computed by integrating |r′(t)|. Think of it as: if you straightened the curve into a line, how long would it be?
Curve of intersection
The space curve formed where two surfaces meet. You find it by solving the surface equations simultaneously and expressing the result as a vector function of one parameter.
A vector function r(t) = f(t)i + g(t)j + h(t)k assigns a position vector to each t.
To sketch the curve, you eliminate t between the component functions or recognise standard shapes.
Example: r(t) = ti + 2 sin(t)j + 4 cos(t)k. The x-component increases linearly with t, while y and z oscillate. The curve is a helix: it winds around an elliptical cylinder (y²/4 + z²/16 = 1) while marching along the x-axis.
To find direction, pick increasing values of t and plot a few points.
Indicating direction: always mark the curve with an arrow showing the direction of increasing t.
Circle or ellipse in a plane: components that pair up as (a cos t, b sin t) trace an ellipse. If a = b, it is a circle.
Helix: a circle/ellipse pair in two coordinates plus a linearly increasing third coordinate. The pitch of the helix depends on the coefficient of t in the linear component.
Plane curves: when h(t) = 0, r(t) lies in the xy-plane. Example: r(t) = (1 + cos t)i + (3 + sin t)j is a circle of radius 1, centred at (1, 3).
Two particles travel along curves r₁(t) and r₂(t). They collide if r₁(t₀) = r₂(t₀) for some single value t₀ (same place, same time).
Their paths intersect if r₁(t₁) = r₂(t₂) for possibly different parameter values t₁ and t₂ (same place, not necessarily the same time).
To test for collision: set the components equal and solve for a common t.
To test for path intersection: use different parameters (say t and s) and solve the resulting system.
When two surfaces in ℝ³ meet, their overlap is typically a space curve. To find a vector parametrisation of that curve, you solve the two surface equations simultaneously, introducing a parameter t to express x, y, and z.
Identify the two surface equations.
Choose one variable to set as the parameter (often whichever makes the algebra simplest).
Solve for the remaining variables using the surface equations.
Write the result as r(t) = ⟨x(t), y(t), z(t)⟩.
To show (x, y, z) = (sin t, cos t, sin²t) is the intersection of z = x² and x² + y² = 1:
Check z = x²: sin²t = (sin t)² ✓
Check x² + y² = 1: sin²t + cos²t = 1 ✓
Both surface equations are satisfied for all t, so the parametric curve lies on both surfaces simultaneously. It is their curve of intersection.
Given z = 4x² + y² and y = x²:
Let x = t. Then y = t² and z = 4t² + (t²)² = 4t² + t⁴.
The intersection curve is r(t) = ⟨t, t², 4t² + t⁴⟩.
The choice of x = t is arbitrary but convenient. You could equally let y = t, but the algebra would be slightly messier.
\mathbf{r}(t) = \langle f(t),\, g(t),\, h(t) \rangle\mathbf{r}'(t) = \langle f'(t),\, g'(t),\, h'(t) \rangleL = \int_a^b |\mathbf{r}'(t)|\, dt = \int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2 + [h'(t)]^2}\, dt\mathbf{a}(t) = \mathbf{r}''(t) = \langle f''(t),\, g''(t),\, h''(t) \rangleThe tangent vector r′(t) is computed component-by-component. Arc length L integrates the speed (the magnitude of the velocity vector) over the interval [a, b]. The acceleration vector a(t) is simply the second derivative of r(t).
Vector functions describe the trajectory of any object moving through space: a satellite in orbit, a baseball in flight, a roller coaster on its track. Engineers parametrise pipe routes and road curves this way to compute lengths, clearances, and curvatures. In computer graphics, parametric curves (Bezier curves, splines) are the building blocks of every smooth shape you see on screen.
Students often think the parameter t must represent time. It does not. It is any real number that indexes the curve; it could be an angle, a proportion, or a completely abstract quantity.
Students often confuse "the particles collide" with "the paths intersect." Collision requires a single shared t-value. Intersection only requires the same point, possibly at different parameter values.
Students often think a curve has only one parametrisation. The same geometric curve can be traced by infinitely many different vector functions (forwards, backwards, faster, slower).
Students often forget to check direction when sketching. A helix winding clockwise versus anticlockwise is a different answer, and the direction depends on which trig function is in which component.
⚠️ Sketching a space curve and marking the direction of increasing t is a classic exam question. Practise identifying helices, circles, and ellipses from their components.
⚠️ "Find a vector function for the curve of intersection" appears frequently. The method is always the same: pick a parameter, substitute, simplify.
⚠️ Collision vs. intersection is a favourite conceptual question. Know the difference cold.
⚠️ Arc length integrals can be computationally heavy. On exams, the integrand is often engineered to simplify into a perfect square under the radical. Look for that before grinding through algebra.
True or False: If r(t) = ⟨cos t, sin t, t⟩, the curve is a helix. (True)
True or False: Two particles whose paths intersect must collide. (False, they can be at the intersection point at different times.)
Fill in the blank: To find arc length, you integrate the ______ of the derivative vector. (magnitude)
True or False: The parametrisation r(t) = ⟨cos 2t, sin 2t, 0⟩ traces a different geometric curve than r(t) = ⟨cos t, sin t, 0⟩. (False, same unit circle, just traced at twice the speed.)
Fill in the blank: If x = sin t and y = cos t, then x² + y² = ______. (1)
Q: A curve is given by r(t) = ⟨t, t², t³⟩. Another particle travels along r₂(t) = ⟨1 + 2t, 1 + 6t, 1 + 14t⟩. Do the particles collide?
A: Set t = 1 + 2t, t² = 1 + 6t, t³ = 1 + 14t. From the first equation, t = -1 does not satisfy the others. No collision.
Q: Do the paths of the two particles above intersect?
A: Use different parameters: t and s. Solve t = 1 + 2s, t² = 1 + 6s, t³ = 1 + 14s. From the first two equations, find candidate (t, s) pairs. Then check each in the third equation. The paths intersect at (1, 1, 1) (with t = 1, s = 0) and at (-1, 1, -1).
Q: Find a vector function for the curve of intersection of the paraboloid z = 4x² + y² and the cylinder y = x².
A: Let x = t. Then y = t², z = 4t² + t⁴. So r(t) = ⟨t, t², 4t² + t⁴⟩.
Q: Find the arc length of r(t) = 12t i + 8t^(3/2) j + 3t² k for 0 ≤ t ≤ 1.
A: r′(t) = ⟨12, 12t^(1/2), 6t⟩. Then |r′(t)|² = 144 + 144t + 36t² = (12 + 6t)² (recognise the perfect square). So |r′(t)| = 12 + 6t, and L = ∫₀¹ (12 + 6t) dt = 12 + 3 = 15.
Q: The curve r(t) = (2 cos t, 2 sin t, eᵗ) for 0 ≤ t ≤ π. At what point is the tangent line parallel to the plane √3 x + y = 7?
A: The plane's normal is n = ⟨√3, 1, 0⟩. The tangent vector r′(t) = ⟨-2 sin t, 2 cos t, eᵗ⟩ must be perpendicular to n: -2√3 sin t + 2 cos t = 0, giving tan t = 1/√3, so t = π/6. The point is (√3, 1, e^(π/6)).
This material connects directly to the next sections of Calculus III: curvature, the TNB frame, and motion in space all build on the tangent vector and arc length concepts introduced here. Curves of intersection reappear when you set up line integrals and surface integrals later in the course. The parametric-thinking habit you build here is also essential for Green's, Stokes', and the Divergence Theorem.
vector-valued function, vector function, parametric curve, space curve, helix, parametric equations, curve of intersection, surface intersection, parametrisation, arc length formula, tangent vector, velocity vector, acceleration vector, r(t), component functions, collision vs intersection, particle paths, Math 215, Calculus III, multivariable calculus