Difficulty: Intermediate | Prerequisites: Calculus I/II (derivatives, integrals), basic vectors and dot/cross product operations.
Space curves extend the idea of a parametric curve from the plane into three dimensions. Instead of tracing a path on a flat sheet, you trace one through space, and the calculus you already know (derivatives, integrals) carries over component by component. This material sits early in most Calculus III courses because it introduces the vector-valued function, the object you will work with for the rest of the term. If you are comfortable differentiating and integrating single-variable functions and can compute dot and cross products, you have what you need.
A space curve is a vector-valued function r(t) = ⟨x(t), y(t), z(t)⟩ whose derivative gives velocity and whose magnitude gives speed. You find arc length by integrating speed, test whether a curve lies in a plane by checking that its cross products stay parallel, and the Bernoulli spiral is the classic example where the angle between the position vector and the tangent vector is constant.
Vector-valued function / space curve
A function r(t) = ⟨x(t), y(t), z(t)⟩ that assigns a position vector in R³ to each value of the parameter t. In simple terms, it is a recipe that traces out a curve in space as t changes.
Tangent vector
r'(t) = ⟨x'(t), y'(t), z'(t)⟩, the derivative of the position vector. Think of it as the velocity of a particle moving along the curve: it points in the direction of travel.
Speed
The magnitude of the tangent vector, |r'(t)|. This is how fast the particle moves along the curve at time t, stripped of direction.
Arc length
The total distance travelled along the curve from t = a to t = b, computed as the integral of speed. Think of it as unrolling the curve into a straight line and measuring it.
Normal vector to a plane
A vector perpendicular to every vector lying in the plane. If a curve lies in a plane, the cross product of any two non-parallel tangent vectors to the curve gives a normal to that plane.
Bernoulli spiral (equiangular spiral / logarithmic spiral)
A plane curve of the form r(t) = ⟨e^t cos(2t), e^t sin(2t)⟩ (or similar) for which the angle between the position vector and the tangent vector is the same at every point. In simple terms, the spiral crosses every radial line at the same angle.
Parametrising a space curve
A space curve is given by r(t) = ⟨x(t), y(t), z(t)⟩ for t in some interval.
Each component is an ordinary function of one variable. You differentiate and integrate component by component.
Showing a curve lies in a plane
The standard test: compute r'(t) and r''(t). Their cross product r'(t) x r''(t) gives a vector normal to the osculating plane. If that cross product is a constant vector (same direction for all t), every point of the curve lies in a single plane.
Worked approach for r(t) = ⟨t² - 1, 6 - 3t + t², 4 - 6t⟩:
Differentiate: r'(t) = ⟨2t, -3 + 2t, -6⟩.
Differentiate again: r''(t) = ⟨2, 2, 0⟩.
Cross product: r'(t) x r''(t) = ⟨0·0 - (-6)·2, (-6)·2 - (2t)·0, (2t)·2 - (-3+2t)·2⟩ = ⟨12, -12, 4t - (-6+4t)⟩ = ⟨12, -12, 6⟩.
This simplifies to a constant vector (no t remaining), confirming the curve lies in a plane.
To find the plane equation, use the normal ⟨12, -12, 6⟩ (or simplify to ⟨2, -2, 1⟩) and a point on the curve, say r(0) = ⟨-1, 6, 4⟩.
Plane: 2(x + 1) - 2(y - 6) + 1(z - 4) = 0, which simplifies to 2x - 2y + z + 10 = 0.
Containing plane for r(t) = ⟨2(t - sin t), 2 cos t, 3 cos t⟩
The y and z components are both multiples of cos t: z = (3/2)y for all t. So the curve lies in the plane 3y - 2z = 0 (or equivalently 2z - 3y = 0). You can verify by checking that 3·(2 cos t) - 2·(3 cos t) = 0 identically.
Parametrising a line segment
The segment from point C to point D is r(t) = (1 - t)C + tD for t in [0, 1]. For C = (1, 3) and D = (2, 11): r(t) = ⟨1 + t, 3 + 8t⟩, and the distance is |D - C| = √(1² + 8²) = √65.
Speed of a parametric curve
For r(t) = ⟨x(t), y(t), z(t)⟩, the speed at time t is:
|r'(t)| = √(x'(t)² + y'(t)² + z'(t)²)
Worked example: r(t) = ⟨2(t - sin t), 2 cos t, 3 cos t⟩
r'(t) = ⟨2(1 - cos t), -2 sin t, -3 sin t⟩.
|r'(t)|² = 4(1 - cos t)² + 4 sin²t + 9 sin²t = 4(1 - 2cos t + cos²t) + 13 sin²t.
Expanding: 4 - 8 cos t + 4 cos²t + 13 sin²t = 4 - 8 cos t + 4 cos²t + 13(1 - cos²t) = 17 - 8 cos t - 9 cos²t.
Using the identity 1 - cos t = 2 sin²(t/2), you can simplify further, but the key result is that you integrate |r'(t)| from 0 to π to get arc length.
Arc length formula
L = ∫ from a to b of |r'(t)| dt
This integral sums up tiny straight-line distances along the curve. For the example above on [0, π], evaluate the integral after simplification.
The curve
r(t) = ⟨e^t cos 2t, e^t sin 2t⟩. This is a logarithmic (Bernoulli) spiral: the distance from the origin grows exponentially while the point rotates.
Tangent vector
r'(t) = ⟨e^t(cos 2t - 2 sin 2t), e^t(sin 2t + 2 cos 2t)⟩.
The constant-angle claim
The angle α between the position vector r(t) and the tangent vector r'(t) satisfies:
cos α = (r · r') / (|r| |r'|)
Computing the dot product:
r · r' = e^(2t)(cos 2t(cos 2t - 2 sin 2t) + sin 2t(sin 2t + 2 cos 2t)) = e^(2t)(cos²2t + sin²2t) = e^(2t).
|r| = e^t.
|r'| = e^t √(1 + 4) = e^t √5.
So cos α = e^(2t) / (e^t · e^t √5) = 1/√5.
Because 1/√5 has no t in it, the angle is the same at every point. The ratio of radial growth to angular growth is constant, which is the geometric reason the angle stays fixed.
This is why answer B is correct on the multiple-choice section: "The angle is constant because the ratio of radial and angular growth is constant."
Tangent vector: r'(t) = ⟨x'(t), y'(t), z'(t)⟩
Speed: |r'(t)| = √(x'² + y'² + z'²)
Arc length: L = ∫_a^b |r'(t)| dt
Cross product test for planarity: If r'(t) × r''(t) = constant vector for all t, the curve lies in a single plane.
Plane equation: Given normal n = ⟨a, b, c⟩ and point P₀ = (x₀, y₀, z₀), the plane is a(x - x₀) + b(y - y₀) + c(z - z₀) = 0.
Line segment parametrisation: r(t) = (1 - t)C + tD, t ∈ [0, 1].
Angle between two vectors: cos α = (u · v) / (|u| |v|).
Space curves describe the path of a satellite in orbit, the trajectory of a charged particle in a magnetic field, or the centreline of a twisted pipe. Arc length is used in engineering whenever you need the actual material length of a curved beam, cable, or roller-coaster track. The constant-angle property of logarithmic spirals appears in the geometry of nautilus shells and in antenna design.
Students often confuse the tangent vector r'(t) with the unit tangent vector T(t) = r'(t)/|r'(t)|. The tangent vector carries speed information; the unit tangent does not.
A constant cross product r' × r'' does not mean the curve is a straight line. It means the curve bends, but all the bending happens inside one plane.
When asked for the plane containing a curve, students sometimes forget to plug in a specific point. The normal vector alone is not enough; you need both a normal and a point on the curve.
For arc length, forgetting to take the square root before integrating is a common algebraic slip. You integrate |r'(t)|, not |r'(t)|².
⚠️ "Show the curve lies in a plane" is a standard exam question. Know the cross-product test cold.
⚠️ Speed vs. velocity: the exam may ask for "the speed" (a scalar) when you have computed r'(t) (a vector). Do not hand in the vector.
⚠️ The plane 3y - 2z = 0 for ⟨2(t - sin t), 2 cos t, 3 cos t⟩ is the quick-inspection answer. Exams reward noticing component relationships before grinding through cross products.
⚠️ The Bernoulli spiral constant-angle result is a prove-it problem. You must show cos α has no t, not merely state it.
True or false: If r'(t) × r''(t) = 0 for all t, the curve lies in a plane. (Answer: False, it means the curve is a straight line or the acceleration is parallel to velocity.)
The speed of r(t) = ⟨3t, 4t, 0⟩ is ___. (Answer: 5)
True or false: Arc length depends on how the curve is parametrised. (Answer: False, arc length is a geometric quantity, independent of parametrisation.)
Fill in the blank: For the Bernoulli spiral, cos α = ___. (Answer: 1/√5)
True or false: The plane containing r(t) = ⟨2(t - sin t), 2 cos t, 3 cos t⟩ passes through the origin. (Answer: True, since 3y - 2z = 0 passes through (0,0,0).)
Q: For r(t) = ⟨t² - 1, 6 - 3t + t², 4 - 6t⟩, show that the curve lies in a plane and find the equation of that plane.
A: Compute r'(t) = ⟨2t, -3+2t, -6⟩ and r''(t) = ⟨2, 2, 0⟩. The cross product r' × r'' = ⟨12, -12, 6⟩, a constant vector with no t. This confirms the curve is planar. Using normal ⟨2, -2, 1⟩ and point r(0) = (-1, 6, 4), the plane is 2x - 2y + z + 10 = 0.
Q: For r(t) = ⟨2(t - sin t), 2 cos t, 3 cos t⟩, which plane contains the curve?
A: Since z(t) = (3/2)y(t) for all t, every point satisfies 3y - 2z = 0. That is the containing plane.
Q: For the Bernoulli spiral r(t) = ⟨e^t cos 2t, e^t sin 2t⟩, compute r'(t) and show the angle between r(t) and r'(t) is constant.
A: r'(t) = ⟨e^t(cos 2t - 2 sin 2t), e^t(sin 2t + 2 cos 2t)⟩. The dot product r · r' = e^(2t). The magnitudes are |r| = e^t and |r'| = e^t√5. So cos α = e^(2t)/(e^(2t)√5) = 1/√5, which is independent of t.
Q: For the curve r(t) = ⟨2(t - sin t), 2 cos t, 3 cos t⟩, which multiple-choice answer gives the containing plane: (A) x+y+z=0, (B) 3y-2z=0, (C) 2y-3z=0, (D) 3x-2z=0?
A: (B) 3y - 2z = 0.
This material feeds directly into curvature and the Frenet-Serret frame (TNB vectors), which appear later in the course. The arc-length integral reappears when you compute line integrals of scalar and vector fields. The cross-product planarity test is a special case of the more general idea that curvature measures how a curve deviates from lying in a plane.
space curve, vector-valued function, parametric curve in 3D, tangent vector, velocity vector, speed, arc length, arc length integral, curve in a plane, planarity test, cross product test, normal to a plane, Bernoulli spiral, logarithmic spiral, equiangular spiral, constant angle property, line segment parametrisation, r(t), r'(t), r''(t), MATH 215, Calculus III, University of Michigan