Vector Functions and Space Curves

Vector functions and space curves go together: a vector function r(t) = ⟨f(t), g(t), h(t)⟩ assigns a 3D vector to each value of t, and the tips of those vectors trace a curve in space. You find limits, derivatives and integrals of r(t) one component at a time, and the derivative r′(t) points along the curve.

Before this lesson: Vector-Valued Functions, Equations of Lines in 3D and Derivatives of Trig Functions

What a space curve is

In Calculus 2, a vector function ⟨x(t), y(t)⟩ traced a path in the plane. Adding a third component lets the path leave the plane. Think of t as time and r(t) as the position of a particle: as t changes, the particle moves and leaves a trail, the space curve.

The classic example is the helix r(t) = ⟨2 cos t, 2 sin t, 3t⟩. The first two components go around a circle of radius 2, while the third keeps climbing. The result is a spring shape wrapped around a cylinder.

The domain of r(t) is every t where all three components are defined. The derivative r′(t) is the velocity of the moving point: it is tangent to the curve and points in the direction of motion.

Calculus of vector functions

\lim_{t \to a} \mathbf{r}(t) = \left\langle \lim_{t \to a} f(t), \; \lim_{t \to a} g(t), \; \lim_{t \to a} h(t) \right\rangle

\mathbf{r}'(t) = \langle f'(t), \; g'(t), \; h'(t) \rangle

\int_a^b \mathbf{r}(t)\,dt = \left\langle \int_a^b f(t)\,dt, \; \int_a^b g(t)\,dt, \; \int_a^b h(t)\,dt \right\rangle

• The tangent line at t = t₀ passes through r(t₀) with direction r′(t₀), provided r′(t₀) ≠ 0.
• An indefinite integral gets a constant vector C = ⟨C₁, C₂, C₃⟩, one constant per component.
• The product rules still hold: (u · v)′ = u′ · v + u · v′ and (u × v)′ = u′ × v + u × v′.

Vector functions and space curves examples

Example 1: Recognizing a helix

Show that the curve r(t) = ⟨2 cos t, 2 sin t, 3t⟩ lies on a cylinder, and find how much it rises in one turn.

x^2 + y^2 = (2\cos t)^2 + (2\sin t)^2

= 4\cos^2 t + 4\sin^2 t

= 4(\cos^2 t + \sin^2 t)

= 4

3(2\pi) - 3(0) = 6\pi

Answer: every point satisfies x² + y² = 4, so the curve lies on the cylinder of radius 2 around the z-axis. It is a helix that rises 6π ≈ 18.85 units per turn.

Example 2: Limit of a vector function

Find the limit as t → 0 of r(t) = ⟨(sin t)/t, e²ᵗ, (t² − t)/t⟩.

\lim_{t \to 0} \dfrac{\sin t}{t} = 1

\lim_{t \to 0} e^{2t} = e^0 = 1

\dfrac{t^2 - t}{t} = t - 1

\lim_{t \to 0} (t - 1) = -1

Answer: the limit is ⟨1, 1, −1⟩.

Example 3: Derivative and tangent line

Find r′(t) for r(t) = ⟨t², 3t − 1, eᵗ⁻¹⟩ and the tangent line at t = 1.

\mathbf{r}'(t) = \langle 2t, \; 3, \; e^{t-1} \rangle

\mathbf{r}(1) = \langle 1^2, \; 3(1) - 1, \; e^0 \rangle = \langle 1, 2, 1 \rangle

\mathbf{r}'(1) = \langle 2(1), \; 3, \; e^0 \rangle = \langle 2, 3, 1 \rangle

x = 1 + 2s, \quad y = 2 + 3s, \quad z = 1 + s

Answer: r′(t) = ⟨2t, 3, eᵗ⁻¹⟩, and the tangent line at t = 1 is x = 1 + 2s, y = 2 + 3s, z = 1 + s.

Example 4: Definite integral of a vector function

Evaluate the integral of r(t) = ⟨cos t, 2t, sin 2t⟩ from t = 0 to t = π/2.

\int_0^{\pi/2} \cos t\,dt = \sin\dfrac{\pi}{2} - \sin 0 = 1

\int_0^{\pi/2} 2t\,dt = \left(\dfrac{\pi}{2}\right)^2 - 0^2 = \dfrac{\pi^2}{4}

\int_0^{\pi/2} \sin 2t\,dt = -\dfrac{\cos \pi}{2} + \dfrac{\cos 0}{2}

= \dfrac{1}{2} + \dfrac{1}{2}

= 1

Answer: ⟨1, π²/4, 1⟩ ≈ ⟨1, 2.47, 1⟩.

Example 5: Position of a drone (word problem)

A drone’s velocity is v(t) = ⟨4, 6t, 3t²⟩ meters per second, and at t = 0 it is at r(0) = ⟨1, 0, 2⟩. Find its position r(t), where it is at t = 2, and its speed then.

\mathbf{r}(t) = \langle 4t + C_1, \; 3t^2 + C_2, \; t^3 + C_3 \rangle

\langle C_1, C_2, C_3 \rangle = \langle 1, 0, 2 \rangle

\mathbf{r}(t) = \langle 4t + 1, \; 3t^2, \; t^3 + 2 \rangle

\mathbf{r}(2) = \langle 9, 12, 10 \rangle

\mathbf{v}(2) = \langle 4, 12, 12 \rangle

\sqrt{16 + 144 + 144} = \sqrt{304} = 4\sqrt{19} \approx 17.44

Answer: r(t) = ⟨4t + 1, 3t², t³ + 2⟩. At t = 2 the drone is at (9, 12, 10) and moving at about 17.44 m/s.

Common mistakes

Related methods

Vector functions in the plane

With two components, the same rules give velocity, speed and acceleration for motion in the plane, as in vector-valued functions. In space, the next lessons find the length of a curve, the unit tangent and normal vectors, and curvature.

Curves of intersection

Where two surfaces meet, you can often write the curve as a vector function. For example, the cylinder x² + y² = 1 and the plane z = 2 − y meet along r(t) = ⟨cos t, sin t, 2 − sin t⟩: the first two components stay on the cylinder, and the third makes z = 2 − y.

Practice problems

1. Find the domain of r(t) = ⟨√(t − 1), 1/(5 − t), t⟩.
2. Find r(2) for r(t) = ⟨t, t², t³⟩.
3. Find the limit as t → 2 of ⟨t², (t² − 4)/(t − 2), 3t⟩.
4. Find r′(t) for r(t) = ⟨sin 3t, t⁴, e⁻ᵗ⟩.
5. Find the tangent line to r(t) = ⟨t, t², t³⟩ at t = 1.
6. Show that r(t) = ⟨3 cos t, 3 sin t, 4t⟩ lies on the cylinder x² + y² = 9.
7. Evaluate the integral of ⟨3t², 4, 2t⟩ from t = 0 to t = 2.
8. Find the indefinite integral of ⟨e²ᵗ, 6t, cos t⟩.
9. Find r(t) if r′(t) = ⟨2t, 1, 3t²⟩ and r(0) = ⟨1, −1, 0⟩.
10. A particle has velocity v(t) = ⟨1, 2t, 2⟩ and starts at r(0) = ⟨0, 0, 5⟩. Find its position and speed at t = 3.

t - 1 \ge 0 \Rightarrow t \ge 1

5 - t \ne 0 \Rightarrow t \ne 5

\langle 2, \; 2^2, \; 2^3 \rangle = \langle 2, 4, 8 \rangle

\lim_{t \to 2} t^2 = 4

\dfrac{t^2 - 4}{t - 2} = \dfrac{(t - 2)(t + 2)}{t - 2} = t + 2

\lim_{t \to 2} (t + 2) = 4

\lim_{t \to 2} 3t = 6

(\sin 3t)' = 3\cos 3t

(t^4)' = 4t^3

(e^{-t})' = -e^{-t}

\mathbf{r}(1) = \langle 1, 1, 1 \rangle

\mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle

\mathbf{r}'(1) = \langle 1, 2, 3 \rangle

x = 1 + s, \quad y = 1 + 2s, \quad z = 1 + 3s

(3\cos t)^2 + (3\sin t)^2 = 9\cos^2 t + 9\sin^2 t

= 9(\cos^2 t + \sin^2 t)

= 9

\int_0^2 3t^2\,dt = 2^3 - 0^3 = 8

\int_0^2 4\,dt = 4(2) - 4(0) = 8

\int_0^2 2t\,dt = 2^2 - 0^2 = 4

\int e^{2t}\,dt = \dfrac{e^{2t}}{2}

\int 6t\,dt = 3t^2

\int \cos t\,dt = \sin t

\mathbf{r}(t) = \langle t^2 + C_1, \; t + C_2, \; t^3 + C_3 \rangle

\langle C_1, C_2, C_3 \rangle = \langle 1, -1, 0 \rangle

\mathbf{r}(t) = \langle t + C_1, \; t^2 + C_2, \; 2t + C_3 \rangle

\langle C_1, C_2, C_3 \rangle = \langle 0, 0, 5 \rangle

\mathbf{r}(t) = \langle t, \; t^2, \; 2t + 5 \rangle

\mathbf{r}(3) = \langle 3, 9, 11 \rangle

\mathbf{v}(3) = \langle 1, 6, 2 \rangle

\sqrt{1 + 36 + 4} = \sqrt{41} \approx 6.40

Vector functions and space curves FAQ

What is a vector function?

A vector function r(t) = ⟨f(t), g(t), h(t)⟩ takes a number t and returns a vector. As t varies, the tip of r(t) traces a curve.

How do you differentiate a vector function?

Differentiate each component separately: r′(t) = ⟨f′(t), g′(t), h′(t)⟩. The result is a tangent vector to the curve.

What is a space curve?

A space curve is the path in three dimensions traced by the tip of a vector function, such as the helix ⟨cos t, sin t, t⟩.

How do you integrate a vector function?

Integrate each component. For an indefinite integral, add a constant vector C; for a definite integral, the result is a vector of three numbers.

Summary

• A vector function r(t) = ⟨f(t), g(t), h(t)⟩ traces a space curve as t varies.
• Its domain is where all three components are defined.
• Limits, derivatives and integrals work one component at a time.
• r′(t) is tangent to the curve; the tangent line at t₀ is r(t₀) + s r′(t₀).
• Indefinite integrals need a constant vector C, found from an initial condition.

Previous lesson: Quadric Surfaces

Next lesson: Arc Length of a Space Curve

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