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Calculus III (Multivariable) Unit 2: Vector-valued functions
A vector-valued function traces a curve in space. Differentiating it gives velocity; integrating gives position.
Covers: Space curves, derivatives & integrals of vector functions, arc length, curvature
Study guide
A vector-valued function r(t) = ⟨x(t), y(t), z(t)⟩ traces a space curve as t changes, like a helix ⟨cos t, sin t, t⟩. Limits, derivatives and integrals are taken component by component.
The derivative r′(t) is tangent to the curve. If r(t) is position, then r′(t) is velocity, |r′(t)| is speed, and r″(t) is acceleration. The unit tangent vector is T(t) = r′(t)/|r′(t)|. Derivative rules carry over, including product rules for dot and cross products. Integrating velocity recovers position, using an initial condition to find the constant vector.
Arc length from t = a to t = b is L = ∫ₐᵇ |r′(t)| dt, the integral of speed. The arc length function s(t) can be used to reparametrize a curve by arc length, so that moving one unit of s means traveling one unit along the curve.
Curvature κ measures how sharply a curve bends: κ = |T′(t)|/|r′(t)| = |r′ × r″|/|r′|³. A straight line has curvature 0, and a circle of radius R has curvature 1/R. The principal unit normal N = T′/|T′| points toward the inside of the bend. Acceleration splits into a tangential part (change in speed) and a normal part (change in direction), with the normal part equal to κ·speed².
The tangential and normal components of acceleration are a_T = (r′·r″)/|r′| and a_N = |r′ × r″|/|r′|. Also, a_N² = |a|² − a_T². If r·r is constant (the motion stays on a sphere centered at the origin), differentiating gives r·r′ = 0, so velocity is perpendicular to position. In general, differentiating a constant dot product produces a perpendicularity condition.
To find extremes of speed, work with |r′(t)|², which avoids the square root. Use trig bounds such as −1 ≤ cos t ≤ 1, or set the derivative of |r′(t)|² to zero.
Key ideas
- r(t) = ⟨x(t), y(t), z(t)⟩. Differentiate and integrate component by component.
- r′(t) is the velocity (tangent) vector; |r′(t)| is speed.
- Arc length integrates speed.
- Unit tangent T = r′/|r′|. Curvature measures how sharply a curve bends.
Formulas and rules
Common mistakes
- Forgetting a constant vector when integrating.
- Confusing speed (a number) with velocity (a vector).
Words to know
- Velocity from position r(t)
- v(t) = r′(t)
- Speed
- |r′(t)|
- Arc length of a space curve
- ∫ |r′(t)| dt
- Unit tangent vector
- T = r′/|r′|
- Curvature formula
- κ = |r′ × r″| / |r′|³
- Vector-valued function
- r(t) = ⟨x(t), y(t), z(t)⟩, tracing a space curve.
- Helix
- A spiral curve like ⟨cos t, sin t, t⟩.
- Acceleration (vector)
- a(t) = r″(t)
- Arc length parametrization
- Reparametrizing by s so each unit of s is one unit of length.
- Principal unit normal (N)
- T′/|T′|: points toward the inside of the bend.
- Curvature of a circle
- 1/R
- Tangential and normal acceleration
- a = (d|v|/dt)T + κ|v|²N
Sample practice questions
11 more questions like these, from level 1 to 5, are in the practice set.
r(t) = ⟨t², 3t, 1⟩. Find r′(t).
- ⟨2t, 3t, 0⟩
- ⟨t, 3, 0⟩
- ⟨2t, 3, 0⟩
- ⟨2t, 3, 1⟩
Show the answer and steps
Answer: C) ⟨2t, 3, 0⟩
- Differentiate each component.
- ⟨2t, 3, 0⟩.
Speed of r(t) = ⟨3 cos t, 3 sin t, 4t⟩.
- 7
- 5
- 25
- 3
Show the answer and steps
Answer: B) 5
- r′ = ⟨−3 sin t, 3 cos t, 4⟩.
- r′
- = √(9 sin²t + 9 cos²t + 16) = √25.
- = 5.
Arc length of that helix for 0 ≤ t ≤ 2π.
- 14π
- 6π
- 10π
- 5
Show the answer and steps
Answer: C) 10π
- Speed is the constant 5.
- Length = 5 × 2π = 10π.