Learning Objectives
Differentiate and integrate a vector function component by component.
Find a tangent line when the derivative vector is nonzero.
Determine a vector constant of integration from an initial condition.
Prerequisites: vector-valued functions and single-variable differentiation and integration, including the chain rule. Angles are in radians. The calculus acts on each coordinate, while vector notation keeps the components together.
1. Differentiate Each Coordinate
The derivative is r′(t) = lim(h → 0) [r(t + h) − r(t)]/h when this limit exists. If r(t) = ⟨f(t), g(t), h(t)⟩ and the component functions are differentiable, then r′(t) = ⟨f′(t), g′(t), h′(t)⟩.
Worked Example: A Derivative Vector
For r(t) = ⟨t², sin(3t), eᵗ⟩, differentiate each component.
The derivative is r′(t) = ⟨2t, 3 cos(3t), eᵗ⟩. The factor 3 in the middle component comes from the chain rule.
At t = 0, r(0) = ⟨0, 0, 1⟩ and r′(0) = ⟨0, 3, 1⟩. Position and direction are different vectors.
2. Use the Derivative as a Tangent Direction
When r′(t₀) ≠ 0 and the curve is differentiable, a tangent line at r(t₀) is L(s) = r(t₀) + sr′(t₀), with s real. Use a new parameter s to avoid confusing the tangent line with the original curve.
Worked Example: A Tangent Line
For the preceding curve at t = 0, use position ⟨0, 0, 1⟩ and direction ⟨0, 3, 1⟩.
The tangent line is L(s) = ⟨0, 0, 1⟩ + s⟨0, 3, 1⟩, or x = 0, y = 3s, z = 1 + s.
A unit tangent vector is T(0) = ⟨0, 3, 1⟩/√10. Normalization changes the length of the direction vector but not the tangent line.
If r′(t₀) = 0, this formula provides no nonzero tangent direction. Do not conclude automatically that the geometric curve has no tangent; a different analysis or parameterization may be needed.
3. Product Rules Preserve the Type of Product
Expression | Derivative |
|---|---|
a(t)u(t), scalar a | a′u + au′ |
u(t) · v(t) | u′ · v + u · v′ |
u(t) × v(t) | u′ × v + u × v′ |
The derivative of a dot product is a scalar. The derivative of a cross product is a vector, and the factor order must be preserved because cross products are not commutative. The scalar product rule also applies when a changing scalar stretches a vector.
4. Integrate with a Vector Constant
Worked Example: Recover a Position Function
Suppose r′(t) = ⟨2t, 3, cos t⟩ and r(0) = ⟨1, −2, 4⟩. Integrating gives r(t) = ⟨t² + C₁, 3t + C₂, sin t + C₃⟩.
Substitution at t = 0 gives C₁ = 1, C₂ = −2, and C₃ = 4.
Thus r(t) = ⟨t² + 1, 3t − 2, sin t + 4⟩. Differentiating and checking the initial position verifies the result.
Definite integrals are also componentwise: integrate each component over the same parameter interval. The result is a vector. For a velocity function, that vector is displacement, not total distance traveled.
Common Mistakes
Apply chain-rule factors inside trigonometric and exponential components.
Use a point on the curve and a direction vector to form a tangent line.
An indefinite vector integral generally needs a separate constant for each coordinate.
5. Practice
1. Differentiate r(t) = ⟨cos(2t), t³, ln t⟩ for t > 0.
2. Find a tangent line to r(t) = ⟨t, t², t³⟩ at t = 1.
3. Evaluate the vector integral of ⟨2t, 1, 3t²⟩ from t = 0 to t = 2.
Worked Solutions
1. r′(t) = ⟨−2 sin(2t), 3t², 1/t⟩.
2. The point is ⟨1, 1, 1⟩ and the derivative is ⟨1, 2, 3⟩. A tangent line is L(s) = ⟨1, 1, 1⟩ + s⟨1, 2, 3⟩.
3. The component integrals are [t²]₀² = 4, [t]₀² = 2, and [t³]₀² = 8. The result is ⟨4, 2, 8⟩.
6. Summary
- Differentiate and integrate coordinates separately.
- A nonzero derivative gives a tangent direction.
- Preserve the order in a cross-product derivative.
- Use initial conditions to determine the vector integration constant.