Source: MATH 163 Abridged Class Notes, Calculus III (University of Washington)
Tags: vector derivative, tangent vector, unit tangent vector, T(t), vector integration, integration by parts, differentiation rules, cross product derivative, Calculus III, multivariable calculus
Difficulty: Intermediate | Prerequisites: Section 13.1 (vector-valued functions, space curves, component-wise limits), Calculus I differentiation rules (chain rule, product rule), Calculus II integration techniques (integration by parts, u-substitution).
Section 13.1 introduced vector-valued functions and how to take their limits. This section extends the two core operations of calculus, differentiation and integration, to those functions. The derivative of a vector-valued function gives you the tangent vector to a space curve at any point, which has immediate physical meaning as velocity. Integration reverses the process, recovering position from velocity. These tools are the foundation for arc length, curvature, and the TNB frame that appear later in the chapter.
Differentiate and integrate vector-valued functions component by component, using exactly the same rules you know from Calculus I and II on each piece. The derivative r'(t) points tangent to the curve (it is the velocity vector), and dividing it by its own magnitude gives the unit tangent vector T. Six product-style differentiation rules carry over from scalars, but order matters for the cross product.
Derivative of a vector-valued function
r'(t) = lim(h→0) [r(t+h) - r(t)] / h = ⟨x'(t), y'(t), z'(t)⟩. Think of it as: differentiate each component on its own, then put the results back into a vector.
Tangent vector
The vector r'(t) at a given value of t. It points in the direction the curve is heading at that moment and is tangent to the space curve. In simple terms, if r(t) traces the path of a moving object, r'(t) is its velocity.
Unit tangent vector, T
T(t) = r'(t) / |r'(t)|. The tangent vector scaled to have magnitude 1, so it captures direction of motion only, stripped of speed information.
Tangent line to a space curve
The line through the point r(t₀) in the direction of r'(t₀). Written in vector form as L(t) = r(t₀) + r'(t₀)·t. Think of it as the 3D equivalent of drawing a tangent line to a graph in Calculus I.
Integral of a vector-valued function
∫ r(t) dt = ⟨∫ x(t) dt, ∫ y(t) dt, ∫ z(t) dt⟩ + C_vec. In simple terms, integrate each component separately, and add a constant vector (not a scalar constant) at the end.
Vector constant of integration, C_vec
The constant of integration for a vector-valued antiderivative is itself a vector ⟨C₁, C₂, C₃⟩, one arbitrary constant per component. This catches students who write a single scalar C instead.
The derivative is built from the limit definition, just as in Calculus I, but applied to the vector as a whole: r'(t) = lim(h→0) [r(t+h) - r(t)] / h.
Because the limit of a vector is the vector of the limits (Section 13.1), the result is simply r'(t) = ⟨x'(t), y'(t), z'(t)⟩. Differentiate each component with whatever scalar rules apply (power rule, chain rule, product rule, etc.).
Example: for r(t) = ⟨t⁷ - 5, √(4 - t³), ln(t²)⟩, differentiate component by component to get r'(t) = ⟨7t⁶, -3t² / (2√(4 - t³)), 2/t⟩. The middle component uses the chain rule on (4 - t³)^(1/2).
Sum/difference: d/dt (v ± w) = v' ± w'
Scalar multiple: d/dt (C·v) = C·v'
Scalar-vector product rule: d/dt [f(t)·v(t)] = f'(t)·v(t) + f(t)·v'(t)
Dot product rule: d/dt (v · w) = v' · w + w' · v
Cross product rule: d/dt (v × w) = v' × w + v × w'. Order matters here because the cross product is not commutative. Swapping the order changes the sign.
Chain rule: d/dt v(f(t)) = f'(t)·v'(f(t))
The derivative r'(t) gives a tangent vector, but its magnitude varies along the curve. To isolate direction alone, normalise it: T(t) = r'(t) / |r'(t)|.
For a general formula this can be extremely messy. A much more common exam problem is to evaluate T at a specific value of t.
Strategy for "find the unit tangent vector at t = a":
Compute r'(t) in full generality.
Substitute t = a to get a numerical vector.
Divide by the magnitude of that numerical vector.
Example: for r(t) = ⟨1 + t³, te^(-t), sin(2t)⟩, we get r'(t) = ⟨3t², e^(-t) - te^(-t), 2cos(2t)⟩. At t = 0 this gives r'(0) = ⟨0, 1, 2⟩, with magnitude √5. So T(0) = ⟨0, 1/√5, 2/√5⟩.
A tangent line needs a point and a direction. The point is r(t₀) and the direction is r'(t₀).
The line equation in vector form: L(t) = r(t₀) + r'(t₀)·t.
When the problem gives you a point (x₀, y₀, z₀) rather than a value of t, match the coordinates against r(t) to find which t₀ produces that point, then proceed.
Example: for r(t) = ⟨2cos(t), sin(t), t⟩ at the point (0, 1, π/2), matching components shows t = π/2. Then r'(π/2) = ⟨-2, 0, 1⟩, so the tangent line is L(t) = ⟨0, 1, π/2⟩ + ⟨-2, 0, 1⟩t.
Integration is also component-wise: ∫ r(t) dt = ⟨∫ x(t) dt, ∫ y(t) dt, ∫ z(t) dt⟩ + C_vec.
Each component integral uses the standard techniques from Calculus II: power rule, u-substitution, integration by parts, trig identities, etc.
The constant of integration is a vector, C_vec = ⟨C₁, C₂, C₃⟩.
Example: for r(t) = ⟨1 + t³, te^(-t), sin(2t)⟩:
x-component: ∫(1 + t³) dt = t + t⁴/4
y-component: ∫ te^(-t) dt uses integration by parts (u = t, dv = e^(-t) dt) to give -te^(-t) - e^(-t)
z-component: ∫ sin(2t) dt = -½ cos(2t)
Result: ⟨t + t⁴/4, -te^(-t) - e^(-t), -½ cos(2t)⟩ + C_vec
\vec{r}\,'(t) = \langle x'(t),\, y'(t),\, z'(t) \rangle\vec{T}(t) = \frac{\vec{r}\,'(t)}{|\vec{r}\,'(t)|}\vec{L}(t) = \vec{r}(t_0) + \vec{r}\,'(t_0)\,t \quad \text{(tangent line)}\int \vec{r}(t)\,dt = \left\langle \int x(t)\,dt,\; \int y(t)\,dt,\; \int z(t)\,dt \right\rangle + \vec{C}\frac{d}{dt}(\vec{v} \times \vec{w}) = \vec{v}\,' \times \vec{w} + \vec{v} \times \vec{w}\,' \quad \text{(order matters)}If r(t) gives the position of an aircraft at time t, then r'(t) is its velocity vector and |r'(t)| is its speed. Air traffic control systems compute these continuously. The unit tangent vector tells a self-driving car which direction to steer, independent of how fast it is travelling. Integration of a velocity vector recovers the position function, which is how inertial navigation systems (used in submarines and spacecraft) reconstruct location from accelerometer data.
Students often try to find the general unit tangent vector T(t) when the problem only asks for T at a specific value of t. Evaluate r'(t₀) first, then normalise the resulting numerical vector. The algebra is vastly simpler.
When differentiating a cross product, students frequently forget that order matters. Writing w' × v instead of v × w' changes the sign. The dot product rule has no such issue because the dot product is commutative.
Students sometimes write a single scalar constant C after integrating a vector-valued function. The constant of integration must be a vector, C_vec = ⟨C₁, C₂, C₃⟩.
When finding a tangent line from a given point, students sometimes skip the step of determining what value of t corresponds to that point. If you use the wrong t, your direction vector will be wrong even if your derivative is correct.
⚠️ "Find the unit tangent vector at t = a" is one of the most common exam problems in this section. Master the evaluate-then-normalise workflow.
⚠️ Tangent line problems require two things: the point r(t₀) and the direction r'(t₀). Make sure you provide both and assemble the line correctly.
⚠️ The cross product differentiation rule (order matters) is a favourite for multiple-choice or true/false questions.
⚠️ Integration by parts appears regularly in the y-component of vector integrals (the class example uses ∫ te^(-t) dt). Be fluent with IBP.
True or false: to differentiate r(t) = ⟨sin(t), t², eᵗ⟩, you differentiate each component independently. True.
Fill in the blank: the unit tangent vector T(t) = r'(t) / ___. |r'(t)| (the magnitude of the derivative).
True or false: d/dt (v × w) = w × v' + w' × v. False. The correct rule is v' × w + v × w'. Order matters.
Fill in the blank: the constant of integration for ∫ r(t) dt is a ___, not a scalar. vector
True or false: if r(t) = ⟨cos(t), sin(t), t⟩, then r'(t) = ⟨-sin(t), cos(t), 1⟩ is tangent to the curve for all t. True.
Q: For r(t) = ⟨eᵗ, t², sin(t)⟩, find r'(t).
A: r'(t) = ⟨eᵗ, 2t, cos(t)⟩. Each component is differentiated independently using standard rules.
Q: For r(t) = ⟨3sin(t), 3cos(t), 4t⟩, find the unit tangent vector at t = 0.
A: r'(t) = ⟨3cos(t), -3sin(t), 4⟩. At t = 0: r'(0) = ⟨3, 0, 4⟩. Magnitude = √(9 + 0 + 16) = 5. So T(0) = ⟨3/5, 0, 4/5⟩.
Q: Find the equation of the tangent line to r(t) = ⟨t², t³, t⟩ at the point (1, 1, 1).
A: Matching r(t) = ⟨1, 1, 1⟩ gives t = 1. r'(t) = ⟨2t, 3t², 1⟩, so r'(1) = ⟨2, 3, 1⟩. The tangent line is L(t) = ⟨1, 1, 1⟩ + ⟨2, 3, 1⟩t.
Q: Compute ∫ ⟨2t, e³ᵗ, cos(t)⟩ dt.
A: ⟨t², (1/3)e³ᵗ, sin(t)⟩ + C_vec. Each component integrated separately; the constant is a vector.
Q: If d/dt (v × w) = v' × w + v × w', what goes wrong if you swap the order in the second term?
A: Writing w' × v instead of v × w' reverses the sign of that cross product, because a × b = -(b × a). You would get the wrong answer.
The derivative r'(t) and the unit tangent vector T are the first pieces of the TNB (Frenet-Serret) frame, which adds the unit normal N and the binormal B in the next section. Arc length (Section 13.3) is computed by integrating |r'(t)|, so comfort with both differentiation and integration of vector functions is essential. These ideas also feed directly into Chapter 16: line integrals weight a scalar or vector field along a parametrised curve, using the same r'(t) as the tangent direction.
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