Defining and Differentiating Vector-Valued Functions: Study Guide (Calculus BC 9.4)

BC-only material. Vector-valued functions are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

If any of these is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Parametric equations, dy/dx = (dy/dt)/(dx/dt)	9.1	The same curves, now in vector form
Parametric second derivative d²y/dx²	9.2	To compare with r″(t)
Product, quotient and chain rules	2.8, 2.9, 3.1	Differentiating each component
Vectors as arrows with components	Background	Reading ⟨a, b⟩ as “a across, b up”

Notation on this page: ⟨a, b⟩ is the vector with horizontal component a and vertical component b. It is the same as a i + b j. Bold letters such as r are vectors.

What is a vector-valued function?

An ordinary function takes a number and returns a number. A vector-valued function takes a number t and returns a vector:

r(t) = ⟨x(t), y(t)⟩ = x(t) i + y(t) j

The functions x(t) and y(t) are its component functions.

Picture r(t) as an arrow from the origin to the point (x(t), y(t)). This is the position vector. As t changes, the tip of the arrow moves and draws a curve. That curve is exactly the curve of the parametric equations x = x(t), y = y(t) from Topic 9.1. The vector form simply packs both equations into one object.

Domain. r(t) is defined only where both components are defined. For example, r(t) = ⟨√(5 − t), 1/t⟩ needs t ≤ 5 (for the root) and t ≠ 0 (for the fraction). Its domain is t ≤ 5 with t ≠ 0.

The derivative: one component at a time

Use the same limit definition as for ordinary functions:

r′(t) = lim as h → 0 of [r(t + h) − r(t)] / h

Subtracting two vectors subtracts their components, and dividing a vector by the number h divides each component by h. So

[r(t + h) − r(t)] / h = ⟨ (x(t + h) − x(t))/h , (y(t + h) − y(t))/h ⟩.

As h → 0, each component becomes an ordinary derivative. That gives the key result:

r′(t) = ⟨x′(t), y′(t)⟩

Other notations: dr/dt, or x′(t) i + y′(t) j. If r(t) is a position, r′(t) is called the velocity vector; Topic 9.6 develops this.

All the usual rules carry over, because each component is an ordinary function:

• Sum: (r + s)′ = r′ + s′
• Constant multiple: (c r)′ = c r′
• Scalar function times a vector function: (f r)′ = f′ r + f r′ (a product rule)
• Chain rule: d/dt [r(g(t))] = g′(t) r′(g(t))

Second derivative. Differentiate again, component by component:

r″(t) = ⟨x″(t), y″(t)⟩

If r is a position, r″(t) is the acceleration vector.

What r′(t) tells you about the curve

Draw r′(t₀) with its tail at the point r(t₀). Then:

• Direction. It lies along the tangent line and points the way the curve is traced as t increases. The signs of its components tell you the direction: x′ > 0 means moving right, x′ < 0 left; y′ > 0 means moving up, y′ < 0 down.
• Slope. If x′(t₀) ≠ 0, the tangent slope is dy/dx = y′(t₀) / x′(t₀), the same rule as Topic 9.1.
• Horizontal tangent where y′(t₀) = 0 and x′(t₀) ≠ 0.
• Vertical tangent where x′(t₀) = 0 and y′(t₀) ≠ 0.
• If both components are 0, r′(t₀) = ⟨0, 0⟩ gives no direction. You need more work to decide what happens there.
• Size. The length of r′(t) is |r′(t)| = √(x′(t)² + y′(t)²). This is exactly the arc length integrand from Topic 9.3, so arc length is ∫ |r′(t)| dt.

Worked example 1: derivatives, tangents and the domain

Question. Let r(t) = ⟨t³ − 3t, 4 ln(t + 1)⟩. No calculator.

(a) State the domain. (b) Find r′(t) and r″(t). (c) Find where the curve has a vertical or a horizontal tangent. (d) Find the tangent line at t = 2.

1. Domain. t³ − 3t is defined for all t. 4 ln(t + 1) needs t + 1 > 0. So the domain is t > −1.
2. First derivative. Differentiate each component:
   • d/dt [t³ − 3t] = 3t² − 3
   • d/dt [4 ln(t + 1)] = 4/(t + 1) (chain rule, inner derivative 1)
   • r′(t) = ⟨3t² − 3, 4/(t + 1)⟩
3. Second derivative.
   • d/dt [3t² − 3] = 6t
   • d/dt [4(t + 1)⁻¹] = −4(t + 1)⁻²
   • r″(t) = ⟨6t, −4/(t + 1)²⟩
4. Vertical tangent. x′(t) = 3t² − 3 = 0 gives t = 1 or t = −1. Only t = 1 is in the domain. There, y′(1) = 4/2 = 2 ≠ 0. So r′(1) = ⟨0, 2⟩: vertical tangent at t = 1, at the point r(1) = ⟨−2, 4 ln 2⟩ ≈ ⟨−2, 2.773⟩. The point is moving straight up.
5. Horizontal tangent. y′(t) = 4/(t + 1) is never 0, so there is no horizontal tangent.
6. Tangent line at t = 2. r(2) = ⟨8 − 6, 4 ln 3⟩ = ⟨2, 4 ln 3⟩ and r′(2) = ⟨12 − 3, 4/3⟩ = ⟨9, 4/3⟩. Slope = (4/3) ÷ 9 = 4/27. Tangent line: y − 4 ln 3 = (4/27)(x − 2).

Checks.

• Sign sense: at t = 2, r′(2) = ⟨9, 4/3⟩ has both components positive, so the curve is moving right and up. A small positive slope (4/27 ≈ 0.148) fits. ✓
• Second derivative at t = 2: r″(2) = ⟨12, −4/9⟩. You can check it by differentiating r′ term by term again. ✓

Worked example 2: a model helicopter

Context. A fictional model helicopter flies in a vertical plane. Its position, in metres, is

r(t) = ⟨10t − t², 6 + 4 sin(πt/6)⟩ for 0 ≤ t ≤ 10, where t is in seconds.

(a) Find r′(4) and describe the direction of motion at t = 4. (b) Find r″(4). (c) When is the helicopter moving straight up or down, and when is it moving horizontally?

1. Differentiate.
   • d/dt [10t − t²] = 10 − 2t
   • d/dt [6 + 4 sin(πt/6)] = 4 · (π/6) cos(πt/6) = (2π/3) cos(πt/6)
   • r′(t) = ⟨10 − 2t, (2π/3) cos(πt/6)⟩ (metres per second)
2. (a) Evaluate at t = 4. cos(4π/6) = cos(2π/3) = −½. r′(4) = ⟨10 − 8, (2π/3)(−½)⟩ = ⟨2, −π/3⟩ ≈ ⟨2, −1.047⟩ m/s. The x-component is positive and the y-component is negative, so at t = 4 the helicopter is moving right and down. The tangent slope there is (−π/3)/2 = −π/6 ≈ −0.524.
3. (b) Second derivative.
   • d/dt [10 − 2t] = −2
   • d/dt [(2π/3) cos(πt/6)] = (2π/3)(−π/6) sin(πt/6) = −(π²/9) sin(πt/6)
   • r″(t) = ⟨−2, −(π²/9) sin(πt/6)⟩, so with sin(2π/3) = √3/2,
   • r″(4) = ⟨−2, −π²√3/18⟩ ≈ ⟨−2, −0.950⟩ m/s².
4. (c) Straight up or down: x′(t) = 0 and y′(t) ≠ 0. 10 − 2t = 0 gives t = 5. y′(5) = (2π/3) cos(5π/6) = −π/√3 ≈ −1.814 ≠ 0. So at t = 5 the helicopter moves straight down, at its furthest point to the right, (25, 8). Horizontally: y′(t) = 0 and x′(t) ≠ 0. cos(πt/6) = 0 gives πt/6 = π/2 or 3π/2, so t = 3 or t = 9. x′(3) = 4 > 0 (moving right, at the top of its path) and x′(9) = −8 < 0 (moving left, at the bottom).

Checks.

• Graph: Figure 1 shows the path turning back at x = 25 when t = 5, the highest point at t = 3 and the lowest at t = 9. ✓
• Units: r is in metres, so r′ is in m/s and r″ in m/s². ✓

Interpretation. At t = 4 seconds the helicopter is moving right at 2 m/s and down at about 1.047 m/s. Its horizontal velocity component is falling by 2 m/s every second, so it will soon stop moving right (at t = 5) and turn back.

r″(t) is not d²y/dx²

Both are called “second derivatives”, but they answer different questions.

• r″(t) is a vector of two rates with respect to t: ⟨x″(t), y″(t)⟩.
• d²y/dx² is a single number about the shape of the curve (concavity), found as in Topic 9.2: differentiate dy/dx with respect to t, then divide by dx/dt.

For the helicopter at t = 4: r″(4) ≈ ⟨−2, −0.950⟩, but dy/dx = (2π/3) cos(πt/6) / (10 − 2t) and the Topic 9.2 method gives d²y/dx² = −π(π√3 + 6)/72 ≈ −0.499 at t = 4. Neither component of r″(4) equals d²y/dx², and dividing y″ by x″ does not give it either (that gives about 0.475).

Common misconceptions

• Adding the components. r′(t) is a vector ⟨x′, y′⟩, not the single number x′ + y′.
• Differentiating only one component, or forgetting the chain-rule factor, such as d/dt [sin(πt/6)] = (π/6) cos(πt/6).
• Product of derivatives. d/dt [t sin t] = sin t + t cos t, not cos t. The product rule still applies inside a component.
• Domain from one component only. Both components must be defined.
• Slope as x′/y′. The tangent slope is y′/x′ (rise over run).
• Treating r″(t) as d²y/dx², or computing d²y/dx² as y″/x″.
• Reading the direction from r(t) instead of r′(t). Where the point is (r) does not tell you which way it is going (r′).
• Declaring a vertical tangent when both x′ and y′ are 0. ⟨0, 0⟩ has no direction; you cannot conclude anything from it alone.

Where this leads

Topic 9.5 reverses this process: given r′(t) and a starting position, you integrate component by component to find r(t). Topic 9.6 then uses r′(t), |r′(t)| and r″(t) as velocity, speed and acceleration in motion problems. Next: the Topic 9.5 study guide. To review arc length and |r′(t)|, go back to the Topic 9.3 study guide.

Try the practice questions now, then use the revision notes and the checklist to consolidate.