Defining and differentiating vector-valued functions

AP Calculus BC· AP Calculus BC CED — Parametric Equations, Polar Coordinates, and Vector-Valued Functions· 14 min read

1. Definition and Domain of Vector-Valued Functions★★☆☆☆⏱ 3 min

A vector-valued function takes a single scalar input (almost always $t$ , typically representing time in motion problems) and outputs a multi-dimensional vector. For AP Calculus BC, you will almost always work with 2D vector-valued functions, with 3D functions following identical rules.

📘 Definition

2D Vector-Valued Function

\vec{r}(t) = \langle x(t), y(t) \rangle = x(t)\hat{i} + y(t)\hat{j}

A function where $x (t)$ and $y (t)$ are scalar functions of the common scalar parameter $t$ . The domain of $r (t)$ is the intersection of the domains of $x (t)$ and $y (t)$ : $t$ is only in the domain if it is valid for both components.

Example:

A parametric curve is the graph of a 2D vector-valued position function.

📐 Worked Example

Find the domain of the vector-valued function $r (t) = ⟨ \frac{1}{t - 2}, 4 - t ⟩$ .

1
First, find the domain of the x-component $x (t) = \frac{1}{t - 2}$ . A rational function is undefined when the denominator is zero, so:
$t - 2 \neq 0 \implies t \neq 2$
2
The domain of $x (t)$ is all real numbers except $t = 2$ , or $(- \infty, 2) \cup (2, \infty)$ . Next, find the domain of the y-component $y (t) = 4 - t$ . The expression inside a square root must be non-negative, so:
$4 - t \geq 0 \implies t \leq 4$
3
The domain of $y (t)$ is $(- \infty, 4]$ . The domain of $r (t)$ is the intersection of the two domains, so we only keep values valid for both components. Excluding $t = 2$ and all $t > 4$ , the final domain is:
$(-\infty, 2) \cup (2, 4]$

Exam tip:

When finding the domain of a vector-valued function, always check each component individually. Common hidden restrictions (denominators, square roots, logarithms) are easy to miss if you only check one component.

2. Component-Wise Differentiation of Vector-Valued Functions★★★☆☆⏱ 4 min

The derivative of a vector-valued function is defined by the same limit structure as scalar functions:

\vec{r}'(t) = \lim_{\Delta t \to 0} \frac{\vec{r}(t + \Delta t) - \vec{r}(t)}{\Delta t}

Because vector operations work component-wise, this limit simplifies to a straightforward rule: differentiate each component function separately using all the single-variable differentiation rules you already know. For $r (t) = ⟨ x (t), y (t)⟩$ , the derivative is:

\vec{r}'(t) = \langle x'(t), y'(t) \rangle

Intuitively, $r^{'} (t)$ is the tangent vector to the curve traced by $r (t)$ at time $t$ , pointing in the direction the curve moves as $t$ increases.

📐 Worked Example

Find $r^{'} (1)$ for the vector-valued function $r (t) = ⟨ t^{3} + sin (2 t), e^{t^{2}} ln t ⟩$ .

1
Differentiate the x-component $x (t) = t^{3} + sin (2 t)$ using the power rule and chain rule for the sine term:
$x'(t) = 3t^2 + 2\cos(2t)$
2
Differentiate the y-component $y (t) = e^{t^{2}} ln t$ using the product rule and chain rule for the exponential term:
$y'(t) = \left(2t e^{t^2}\right)\ln t + e^{t^2} \cdot \frac{1}{t} = e^{t^2} \left(2t \ln t + \frac{1}{t}\right)$
3
Substitute $t = 1$ into both derivatives:
$x'(1) = 3(1)^2 + 2\cos(2(1)) = 3 + 2\cos 2 \approx 2.168 \\ y'(1) = e^{1^2} \left(2(1)\ln 1 + \frac{1}{1}\right) = e(0 + 1) = e \approx 2.718$
4
The resulting derivative vector at $t = 1$ is:
$\vec{r}'(1) = \langle 3 + 2\cos 2, e \rangle \approx \langle 2.168, 2.718 \rangle$

Exam tip:

Always differentiate each component separately, don't try to treat the entire vector as a single scalar expression. If you forget that differentiation is component-wise, you will almost certainly make an error on components with products or composite functions.

3. Velocity, Acceleration, and Speed from Position Vectors★★★☆☆⏱ 3 min

When $r (t)$ is the position vector of an object moving in the plane at time $t$ , its first and second derivatives have standard physical interpretations that are frequently tested on the AP exam:

\begin{aligned} \text{Position: } \vec{r}(t) &= \langle x(t), y(t) \rangle \\ \text{Velocity: } \vec{v}(t) &= \vec{r}'(t) = \langle x'(t), y'(t) \rangle \\ \text{Acceleration: } \vec{a}(t) &= \vec{r}''(t) = \langle x''(t), y''(t) \rangle \end{aligned}

Velocity is a vector that gives the direction of motion and the rate of movement. Speed is the scalar magnitude (length) of the velocity vector, calculated as:

\text{speed} = |\vec{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}

📐 Worked Example

A particle moves in the plane with position vector $r (t) = ⟨ 3 cos t, 4 sin t ⟩$ for $t \geq 0$ . Find the velocity vector, acceleration vector, and speed at $t = \frac{π}{3}$ .

1
Differentiate position to get velocity, then substitute $t = \frac{π}{3}$ . We know $sin (\frac{π}{3}) = \frac{3}{2}$ and $cos (\frac{π}{3}) = \frac{1}{2}$ :
$\vec{v}(t) = \langle -3\sin t, 4\cos t \rangle \implies \vec{v}\left(\frac{\pi}{3}\right) = \left\langle -\frac{3\sqrt{3}}{2}, 2 \right\rangle$
2
Differentiate velocity to get acceleration, then substitute $t = \frac{π}{3}$ :
$\vec{a}(t) = \langle -3\cos t, -4\sin t \rangle \implies \vec{a}\left(\frac{\pi}{3}\right) = \left\langle -\frac{3}{2}, -2\sqrt{3} \right\rangle$
3
Calculate speed as the magnitude of velocity:
$\text{speed} = \sqrt{\left(-\frac{3\sqrt{3}}{2}\right)^2 + (2)^2} = \sqrt{\frac{27}{4} + 4} = \sqrt{\frac{43}{4}} = \frac{\sqrt{43}}{2} \approx 3.279$

Exam tip:

When a question asks for speed, remember it is a scalar magnitude, not a vector. If you give the velocity vector instead of its magnitude on an FRQ, you will lose points. Always check the question wording for "velocity" vs "speed".

4. Chain Rule for Composite Vector-Valued Functions★★★★☆⏱ 3 min

If $r$ is a function of $u$ , and $u$ is itself a function of $t$ , we have a composite vector-valued function $r (u (t))$ . The chain rule for this composite follows the same structure as for scalar functions, and still operates component-wise:

\frac{d\vec{r}}{dt} = \frac{d\vec{r}}{du} \cdot \frac{du}{dt}

For $r (u) = ⟨ x (u), y (u)⟩$ , this expands to:

\frac{d\vec{r}}{dt} = \langle x'(u(t)) \cdot u'(t), y'(u(t)) \cdot u'(t) \rangle

This rule commonly appears in motion problems where a path parameter is a function of time, or when reparameterizing a curve.

📐 Worked Example

Let $r (u) = ⟨ u^{2}, 3 u + 1 ⟩$ , where $u (t) = e^{2 t}$ . Find $\frac{d r}{d t}$ when $t = 0$ .

1
First, find $\frac{d r}{d u}$ by differentiating each component of $r$ with respect to $u$ :
$\frac{d\vec{r}}{du} = \langle 2u, 3 \rangle$
2
Next, find $\frac{d u}{d t}$ by differentiating $u (t) = e^{2 t}$ with respect to $t$ :
$\frac{du}{dt} = 2e^{2t}$
3
Apply the chain rule to get the general derivative:
$\frac{d\vec{r}}{dt} = \frac{d\vec{r}}{du} \cdot \frac{du}{dt} = \langle 2u \cdot 2e^{2t}, 3 \cdot 2e^{2t} \rangle = \langle 4u e^{2t}, 6 e^{2t} \rangle$
4
Substitute $t = 0$ to get the derivative at the requested point. Note that $u (0) = e^{0} = 1$ :
$\frac{d\vec{r}}{dt}\bigg|_{t=0} = \langle 4(1)(1), 6(1) \rangle = \langle 4, 6 \rangle$

Exam tip:

When working with composite vector-valued functions, don't forget to multiply by the derivative of the inner function. It's easy to drop the $\frac{d u}{d t}$ term because you're focused on remembering component-wise differentiation.

5. Concept Check★★★☆☆⏱ 1 min

✓ Quick check

Test your understanding of component-wise differentiation with this AP-style multiple choice question:

Which of the following is equal to the derivative of $r (t) = ⟨ t^{2} e^{t^{3}}, ln (cos t)⟩$ ?
- $⟨ 3 t^{4} e^{t^{3}}, - tan t ⟩$
- $⟨ 2 t e^{t^{3}} + 3 t^{4} e^{t^{3}}, - tan t ⟩$
- $⟨ 2 t e^{t^{3}} + 3 t^{2} e^{t^{3}}, tan t ⟩$
- $⟨ 6 t^{2} e^{t^{3}}, - \frac{1}{s i n t} ⟩$
Reveal answer
1 —
Correct! You applied the product rule and chain rule to the x-component, and the chain rule to the y-component to get the right result.

6. Common Pitfalls

Wrong move:

Writing the domain of $r (t) = ⟨ 1/ t, t ⟩$ as $[0, \infty)$ , ignoring the $t = 0$ restriction on the x-component.

Why:

Students only check the last component and forget that all components must be defined for $t$ to be in the domain.

Correct move:

Always write down the domain of each component separately, then compute their intersection to get the domain of the vector-valued function.

Wrong move:

Differentiating $r (t) = ⟨ t sin t, t e^{t} ⟩$ as $r^{'} (t) = ⟨ cos t, e^{t} ⟩$ , forgetting the product rule on each component.

Why:

Students think vector differentiation adds new rules that override single-variable rules, or forget that every component needs full single-variable differentiation.

Correct move:

For each component, apply all relevant single-variable differentiation rules (product, quotient, chain) as if it were a standalone scalar function.

Wrong move:

When asked for speed of a particle with velocity $v (t) = ⟨ 2, 3 ⟩$ , answering $⟨ 2, 3 ⟩$ instead of $13$ .

Why:

Students confuse velocity (a vector quantity) with speed (the scalar magnitude of velocity).

Correct move:

Always check the question wording: if it asks for speed, compute the magnitude of the velocity vector after finding the components.

Wrong move:

Differentiating $r (t) = ⟨ sin (t^{2}), cos (t^{2})⟩$ as $r^{'} (t) = ⟨ cos (t^{2}), - sin (t^{2})⟩$ , forgetting the chain rule on the inner $t^{2}$ term.

Why:

Students focus on remembering that differentiation is component-wise and miss chain rule steps inside individual components.

Correct move:

After writing the derivative of the outer function in each component, always check if the argument of the function is anything other than $t$ , and multiply by the derivative of the argument if needed.

Wrong move:

For the composite vector function $r (u (t))$ , writing the derivative as $⟨ x^{'} (u (t)), y^{'} (u (t))⟩$ , omitting the factor of $u^{'} (t)$ on both components.

Why:

Students forget that the chain rule applies to the whole vector, not just individual components, and miss the common outer factor.

Correct move:

Always apply the chain rule explicitly: write $\frac{d r}{d t} = \frac{d r}{d u} \cdot \frac{d u}{d t}$ before substituting values.

7. Quick Reference Cheatsheet

Category	Formula	Notes
General 2D Vector-Valued Function	$r (t) = ⟨ x (t), y (t)⟩ = x (t) \hat{i} + y (t) \hat{j}$	Input is scalar $t$ , output is vector; domain is intersection of domains of $x (t), y (t)$
Derivative of Vector-Valued Function	$r^{'} (t) = ⟨ x^{'} (t), y^{'} (t)⟩$	Differentiate each component separately; $r^{'} (t)$ is the tangent vector to the curve
Velocity Vector (from position)	$v (t) = r^{'} (t) = ⟨ x^{'} (t), y^{'} (t)⟩$	Vector describing direction and rate of motion
Acceleration Vector	$a (t) = r^{''} (t) = ⟨ x^{''} (t), y^{''} (t)⟩$	Second derivative of the position vector
Speed (scalar)	$speed = ∣ v (t) ∣ = (x^{'} (t))^{2} + (y^{'} (t))^{2}$	Magnitude of the velocity vector; always a non-negative scalar
Chain Rule for Composite	$\frac{d r}{d t} = \frac{d r}{d u} \cdot \frac{d u}{d t}$	For $r (u (t))$ , applies component-wise to all dimensions

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Find domain of vector-valued function
2023 · FRQ
Calculate speed from position vector

What's Next

Now that you have mastered defining and differentiating vector-valued functions, you are ready to extend these concepts to integrating vector-valued functions, which is used to find position from velocity and calculate displacement along a parametric curve. This topic is the foundation for analyzing motion in the plane, a common AP Calculus BC FRQ topic that ties together parametric equations, vector functions, and derivatives of scalar functions. Understanding vector-valued functions also prepares you for arc length calculation of parametric curves and the calculus of polar functions, which are other key topics in Unit 9 of the AP Calculus BC CED.