In AP Calculus BC, the velocity vector v(t) is the derivative of a particle's position vector r(t), found by differentiating each component. It tells you both how fast the particle moves and in what direction, and its magnitude |v(t)| is the particle's speed.
The velocity vector is what you get when you take the derivative of a vector-valued position function. If a particle's position is r(t) = ⟨x(t), y(t)⟩, then its velocity is v(t) = r'(t) = ⟨x'(t), y'(t)⟩. You just differentiate each component separately, exactly the way you would for any real-valued function (that's the whole point of EK 9.4.A). The x-component tells you how fast the particle moves horizontally, the y-component tells you how fast it moves vertically, and together they point in the direction the particle is actually heading at that instant.
Here's the part that trips people up. Velocity is a vector, but speed is a number. Speed is the magnitude of the velocity vector, |v(t)| = √((x'(t))² + (y'(t))²). So a particle can have velocity ⟨-3, 4⟩ (moving left and up) with speed 5. Per FUN-8.B.1, derivatives of position give you velocity, speed, and acceleration for planar motion. The velocity vector is the middle link in the chain r(t) → v(t) → a(t).
The velocity vector lives in Unit 9 (Parametric Equations, Polar Coordinates, and Vector-Valued Functions), which is BC-only, so AB students can skip it. It directly supports two learning objectives. LO 9.4.A asks you to calculate derivatives of vector-valued functions, and v(t) = r'(t) is the most common reason you'd ever do that. LO 9.6.A asks you to determine positions and rates of change in planar motion problems, and the velocity vector is the engine of every one of those problems. FUN-8.B.2 gives it a second job going the other direction. Integrate the velocity vector over a time interval and you get displacement, the net change in position. Integrate its magnitude (speed) and you get total distance traveled. If you understand what v(t) is and isn't, the entire Unit 9 motion package falls into place.
Keep studying AP Calculus Unit 9
Visual cheatsheet
view galleryAcceleration Vector (Unit 9)
Differentiate the velocity vector and you get the acceleration vector, a(t) = v'(t) = r''(t). Motion problems are really just a derivative ladder, and velocity is the middle rung between position and acceleration.
Displacement Vector (Unit 9)
Integrating the velocity vector component-by-component over a time interval gives displacement, the net change in position (FUN-8.B.2). Add displacement to a known starting position and you can recover where the particle ends up.
Distance Traveled (Unit 9)
Total distance is the integral of speed, which is the magnitude of the velocity vector. Integrating |v(t)| measures every inch of the path, while integrating v(t) itself only measures net change. That difference is a classic exam trap.
v(t) in straight-line motion (Unit 4)
You already met velocity in Unit 4, where a particle moves along a line and v(t) is a single function. The velocity vector is the 2D upgrade of that same idea. The sign of v(t) on a line becomes the direction of the vector in the plane.
Velocity vector questions show up in BC multiple choice and FRQs as planar motion problems. The most common tasks: differentiate a position function r(t) component-by-component to find v(t), then maybe differentiate again for a(t). Practice questions in this style hand you something like r(t) = ⟨t² + 4t, 4t³ - 3t⟩ and ask for the velocity or acceleration vector-valued function. Direction questions are also fair game, like finding when a particle moves in the direction of increasing x, which just means solving x'(t) > 0 using the first component of v(t). On the calculator-active FRQ, expect to compute speed as |v(t)| at a specific time, integrate v(t) to find a position, or integrate |v(t)| for total distance. Know which integral answers which question, because the exam loves to offer both as answer choices.
Velocity is a vector with both magnitude and direction, written as ⟨x'(t), y'(t)⟩. Speed is a scalar, just the magnitude |v(t)| = √((x'(t))² + (y'(t))²), with no direction attached. This matters for integrals too. The definite integral of the velocity vector gives displacement (net change), while the definite integral of speed gives total distance traveled. If a particle loops back to where it started, displacement is zero but distance traveled isn't.
The velocity vector is the derivative of the position vector, v(t) = r'(t), found by differentiating each component separately.
Speed is the magnitude of the velocity vector, |v(t)| = √((x'(t))² + (y'(t))²), and it is a scalar, not a vector.
Integrating the velocity vector over a time interval gives displacement, and integrating speed gives total distance traveled (FUN-8.B.2).
The sign of each component of v(t) tells you direction, so the particle moves toward increasing x exactly when x'(t) > 0.
Differentiating the velocity vector gives the acceleration vector, completing the chain r(t) → v(t) → a(t).
Velocity vectors are tested only on the BC exam, in Unit 9 planar motion problems.
It's the derivative of a particle's position vector. If r(t) = ⟨x(t), y(t)⟩, then v(t) = ⟨x'(t), y'(t)⟩, which captures both how fast the particle moves and which direction it's headed at time t.
No. Velocity is a vector with direction, while speed is the scalar magnitude |v(t)|. A particle with velocity ⟨-3, 4⟩ has speed 5, and the exam regularly checks whether you know the difference, especially in integral questions where ∫v(t) dt gives displacement but ∫|v(t)| dt gives distance.
No, velocity vectors are BC-only. They live in Unit 9 (Topics 9.4 and 9.6) with parametric and vector-valued functions. AB covers velocity only for straight-line motion, where v(t) is a single function rather than a vector.
Differentiate each component. For r(t) = ⟨t² + 4t, 4t³ - 3t⟩, the velocity vector is v(t) = ⟨2t + 4, 12t² - 3⟩. All your usual derivative rules carry over component-by-component, which is exactly what LO 9.4.A tests.
Velocity is an instantaneous rate of change of position, while displacement is the net change in position over an interval. You get displacement by integrating the velocity vector from one time to another, so velocity is the derivative side and displacement is the integral side of the same relationship.