Topic 9.4 Defining and Differentiating Vector-Valued Functions: define a vector-valued function and differentiate it component by component to find velocity and acceleration (BC).

What this topic is asking

The College Board (Topic 9.4, BC only) introduces vector-valued functions, which package a particle's $x$- and $y$-coordinates into a single object $\langle x(t), y(t)\rangle$. This is really parametric motion in vector clothing, and it lets you talk about the velocity vector and acceleration vector of a moving point, not just slopes.

Definitions and component-wise differentiation

Vector velocity versus scalar speed

The single most important distinction in this topic is velocity (a vector) versus speed (a scalar). The velocity vector $\langle x'(t), y'(t)\rangle$ carries both how fast and in which direction the particle moves. The speed is the magnitude of that vector, $\sqrt{(x')^2 + (y')^2}$, a single nonnegative number with no direction. A question asking "find the velocity at $t = 2$" wants the vector $\langle x'(2), y'(2)\rangle$; one asking "find the speed at $t = 2$" wants the number $\sqrt{(x'(2))^2 + (y'(2))^2}$. Reporting a vector when a scalar is wanted (or vice versa) loses marks even when the components are correct, so read the question's noun carefully.

A worked velocity and acceleration

The direction of motion

Beyond magnitude, the velocity vector tells you the direction of motion at any instant: the particle moves in the direction of $\mathbf{v}(t)$, which is tangent to the path. This connects to Topic 9.1, since the slope of the path is $\frac{dy}{dx} = \frac{y'(t)}{x'(t)}$, exactly the ratio of the velocity components. So the velocity vector encodes both the parametric slope (through the ratio of its components) and the speed (through its magnitude). When $x'(t) > 0$ the particle moves rightward; when $y'(t) < 0$ it moves downward; the signs of the components describe the motion qualitatively. Acceleration, in turn, describes how the velocity vector itself is changing, in both magnitude and direction.

Connecting to one-dimensional motion

Vector-valued functions generalize the straight-line motion of Unit 4 from one dimension to two. In 1D, position $s(t)$ gives velocity $s'(t)$ and acceleration $s''(t)$, all scalars, and speed is $|s'(t)|$. In 2D, position is the vector $\langle x(t), y(t)\rangle$, and velocity and acceleration become vectors found by differentiating each coordinate, while speed is the magnitude of the velocity vector. The mental model is two independent 1D motions, one in $x$ and one in $y$, running on the same clock $t$. This is why every technique here is just the familiar derivative applied twice, once per component, and why the next topic (integration of vector-valued functions) recovers velocity and position by integrating each component back.