Topic 4.9 Notes – Vector-Valued Functions

What a Vector-Valued Position Function Is

A vector-valued position function gives the particle’s location in the plane at time $t$.

$$\mathbf p(t)=(x(t),y(t)) \qquad \text{or} \qquad \mathbf p(t)=x(t)\mathbf i+y(t)\mathbf j$$

Those are the same thing written two ways. It is also the same motion as the parametric equations $x=x(t)$, $y=y(t)$. Vector form just packages both coordinates into one vector.

• $\mathbf p(t)$ tells you where the particle is
• the point is $(x(t),y(t))$
• the position vector is drawn from the origin to that point
• a restricted interval like $0\le t\le 4$ gives the time of the motion, so it also determines the start and end points

The graph here is the path in the coordinate plane. It is not a graph with $t$ on an axis. Arrows on the path show which way the particle moves as $t$ increases.

Parametric path of a particle in the plane

Position, Distance from the Origin, Velocity, and Speed

These four ideas get mixed up a lot, so keep them separate.

• Position
  $\mathbf p(t)$ tells the particle’s location.
• Distance from the origin
  $$\|\mathbf p(t)\|=\sqrt{x(t)^2+y(t)^2}$$ This is the straight-line distance from $(0,0)$ to the particle.
• Velocity
  $$\mathbf v(t)=\mathbf p'(t)=(x'(t),y'(t))=x'(t)\mathbf i+y'(t)\mathbf j$$ Differentiate component by component.
• Speed
  $$\|\mathbf v(t)\|=\sqrt{(x'(t))^2+(y'(t))^2}$$ Speed is the magnitude of velocity.

Example with $\mathbf p(t)=(2t-1,\ t^2-4t+3)$

• position at $t=3$
  $\mathbf p(3)=(5,0)$
• distance from origin at $t=3$
  $\|\mathbf p(3)\|=5$
• velocity
  $\mathbf v(t)=(2,2t-4)$
• speed at $t=3$
  $\|\mathbf v(3)\|=\sqrt{2^2+2^2}=2\sqrt2$

Two common traps:

• $\|\mathbf p(t)\|$ is not total distance traveled along the path
• speed can never be negative, even if one or both velocity components are negative

How to Read Direction of Motion

Direction comes from the velocity components, not from where the particle is located.

Horizontal motion

Use $x'(t)$.

• $x'(t)>0$ means moving right
• $x'(t)<0$ means moving left
• $x'(t)=0$ means no horizontal motion at that instant

Vertical motion

Use $y'(t)$.

• $y'(t)>0$ means moving up
• $y'(t)<0$ means moving down
• $y'(t)=0$ means no vertical motion at that instant

Combined direction

• $(+,+)$ right and up
• $(+,-)$ right and down
• $(-,+)$ left and up
• $(-,-)$ left and down

If one component is zero, the motion is purely horizontal or vertical. If both are zero, the particle is instantaneously at rest.

A point like $(-2,3)$ only tells location. It does not tell left, right, up, or down.

Working from Different Representations

From a formula for $\mathbf p(t)$:

1. evaluate $\mathbf p(t)$ for location
2. compute $\|\mathbf p(t)\|$ for distance from the origin
3. differentiate for $\mathbf v(t)$
4. use the signs of $x'(t)$ and $y'(t)$ for direction
5. compute $\|\mathbf v(t)\|$ for speed

From a formula or table for $\mathbf v(t)$:

• use component signs for direction
• use magnitude for speed
• support your answer with the actual values or signs

From a graph of the path:

• you can identify the shape of the path
• you need arrows, time labels, or extra information to know direction or speed
• the same path can come from different motions

Common Mix-Ups

• confusing position $\mathbf p(t)$ with velocity $\mathbf v(t)$
• using signs of $x(t)$ and $y(t)$ instead of signs of $x'(t)$ and $y'(t)$
• saying $\|\mathbf p(t)\|$ means distance traveled
• forgetting speed is $\|\mathbf v(t)\|$, so it is never negative
• treating the path as if $t$ were one of the graph axes
• giving a vague answer when the question asks for a specific quantity like location, velocity, direction, or speed