AP Calculus BC: Speed and Distance in Parametric Motion

Understanding how objects move along curved paths is fundamental in physics, engineering, and computer graphics. When motion is defined parametrically, we describe a particle's $x$ and $y$ coordinates as separate functions of a third variable, usually time $t$. This allows us to model complex trajectories that a single function $y=f(x)$ cannot. This article focuses on the core calculus skills of finding a particle's instantaneous speed, its total distance traveled, and the crucial distinction between that and its net displacement.

The Foundation: From Velocity to Speed

When a particle's position is given by parametric equations $x(t)$ and $y(t)$, its horizontal and vertical velocity components are simply the derivatives: $x^{\prime }(t)=\frac{dx}{dt}$ and $y^{\prime }(t)=\frac{dy}{dt}$. These components form a velocity vector $⟨x^{\prime }(t),y^{\prime }(t)⟩$. While this vector tells us direction and the rate of change in each component, we often need the pure magnitude of how fast the particle is moving, irrespective of direction. This is its speed.

Speed is the magnitude of the velocity vector. Using the Pythagorean theorem in the context of its perpendicular $x$ and $y$ components, we derive the fundamental formula: $$\text{Speed}(t)=\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}$$

For example, consider a particle with position $x(t)=3\cos (t)$ and $y(t)=2\sin (t)$. Its velocity components are $dx/dt=-3\sin (t)$ and $dy/dt=2\cos (t)$. Therefore, its speed at any time $t$ is $\sqrt{(-3\sin (t){)}^2+(2\cos (t){)}^2}=\sqrt{9{\sin }^2(t)+4{\cos }^2(t)}$.

Calculating Speed at a Specific Time

The process for finding speed at a given instant is a direct application of the formula. It is a common exam question. The methodical steps are:

1. Differentiate: Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
2. Construct: Plug the derivatives into the speed formula.
3. Evaluate: Substitute the given time $t$ into the resulting expression.

Let's apply this to a concrete problem. A particle moves so that $x(t)=t^2-4t$ and $y(t)=t^3-6t$ for $t\ge 0$. Find the particle's speed at $t=2$.

1. Differentiate: $dx/dt=2t-4$ and $dy/dt=3t^2-6$.
2. Construct: Speed$(t)=\sqrt{(2t-4{)}^2+(3t^2-6{)}^2}$.
3. Evaluate at $t=2$:

Speed$(2)=\sqrt{(2(2)-4{)}^2+(3(2{)}^2-6{)}^2}=\sqrt{(0{)}^2+(6{)}^2}=\sqrt{36}=6$.

Thus, at precisely $t=2$ seconds, the particle is moving at a speed of 6 units per second.

Total Distance Traveled: Integrating Speed

Finding the total distance traveled over a time interval is different from finding the change in position (displacement). Distance accumulates regardless of any backtracking, while displacement is simply the straight-line distance from the starting point to the ending point. To find total distance, you must add up every bit of movement, which is the classic "arc length" application. The total distance traveled from time $t=a$ to $t=b$ is found by integrating the speed function: $$\text{Total Distance}={\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$$

Let's calculate the total distance traveled by the particle in the previous example from $t=0$ to $t=3$. We already have Speed$(t)=\sqrt{(2t-4{)}^2+(3t^2-6{)}^2}$. $$\text{Distance}={\int }_0^3\sqrt{(2t-4{)}^2+(3t^2-6{)}^2}dt$$ Simplify inside the square root: $(2t-4{)}^2=4t^2-16t+16$ and $(3t^2-6{)}^2=9t^4-36t^2+36$. Summing gives $9t^4-32t^2-16t+52$. Thus, the integral becomes ${\int }_0^3\sqrt{9t^4-32t^2-16t+52}dt$.

This integral is best evaluated with a calculator in the AP exam context. The numerical approximation is roughly 22.067 units. This is the precise length of the path the particle traversed, even if it doubled back on itself.

Displacement vs. Distance: A Critical Distinction

It is vital to distinguish the scalar distance traveled from the vector displacement. Displacement is the net change in position: $⟨x(b)-x(a),y(b)-y(a)⟩$. Its magnitude is the straight-line distance between the start and end points. Distance traveled, as calculated above, is always greater than or equal to the magnitude of displacement. They are only equal if the particle moves in a straight line without reversing direction.

For our example from $t=0$ to $t=3$:

• Displacement: $x(3)-x(0)=(9-12)-(0)=-3$ and $y(3)-y(0)=(27-18)-(0)=9$. So the displacement vector is $⟨-3,9⟩$.
• Magnitude of Displacement: $\sqrt{(-3{)}^2+(9{)}^2}=\sqrt{90}\approx 9.487$.
• Total Distance Traveled: $\approx 22.067$ (as computed).

The distance (22.067) is significantly larger than the displacement magnitude (9.487), confirming the particle took a curved or reversing path between its start and end points.

Common Pitfalls

1. Confusing Speed with a Velocity Component: A common error is to mistake $dx/dt$ or $dy/dt$ alone for the speed. Remember, speed is the magnitude of the combined velocity vector. You must use the Pythagorean square root formula every time.
2. Using Position Instead of Velocity: When setting up the integral for total distance, the integrand must be the speed function, which uses the derivatives $dx/dt$ and $dy/dt$. Plugging the original position functions $x(t)$ and $y(t)$ into the square root is a critical mistake.
3. Assuming Distance Equals Displacement: In kinematics problems, the question "How far did it travel?" asks for total distance, which requires integration of speed. The question "What is its distance from the starting point?" typically asks for the magnitude of displacement. Misreading the prompt leads to using the wrong calculation.
4. Forgetting the Absolute Value (in Rectilinear Motion Context): In single-variable motion along a line, distance is $\int |v(t)|dt$. In parametric motion, the square root $\sqrt{(dx/dt{)}^2+(dy/dt{)}^2}$ inherently produces a non-negative value (speed), so no separate absolute value is needed. The pitfall is incorrectly trying to apply the single-variable absolute value method to the parametric components individually.

Summary

• The instantaneous speed of a particle defined by $x(t)$ and $y(t)$ is the magnitude of its velocity vector: $\text{Speed}(t)=\sqrt{(dx/dt{)}^2+(dy/dt{)}^2}$.
• To find speed at a specific time $t=c$, compute the derivatives, construct the speed formula, and evaluate it at $t=c$.
• The total distance traveled over the interval $a\le t\le b$ is calculated by integrating the speed function: ${\int }_a^b\text{Speed}(t)dt$.
• Distance traveled is a scalar total of all movement, while displacement is a vector representing the net change in position. The magnitude of displacement is generally less than or equal to the total distance.
• Always double-check that you are using the derivative functions (velocities) in your speed and distance calculations, not the original position functions.