Parametric Motion, Speed vs. Velocity & Proving the Circumference
You have known the formula \(C = 2\pi r\) for the circumference of a circle since elementary school – but have you ever seen a real proof of why it is true? In this lesson, you will use parametric equations to describe a point moving around a circle, compute its speed using derivatives, and then prove the circumference formula from scratch using calculus. Along the way, you will also learn how to recover an object’s position from its velocity by running differentiation in reverse.
Parametric equations describe how objects actually move through space over time:
- GPS navigation: your phone tracks your position as separate \(x(t)\) and \(y(t)\) coordinates to plot your route on a map
- Video games: every character and projectile on screen moves using parametric equations that update each frame
- Weather tracking: hurricane paths are modeled as parametric curves, with meteorologists computing speed and direction at each moment
- Robotics: a robot arm traces a path through space, and engineers need to know both how fast and in what direction each joint is moving
- Roller coasters: designers use parametric curves to ensure the speed and direction of the car produce thrilling but safe forces on riders
Topics Covered
- Parametric equations for circular motion: \(x(t) = r\cos(2t)\), \(y(t) = r\sin(2t)\)
- Velocity as the derivative of position: \(\vec{v} = \left(\frac{dx}{dt},\, \frac{dy}{dt}\right)\)
- Speed as the magnitude of velocity: \(|\vec{v}| = \sqrt{\dot{x}^2 + \dot{y}^2}\)
- Proving the circumference of a circle: \(C = 2\pi r\)
- Distance traveled vs. displacement
- Recovering position from velocity via anti-differentiation
- Connection to the Fundamental Theorem of Calculus
Lecture Video
Key Frames from the Lecture
What You Need to Know First
Instead of writing \(y\) as a function of \(x\), we describe a curve by giving both coordinates as functions of a third variable, usually time \(t\):
\[x = f(t), \quad y = g(t)\]
As \(t\) changes, the point \((x, y)\) traces out a path called a parametric curve or trajectory.
For example, \(x = \cos t\), \(y = \sin t\) traces the unit circle as \(t\) goes from \(0\) to \(2\pi\).
- Velocity is a vector – it has both magnitude and direction. In 2D, \(\vec{v} = (v_x, v_y)\).
- Speed is a scalar – it is the magnitude of velocity: \(|\vec{v}| = \sqrt{v_x^2 + v_y^2}\).
A car driving at 60 mph north has a velocity of 60 mph north and a speed of 60 mph. If it turns around and drives south at 60 mph, the velocity changed (different direction) but the speed stayed the same.
When you have a function inside another function, the chain rule tells you how to differentiate:
\[\frac{d}{dt}\, f\!\bigl(g(t)\bigr) = f'\!\bigl(g(t)\bigr) \cdot g'(t)\]
For example: \(\frac{d}{dt}\cos(2t) = -\sin(2t) \cdot 2 = -2\sin(2t)\).
You differentiate the outer function, then multiply by the derivative of the inner function.
- Displacement is the straight-line vector from start to finish. If you walk 3 blocks east then 3 blocks west, your displacement is zero.
- Distance traveled is the total length of the path you actually walked. In the example above, the distance is 6 blocks.
Distance is always greater than or equal to the magnitude of displacement.
An anti-derivative (or indefinite integral) reverses differentiation. If \(\frac{d}{dt} F(t) = f(t)\), then \(F(t)\) is an anti-derivative of \(f(t)\).
For example, since \(\frac{d}{dt}\left(\frac{3}{2}t^2\right) = 3t\), the anti-derivative of \(3t\) is \(\frac{3}{2}t^2 + C\), where \(C\) is an arbitrary constant.
From a Circle to Motion: Parametric Equations
We all know the equation of a circle centered at the origin with radius \(r\):
\[x^2 + y^2 = r^2\]
But this is just an orbit – a static shape. To turn it into motion, we introduce time. Using the definitions of cosine and sine on a circle of radius \(r\):
\[x(t) = r\cos(2t), \quad y(t) = r\sin(2t)\]
The factor of \(2\) inside the argument controls how fast the point moves around the circle. Different choices would give different speeds but trace the same path.
Explore the parametric circle – drag the slider to watch the point move:
Finding the Velocity Vector
Velocity is the rate of change of position. We differentiate each coordinate separately:
\[\vec{v}(t) = \left(\frac{dx}{dt},\; \frac{dy}{dt}\right)\]
Applying the chain rule to each component:
\[\frac{dx}{dt} = \frac{d}{dt}\bigl[r\cos(2t)\bigr] = -r\sin(2t) \cdot 2 = -2r\sin(2t)\]
\[\frac{dy}{dt} = \frac{d}{dt}\bigl[r\sin(2t)\bigr] = r\cos(2t) \cdot 2 = 2r\cos(2t)\]
So the velocity vector is:
\[\vec{v}(t) = \bigl(-2r\sin(2t),\; 2r\cos(2t)\bigr)\]
Notice that the velocity is always tangent to the circle – it points in the direction of motion, perpendicular to the radius.
Speed: The Magnitude of Velocity
Speed is the magnitude of the velocity vector:
\[|\vec{v}| = \sqrt{\dot{x}^2 + \dot{y}^2} = \sqrt{(-2r\sin 2t)^2 + (2r\cos 2t)^2}\]
Pull out common factors first to simplify the Pythagorean sum (always a good habit!):
\[= \sqrt{(2r)^2\bigl(\sin^2(2t) + \cos^2(2t)\bigr)}\]
Since \(\sin^2\theta + \cos^2\theta = 1\) for any angle \(\theta\):
When a point travels around a circle, the sine and cosine terms in its velocity cancel out thanks to the Pythagorean identity, leaving a speed that never changes. The speed depends only on the radius and how fast the angle is changing.
\[|\vec{v}| = 2r\]
The speed is constant – the point moves around the circle at the same rate at all times. This is uniform circular motion.
Finding the Period
At \(t = 0\), the point starts at:
\[P_0 = \bigl(r\cos(0),\, r\sin(0)\bigr) = (r,\, 0)\]
To complete one full revolution, the angle \(2t\) must increase by \(2\pi\) (the radian measure of a full turn):
\[2t = 2\pi \implies t = \pi\]
So the period – the time for one complete circle – is \(T = \pi\).
Time cannot equal a number of degrees. The equation \(2t = 2\pi\) works because \(2\pi\) is a dimensionless number (radians are unitless). Writing \(t = 180°\) would be a unit mismatch. This is one of the deep reasons calculus always uses radians.
Proving the Circumference: \(C = 2\pi r\)
We know the speed \(|\vec{v}| = 2r\) is constant, and the period is \(T = \pi\). The total distance traveled in one revolution is:
Distance equals speed times time. Since the speed around the circle is constant at \(2r\) and one full trip takes time \(\pi\), the total distance – the circumference – is simply their product.
\[C = \int_0^{\pi} |\vec{v}|\, dt = \int_0^{\pi} 2r\, dt = 2r \cdot \pi = 2\pi r\]
This is just the familiar formula for circumference, but now we have proven it from calculus rather than stating it as a fact.
Think of it this way: if you drive at a constant speed of \(2r\) for a duration of \(\pi\), you cover a distance of \(2r \times \pi\). It is the area of the rectangle under the constant velocity graph.
See the velocity graph – the area of the rectangle is the circumference:
Distance vs. Displacement in General Motion
The circumference proof highlights an important distinction:
| Quantity | Definition | Formula |
|---|---|---|
| Displacement | Vector from start to finish | \(\Delta\vec{r} = \vec{r}(t_f) - \vec{r}(t_0)\) |
| Distance traveled | Total path length (always positive) | \(s = \int_{t_0}^{t_f} |\vec{v}(t)|\, dt\) |
For the full circle, the displacement is \(\vec{0}\) (you return to the starting point), but the distance traveled is \(2\pi r\).
The key formula connecting speed and distance is:
\[ds = |\vec{v}|\, dt \quad \implies \quad s = \int_{t_0}^{t_f} |\vec{v}|\, dt\]
This is the arc length integral – it adds up all the tiny distance elements \(ds\) along the path.
Worked Example: Non-Uniform 2D Motion
Now consider a more general velocity vector that is not constant:
\[\vec{v}(t) = (3t,\; 1 - t^2)\]
Starting from the origin \(A = (0, 0)\), find the position after \(t = 100\) seconds.
Step 1: Find \(x(t)\) by anti-differentiating \(\dot{x} = 3t\)
\[x(t) = \int 3t\, dt = \frac{3}{2}t^2 + C_1\]
Since \(x(0) = 0\), we get \(C_1 = 0\), so \(x(t) = \frac{3}{2}t^2\).
Step 2: Find \(y(t)\) by anti-differentiating \(\dot{y} = 1 - t^2\)
\[y(t) = \int (1 - t^2)\, dt = t - \frac{t^3}{3} + C_2\]
Since \(y(0) = 0\), we get \(C_2 = 0\), so \(y(t) = t - \frac{t^3}{3}\).
Step 3: Evaluate at \(t = 100\)
\[x(100) = \frac{3}{2}(100)^2 = 15{,}000\]
\[y(100) = 100 - \frac{100^3}{3} = 100 - \frac{1{,}000{,}000}{3} \approx -333{,}233\]
The position at \(t = 100\) is \(B \approx (15{,}000,\; -333{,}233)\).
When computing a change in position, the constant drops out:
\[\Delta x = x(t_f) - x(t_0) = \left[\frac{3}{2}t_f^2 + C\right] - \left[\frac{3}{2}t_0^2 + C\right] = \frac{3}{2}(t_f^2 - t_0^2)\]
The constant \(C\) represents an unknown shift of the coordinate system. Since we only care about the change, it cancels. When we set the starting point at the origin, we effectively choose \(C = 0\).
Step 4: Distance traveled (the harder question)
The distance is not the same as the displacement magnitude. We need the arc length integral:
\[s = \int_0^{100} |\vec{v}(t)|\, dt = \int_0^{100} \sqrt{(3t)^2 + (1 - t^2)^2}\, dt = \int_0^{100} \sqrt{9t^2 + 1 - 2t^2 + t^4}\, dt\]
\[= \int_0^{100} \sqrt{t^4 + 7t^2 + 1}\, dt\]
This integral does not simplify to a nice closed form, which is typical for real-world motion. In practice, we would evaluate it numerically.
Explore the trajectory – drag the slider to watch the point trace out the path:
The Fundamental Theorem Connection
The process we used – going from velocity back to position via anti-differentiation – is the Fundamental Theorem of Calculus in action:
\[x(t_f) - x(t_0) = \int_{t_0}^{t_f} \dot{x}(t)\, dt\]
In words: the integral of a rate of change gives the total change. This is the same principle that says the area under a velocity-vs-time graph equals the displacement.
For distance (not displacement), we integrate the speed (the magnitude of velocity):
\[s = \int_{t_0}^{t_f} |\vec{v}(t)|\, dt\]
Cheat Sheet
| Formula | Notes |
|---|---|
| \(x(t) = r\cos(\omega t),\; y(t) = r\sin(\omega t)\) | Parametric circle, radius \(r\), angular speed \(\omega\) |
| \(\vec{v}(t) = \left(\frac{dx}{dt},\, \frac{dy}{dt}\right)\) | Velocity = time derivative of position |
| \(\lvert\vec{v}\rvert = \sqrt{\dot{x}^2 + \dot{y}^2}\) | Speed = magnitude of velocity |
| \(T = \frac{2\pi}{\omega}\) | Period of circular motion |
| \(C = \lvert\vec{v}\rvert \cdot T = 2\pi r\) | Circumference from speed times time |
| \(s = \int_{t_0}^{t_f} \lvert\vec{v}(t)\rvert\, dt\) | Arc length / distance traveled |
| \(\Delta\vec{r} = \int_{t_0}^{t_f} \vec{v}(t)\, dt\) | Displacement from velocity |
Key Identities Used
\[\sin^2\theta + \cos^2\theta = 1\]
\[\frac{d}{dt}\cos(\omega t) = -\omega\sin(\omega t), \qquad \frac{d}{dt}\sin(\omega t) = \omega\cos(\omega t)\]
Anti-Derivative Rules
\[\int t^n\, dt = \frac{t^{n+1}}{n+1} + C \quad (n \neq -1)\]