Tubegraph

Graphical representation of motion distance and velocity

7 min read

From: 3blue1brown

The concept of a derivative, while subtle and potentially paradoxical, is fundamentally about measuring rates of change, particularly as it relates to phenomena like motion [00:00:19]. A common, yet oxymoronic, phrase used to describe a derivative is “instantaneous rate of change” [00:00:31]. Change inherently occurs between separate points in time, making the idea of change at a single instant nonsensical [00:00:40]. Understanding this paradox helps in appreciating the cleverness of calculus in capturing this idea with the derivative [00:01:00].

Modeling Car Motion

To illustrate the derivative, consider a car that starts at point A, speeds up, and then slows to a stop at point B, 100 meters away, all within 10 seconds [00:01:07].

Distance vs. Time Graph (s(t))

The car’s motion can be graphed by letting the vertical axis represent the distance traveled and the horizontal axis represent time [00:01:23]. The height of the graph at any time t indicates the total distance the car has traveled after that amount of time [00:01:38]. This function is typically named s(t) [00:01:46].

On this graph:

  • Initially, when the car is slow to start, the curve is shallow, as the distance traveled during the first second doesn’t change much [00:01:56].
  • As the car speeds up, the distance traveled per second increases, which corresponds to a steeper slope in the graph [00:02:04].
  • Towards the end, when the car slows down, the curve shallows out again [00:02:13].

This demonstrates a relationship between velocity and distance via the slope of the distance-time graph.

Velocity vs. Time Graph (v(t))

The car’s velocity, measured in meters per second, can also be plotted as a function of time [00:02:20]. This graph might look like a bump:

  • At early times, the velocity is very small [00:02:27].
  • Up to the middle of the journey, the car builds up to a maximum velocity, covering a relatively large distance each second [00:02:30].
  • Then, it slows back down towards a speed of zero [00:02:37].

These two curves (s(t) and v(t)) are inherently related; changing one implies a different form for the other [00:02:41]. Understanding the specific nature of this relationship is key [00:02:51].

The Paradox of Instantaneous Velocity

Intuitively, we understand velocity at a given moment as what a speedometer shows [00:03:08]. It seems logical that higher velocity corresponds to a steeper distance function, meaning more distance covered per unit time [00:03:17].

However, the concept of “velocity at a single moment” poses a paradox [00:03:26]. To compute velocity, one needs two separate points in time to calculate the change in distance divided by the change in time [00:03:39]. Yet, the velocity function v(t) takes only a single value of t [00:03:55]. This conflict between desiring velocity at a single point and needing two points for computation was a challenge for the founders of calculus <a class=“yt=“yt-timestamp” data-t=“00:04:04”>[00:04:04].

Speedometer’s Solution

A physical speedometer bypasses this paradox [00:05:02]. Instead of calculating speed at a single instant, it measures the distance traveled over a very small, finite amount of time, for example, between 3 seconds and 3.01 seconds [00:04:41]. The speed is then computed as that tiny distance divided by that tiny time [00:04:53].

Let dt represent a tiny change in time (e.g., 0.01 seconds) and ds represent the resulting tiny change in distance [00:05:13]. The velocity is then ds divided by dt [00:05:22]. Graphically, dt is a small step to the right on the distance-time graph, and ds is the corresponding change in height [00:05:31]. Thus, ds/dt can be seen as the rise-over-run slope between two very close points on the graph [00:05:51]. This ratio ds/dt can be considered a function of t [00:06:03].

To compute the velocity function, a computer might choose a small dt (e.g., 0.01), then for various times t, calculate (s(t + dt) - s(t)) / dt [00:06:22]. This gives the velocity around each point in time [00:06:58].

The Pure Mathematical Derivative

The concept of ds/dt—a tiny change in s divided by a tiny change in t—is almost the derivative [00:07:27]. However, in pure mathematics, the derivative is not this ratio for a specific dt, but rather what that ratio approaches as dt approaches 0 [00:07:51].

Visualizing the Derivative

Visualizing this limit is crucial:

  • For any specific dt, ds/dt is the slope of a line passing through two separate points on the graph [00:08:02].
  • As dt approaches 0, these two points approach each other, and the slope of the line approaches the slope of a line that is tangent to the graph at the specific point t being examined [00:08:17].

Therefore, the true derivative is the slope of a line tangent to the graph at a single point [00:08:30]. It’s important to note that this does not mean dt becomes “infinitely small” or that dt is set to 0; dt is always a finitely small, non-zero value that is approaching 0 [00:08:42]. This is a “sneaky backdoor” way to reasonably discuss the rate of change at a single point in time, even though change in an instant makes no literal sense [00:09:05].

The healthiest way to think of this slope is not as an instantaneous rate of change, but as the best constant approximation for a rate of change around a point [00:09:30].

Notation

In calculus, when letters like d are used (e.g., ds/dt), it signifies the intention that dt will eventually approach 0 [00:10:01]. The pure mathematical derivative is written as ds/dt, even though it’s technically not a simple fraction, but rather the limit of that fraction as dt shrinks [00:10:11].

Example: Deriving Velocity from s(t) = t^3

Consider a distance function s(t) = t^3 [00:10:38]. To compute the velocity ds/dt at a specific time, say t=2:

  1. Calculate the change in distance: s(2 + dt) - s(2) [00:11:17].
  2. Since s(t) = t^3, this becomes (2 + dt)^3 - 2^3 [00:11:28].
  3. Expand (2 + dt)^3: 2^3 + 3 * 2^2 * dt + 3 * 2 * (dt)^2 + (dt)^3 [00:11:42].
  4. Subtract 2^3 and divide by dt: [ (2^3 + 3 * 2^2 * dt + 3 * 2 * (dt)^2 + (dt)^3) - 2^3 ] / dt = [ 3 * 2^2 * dt + 3 * 2 * (dt)^2 + (dt)^3 ] / dt [00:11:58] = 3 * 2^2 + 3 * 2 * dt + (dt)^2 [00:12:14]
  5. As dt approaches 0, the terms containing dt become negligible [00:12:25].
  6. The result is 3 * 2^2 = 12 [00:12:43].

This means the slope of the tangent line to the graph at t=2 is 12 [00:12:48]. More generally, the derivative of t^3 as a function of t is 3t^2 [00:13:01]. This algebraic simplification by letting dt approach 0 is a core reason why calculus is so powerful [00:14:05].

Readdressing the Paradox: The Car at t=0

Using the s(t) = t^3 example, consider the car’s motion at t=0 [00:14:30].

  • The derivative 3t^2 at t=0 is 3 * 0^2 = 0 [00:14:45].
  • Visually, the tangent line to the graph at t=0 is perfectly flat [00:14:54]. This suggests the “instantaneous velocity” is 0, implying the car is not moving [00:15:03].
  • However, if it doesn’t start moving at t=0, then when does it start moving? This leads to the paradox [00:15:07].

The issue lies in the question itself, as “change in a moment” doesn’t exist [00:15:24]. The derivative being 0 means that the best constant approximation for the car’s velocity around that point is 0 m/s [00:15:33].

For instance, between t=0 and t=0.1 seconds, the car does move (0.001 m), resulting in an average speed of 0.01 m/s [00:15:47]. The derivative approaching 0 means that as dt becomes smaller, this average speed also approaches 0 [00:16:03]. This doesn’t mean the car is static; approximating its movement with a constant velocity of 0 is merely an approximation [00:16:14].

Thus, “instantaneous rate of change” should be understood as a conceptual shorthand for “the best constant approximation for rate of change around a point” [00:16:24]. This perspective resolves the paradox and provides a robust understanding of the derivative.

Graph View

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