In this video, speed, defined as the distance traveled in one unit of time, is represented by rectangles on the distance-time (distance as a function of time) graph. At first, the examples stick to a constant speed per unit of time, and illustrate that within each of these fixed-length intervals, the slope of the line segment and the area of the rectangle (defined to have unit width, and the difference between the outputs from the rightmost and leftmost input values as height) are equal, up to sign. For example, if a unit width rectangle has a height of $3$, then both its area, and the unsigned slope of the line segment connecting two of its diagonal corners must equal $3$.

Of course, realistic graphs rarely consist of these constant outputs per fixed-length interval. This is shown in the video at 7:42 - 8:39, although their continuous function still seems artificially well-suited to the use of unit length rectangles. In general, rectangle length can’t remain unit; we must shrink it indefinitely in order to produce indefinitely good curve/area approximations.

While I’m optimistic about using this illustration as an intuition for the fundamental theorem of calculus, associating slope to area on a single graph prior to moving the area to a speed-time graph, I am puzzled by some apparent slight of hand. A slope of $3$ is not actually identical to an area of $3$; it only appears this way by ignoring units. The slope of a distance-time graph would be measured in distance units/time units (i.e., miles-per-hour), but the area under a speed-time graph would be measured in distance units (i.e., miles).

A second issue is that the association seems to fail when the rectangles are made non-unit length, but in general we need to shrink them to length $dx$. Suppose a distance of $10$ is traversed between $x = 2$ and $x = 4$. The average slope would be $5$, yet the area of the rectangle encompassing the region would be $20$. Or suppose a distance of $10$ is traversed between $x = 1$ and $x = 1.5$. Here the average slope would be $20$ while the corresponding area would be $5$. In general, a slope equals the change in $y$ divided by the change in $x$, while the area of our rectangle equals the change in $y$ times the change in $x$. These quantities only coincide in the video because the change in $x$ is set to the multiplicative identity, which is its own reciprocal. Otherwise, the association between slope and area, while core to the subject of calculus, appears to break down.

How are these apparent difficulties avoided?

  • $\begingroup$ I'm not sure how much I like this video, maybe try looking at other sources. $\endgroup$ – littleO Jun 13 '18 at 0:39
  • $\begingroup$ I've tried several sources, including the linked video, 3Blue1Brown's series, and the MIT video titled "Big Picture of Calculus," and more. I'm still having a hard time getting the standard derivative intuition (slope of y becomes value of y' at all inputs) and the standard integral intuition (finding area under a curve) to feel as related as they should. I only chose the linked video because it seemed to have the most potential, offering a direct association between slope and area, even showing both simultaneously on the same graph. Perhaps the clarity comes at the price of accuracy? $\endgroup$ – user10478 Jun 13 '18 at 1:20
  • $\begingroup$ The geometric interpretation of the integral as the area under the curve is only one interpretation. If you try to compute the total distance a car has traveled, based on knowing the car's velocity at each point in time, you will find yourself doing an integral without ever visualizing an area under a curve. If $f'(t)$ is the velocity at time $t$, then $f'(t) \Delta t$ is (approximately) the distance the car traveled during a very short time interval of duration $\Delta t$. Just add up all these short distances to get the total distance traveled. $\endgroup$ – littleO Jun 13 '18 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.