How to interpret $xdx$ intuitively if $x$ represents time?

Question

From the book Calculus Made Easy

Now in the calculus we write $dx$ for a little bit of $x$. These things such as $dx$ (...) are called “differentials”. If $dx$ is a small bit of $x$, and relatively small of itself, it does not follow that such quantities as $x·dx$, or $x^2dx$, or $a^xdx$ are negligible. But $dx\times dx$ would be negligible, being a small quantity of the second order.

Previously, the author gives as examples for $x$ and $dx$ hours and minutes respectively, and encourages the reader to think of $dx$ as a little bit of $x$. It is not clear to me how to interpret the meaning of $xdx$ in this interpretation, what does multiplying time mean?

Note: i am aware of thinking of it in terms of rectangles, but it does not satisfy me in this case because of it using time.

Answer 1

You can split a square of side $x+dx$ into two squares of sides $x$ and $dx$, and two $x\times dx$ rectangles. This gives an intuitive reason why $(x^2)^\prime=2x$.

Answer 2

If you have a function for example $x^2$, then $x^2\text{d}x$ is an area according to the picture. Height is real (=function height), but width $\text{d}x$ is infinitesimally small (you limit this size to zero so that the sum of the rectangles in the limit approaches the area below the curve). $x^2\text{d}x$ or $\text{d}x$ are small quantities of the first order, $\text{d}x^2$ is a small quantity of the second order etc.

Analogy with time: let $x^2$ be your velocity, $\text{d}x$ small time instant, then $x^2\text{d}x$ is a trajectory, which you pass in time $\text{d}x$.

Answer 3

Based on the comments, it sounds that you're really asking about dimensional analysis, and "what does time multiplied by time signify?".

What usually happens in practice when integrating with respect to time is, velocity $v$ or acceleration $a$ is given as a function of time, so:

• An integrand such as $v(t)\, dt$ has units of velocity multiplied by time and is therefore a distance.
• An integrand $a(t)\, dt$ has units of acceleration multiplied by time and is therefore a velocity.

When you see velocity equations such as $v(t) = -9.8t + v_{0}$, you may want to think of the coefficient, here $-9.8$, as having units of acceleration, so that if $t$ has units of time, then $-9.8t$ is a velocity.

Mathematicians tend not to worry about units (because they always work out). Modern calculus books often introduce integration by plotting the velocity of an object as a function of time and noting that over a time interval of duration $dt$, the velocity $v(t)$ is nearly constant, so $v(t)\, dt$ is the distance traveled over the interval. This numerical quantity also happens to be the signed area of a thin rectangle enclosed by the velocity graph. The curious result is that the signed area enclosed by the velocity graph over some interval is the net displacement of the object over that time interval. The habit of ignoring units allows us to speak as if a displacement is an area.