We've been shown in class the standard parametric form of the parabola $$x(t)=2at$$ $$y(t)=at^2$$

My question is: why is $x(t)$ the derivative of $y(t)$? Is it just a coincidence?

  • 1
    $\begingroup$ Basically, yes. $\endgroup$ – user228113 Oct 24 '16 at 16:46
  • $\begingroup$ Well, that's quite underwhelming then $\endgroup$ – Antrikos Oct 24 '16 at 16:49

It's good to be attentive to some coincidences that sometimes lead to discoveries. Is this case, as G. Sassatelli has said, it looks a mere coincidence.

But in the same vein, I remember having made a similar remark about the fact that the perimeter $p(r):=2 \pi r$ of a circle look as the derivative of the formula for its area $a(r):=\pi r^2$, and it is only some years later that I understood that it was by no means a coincidence.

Explanation: Taylor expansion amounts to the description of what happens in terms of augmentation of area:

$$\tag{1}a(r+h)=a(r)+a'(r)*h+ \text{terms in} \ h^2 \cdots \text{that are neglected}.$$

to be compared to:

$$\tag{2}\pi*(r+h)^2=\pi*r^2 + 2*\pi*r*h (+ \pi h^2 \ \text{that we neglect}$$

Identification between (1) and (2) give $a'(r)*h=2*\pi*r*h$ (giving $a'(r)=2*\pi*r$) and geometricaly interpreted as the augmentation of area that, up to an error of order $h^2$, is equivalent to the area of a rectangle with length $2 \pi r$ and width $h$. See for example (http://www.askamathematician.com/2013/02/q-is-it-a-coincidence-that-a-circles-circumference-is-the-derivative-of-its-area-as-well-as-the-volume-of-a-sphere-being-the-antiderivative-of-its-surface-area-what-is-the-explanation-for-this/) (what a title !) or(Why is the derivative of a circle's area its perimeter (and similarly for spheres)?)


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