Derivation of the second derivative of distance wrt time

Question

The formula for the second derivative of distance with respect to time is

$a = \dfrac{d}{dt}\left(\dfrac{ds}{dt}\right)$

$= \dfrac{d^2s}{dt^2}$ (just the $t$ is squared - it's not $dt$ all squared).

The question is why is $dt$ multiplied by $dt$ equal to $dt^2$ with just the $t$ squared. Why is not the $d$ also squared? As we can see $d . ds$ becomes $d^2 s$.

Apologies for this probably simple question, I am at the beginning of a very long road in maths.

I think it's just a matter of notation. I wouldn't worry about it. — Matti P., May 12 '17 at 08:22
I have used LaTeX to make your question more readable. But you should check that it says what you intended — Henry, May 12 '17 at 08:30
Notation is justifiable too, as $ \frac{d^2s}{dt^2} \neq \frac{1}{\frac{d^2t}{ds^2}}$ — jonsno, May 12 '17 at 08:42

score 2 · Answer 1 · answered May 12 '17 at 08:53

I would say that what looks like the denominator in $\dfrac{d^2s}{dt^2}$ represents $dt$ all squared.

Let's suppose you have the following data with constant acceleration and try to calculate it:

s   35  75  135  215  315   
t   10  12   14   16   18

Take the differences and you get

ds    40   60   80  100
dt     2    2    2    2

suggesting velocities of $\dfrac{ds}{dt}=20,30,40,50$ measured at intervals $2$ seconds apart

Now take the difference of ds to give d^2 s, but square dt to give (dt)^2

d^2 s   20   20   20  
(dt)^2   4    4    4

suggesting that the acceleration is $\dfrac{20}{4} = 5$. This is consistent with the original example data for $s$ and $t$, suggesting the $d^2s$ could be understood as the second difference in distance, i.e. $d(ds)$. It also suggests that the $dt^2$ could be understood as the square of the first difference of time, i.e $dt$ all squared.

In other cases the acceleration will change with time and you are interested in limits as the time intervals $dt$ tend to zero. But while taking that limit you are in effect looking at $(dt)^2$ rather than $d(t^2)$ or $(dt)t$

This is clearer to me now, I am very grateful for your efforts, thank you. — user3065770, May 12 '17 at 19:35

k.Vijay · Answer 2 · 2017-05-12T09:34:43.853

Although the notation of derivative $\dfrac{d}{dt}\left(\dfrac{ds}{dt}\right)$ looks like a fraction but it's not a fraction; that's why it doesn't implies $d$ is multiplied by $ds$ and $dt$ is multiplied by $dt$.

$\dfrac{ds}{dt}$ means derivative of $s$ with respect to $t$ and similarly $\dfrac{d}{dt}\left(\dfrac{ds}{dt}\right)$ means derivative of $\dfrac{ds}{dt}$ with respect to $t$.

So the notation for derivative of $\dfrac{ds}{dt}$ with respect to $t$ is $\dfrac{d}{dt}\left(\dfrac{ds}{dt}\right)$ which is also expressed by $\dfrac{d^2s}{dt^2}$.

For more details you can follow the answers of this question: Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?

Derivation of the second derivative of distance wrt time

2 Answers2