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The second law of Newton is

$$a = \frac{F}{m}$$

and so in a constant gravitational field with gravity $g$ I have

$$a = \frac{d^2x}{dt^2} = g$$

To solve this I multiply with $dt^2$ and get

$$d^2x=g\,dt^2$$

and integrate

$$\iint d^2x = g\,\iint dt^2$$

and so I get

$$\frac{1}{2}x^2=\frac{1}{2}g\,t^2$$

which is wrong! The solution must be $x=\frac{1}{2}g\,t^2$. What am I missing? And what is the difference between $dx^2$ and $d^2x$?

I know it is a trivial question but I got a bit rusty with integrations...

Asaf Karagila
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Alex
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    you did integration wrong because there is no such thing as $d^2x$ integration. You must integrate by $dt$ one by one till you reach $dx$. – CaptainPotato Jan 07 '23 at 16:49

3 Answers3

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Second derivatives and higher are where you cannot treat the differentials as fractions anymore. This is because the first derivative separation of variables

$$\frac{dy}{dx} = f(x) \implies dy = f(x) dx$$

is a shorthand for the chain rule

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = f(x(t)) \implies \frac{dy}{dt} = f(x(t))\frac{dx}{dt}$$

where both sides can now be integrated with respect to the parametric variable

$$\int \frac{dy}{dt}dt = \int f(x(t))\frac{dx}{dt}dt \implies \int dy = \int f(x) dx$$

which is exactly where the shorthand consistently lands and lines up with the rigorous treatment by simple $u$ substitution. However, the chain rule for the second derivative even looks like

$$\frac{d^2y}{dx^2} = \frac{1}{\frac{dx}{dt}}\frac{d}{dt}\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right) = \frac{\frac{dx}{dt}\frac{d^2y}{dt^2}-\frac{dy}{dt}\frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^3}$$

which cannot be multiplied like a simple fraction as the original post does. This only gets more complicated for higher derivatives, so the same heuristic does not work.

Ninad Munshi
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To avoid confusions, note that in this case Newtons second law can be reformulated as $$\dfrac{dv}{dt}=g, \qquad \dfrac{dx}{dt}=v$$

Where $v$ is the velocity of the object.

Now you can integrate the velocity to obtain $v(t)=gt+v_0$.

Where $v_0$ is the initial velocity of the object.

Substituting $v(t)$, you have that $\dfrac{dx}{dt}=gt+v_0$

Now integrate to obtain $$x(t)=x_0+v_0t+\dfrac{1}{2}gt^2$$

Where $x_0$ is the initial position of the object.

Also, note that you are measuring downward.

Evil Witch
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  • But in the case where $g$ is not constant but depends on $x$ like $g=\frac{GM}{x^2}$ then I guess I cannot integrate the velocity first? As the velocity depends on $x$ as well... – Alex Jan 07 '23 at 16:47
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    There are several tricks to solve that differential equations.

    In your example, where the acceleration depends on the position, you can obtain the velocity as a function of position using the chain rule:

    $$\dfrac{d^2x}{dt^2}=\dfrac{dv}{dt}=\dfrac{dv}{dx}\dfrac{dx}{dt}=\dfrac{dv}{dx}v$$

    Then you will have to solve

    $$v\dfrac{dv}{dx}=GMx^2$$

    Which is much simpler.

     – Evil Witch Jan 07 '23 at 16:51
  • Here I get $v^2=-\frac{GM}{x}$ which also looks very strange . Also, I want a relationship with the distance and the time, not with the velocity. Note: you seem to have a typo in your last equation in your comment. – Alex Jan 08 '23 at 09:21
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Consider

$$\iint \dfrac{\text{d}^2x(t)}{\text{d}t^2}\ \text{d}t = \iint g\ \text{d}t$$

$$\int \dfrac{\text{d}x(t)}{\text{d}t} = \int gt \ \text{d}t$$

$$x(t) = \dfrac{1}{2}gt^2 + C$$

Just use the fundamental theorem of calculus.

Enrico M.
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  • So $\int\frac{dx}{dt}=x$ and $\int dx=x$? – Alex Jan 08 '23 at 09:28
  • Be careful: when you write $\int \text{d}x$ you're meaning $\int 1\cdot \text{d}x$ which gives you $x + C$ indeed. It's just integrating a constant ($1$). Whereas $\int \frac{\text{d}x}{\text{d}t}$ has no meaning if there is no measure of integration. You shall write

    $$\int \dfrac{\text{d}x}{\text{d}t}\ \text{d}t = x(t)$$

     – Enrico M. Jan 08 '23 at 11:00