# Deriving Torricelli's equation using calculus

1. May 19, 2015

### Mr Davis 97

I am trying to derive Torricelli's equation, i.e. $v_f^{2} = v_i^{2} + 2a\Delta x$, using calculus.

There are two different ways I have seen. First we start with $\displaystyle \frac {dv}{dt} = a$. Next, we multiply both sides by velocity, $\displaystyle v\frac {dv}{dt} = a\frac {dx}{dt}$. From here are where the two ways diverge. One author "cancels" out both $dt$ terms and proceeds to integrate with respect to the differential on each side (dv and dx). However, how is this justified? Doesn't one need to always introduce a variables of integration to the equation when they integrate both sides? I don't see how it's justified to just use the differentials that are already there. Another author does it a different way. He takes $\displaystyle v\frac {dv}{dt} = a\frac {dx}{dt}$ and integrates both sides with respect to time. However, I think it gets fishy when he cancels out the $dt$ in the derivatives with the $dt$ that he introduces as a variable of integration, rendering the variables of integration for each side as $v$ and $x$. Also, the bounds of integration are $t_{1}$ and $t_{2}$ for both integrals, so how do the bounds of integration somehow become in terms of $x$ and $v$ respectively? If someone could answer my questions for both cases, I would be happy.

2. May 19, 2015

### SteamKing

Staff Emeritus
In the same manner that one goes from:

y = f(x)

y' = dy/dx = f'(x)

dy = f'(x) dx

∫ dy = ∫ f'(x) dx = f(x) + C

It's like a Fundamental Theorem of Calculus, or something.

Under certain conditions, differentials can be manipulated as if they were the ratio of two numbers.

I think you mean constants of integration, here.

It's not clear what you mean here.

It's not clear how this approach is any different from the first approach shown above.

It's still not clear what you are saying here.

Do you have some links to these two separate derivations?

3. May 20, 2015

### Noctisdark

Hey, This forumla only work for constant acceleration, that why there's delta x, it's also useful for approximation, let's prove it
Suppose a is constant, then V = at and X = (At^2)/2
Vf^2 - Vi^2 = k*k*(Tf^2 - Ti^2)
but remember that Xf^2 - Xi^2 = a(Tf^2 - Ti^2)/2 ?
If so then DT = 2Dx/a, and Vf^2 - Vi^2 reduces to a*a*2Dx/a = 2aDx.

4. May 22, 2015

### theodoros.mihos

$$\int_{t_1}^{t_2}v\frac{dv}{dt}dt = \int_{t_1}^{t_2}a\frac{dx}{dt}dt \Rightarrow \int_{v(t_1)}^{v(t_2)}v\,dv = \int_{x(t_1)}^{x(t_2)}a\,dx$$