Deriving Torricelli's equation using calculus

[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.

 

[SteamKing]

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?

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.

Doesn't one need to always introduce a variables of integration to the equation when they integrate both sides?

I think you mean constants of integration, here.

I don't see how it's justified to just use the differentials that are already there.

It's not clear what you mean here.

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 $x$.

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

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.

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

Do you have some links to these two separate derivations?

 

[Noctisdark]

Hey, This formula 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.

 

[theodoros.mihos]

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