# Deriving Torricelli's equation using calculus

• Mr Davis 97
In summary, the two different derivations for Torricelli's equation use different methods for canceling out the derivatives. One author integrates with respect to the differential on each side, while the other author takes the derivative and integrates with respect to time. The bounds of integration for each method become in terms of the variables of integration.
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.

Mr Davis 97 said:
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?

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.

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

## 1. What is Torricelli's equation?

Torricelli's equation is a mathematical equation that relates the speed of a fluid flowing out of a hole at the bottom of a container to the height of the fluid above the hole. It is named after Italian physicist Evangelista Torricelli who first derived this equation in the 17th century.

## 2. How is Torricelli's equation derived using calculus?

To derive Torricelli's equation using calculus, we use the principles of fluid mechanics and apply the laws of motion to a small element of fluid flowing out of the hole. This involves setting up an equation for the conservation of energy and then using differentiation and integration techniques to solve for the velocity of the fluid.

## 3. What are the assumptions made in deriving Torricelli's equation using calculus?

The derivation of Torricelli's equation using calculus assumes that the fluid is incompressible, non-viscous, and inviscid. It also assumes that the flow is steady and that there is no external force acting on the fluid.

## 4. What are the applications of Torricelli's equation?

Torricelli's equation has various applications in fluid mechanics, such as in the design of pipelines, water pumps, and sprinkler systems. It is also used in meteorology to predict the speed of wind and in aviation for calculating the fuel consumption of aircraft.

## 5. Are there any limitations to Torricelli's equation?

Torricelli's equation has limitations in its applicability, as it only applies to idealized situations and does not take into account factors such as turbulence, friction, and changing fluid density. It also assumes that the fluid is flowing out of a small hole, which may not always be the case in real-world scenarios.

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