Need help with an integral -- How to integrate velocity squared?

In summary, the conversation discusses a problem with an implicit function integral and the difficulties in solving it without knowing the initial function. The chain rule is mentioned as a potential solution, but it only works for systems with a first order ODE. It is suggested to approach the problem by considering the energy dissipation term.
  • #1
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TL;DR Summary
Problem with a implicit function integral.
The integral is this one:

##\int (\dot x)^2 \, dt,##

With ##x=x(t). ##

I don't know how to solve that integral and I haven't find nothing to read about on how to proceed with this kind of (implicit function?) integrals without having the initial function.
 
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  • #3
Tomder said:
TL;DR Summary: Problem with a implicit function integral.

The integral is this one:

##\int (\dot x)^2 \, dt,##

With ##x=x(t). ##

I don't know how to solve that integral and I haven't find nothing to read about on how to proceed with this kind of (implicit function?) integrals without having the initial function.

By the chain rule: [tex]\int \dot x^2\,dt = \int \dot x \frac{dx}{dt}\,dt = \int \dot x \,dx.[/tex] But this only helps you if you know [itex]\dot x[/itex] in terms of [itex]x[/itex], ie. your system satisfies a first order ODE. But you originally asked this question in the context of a second-order nonlinear ODE [tex]
\ddot x = -f(x) - k\dot x.[/tex] In that case you do not have [itex]\dot x[/itex] in terms of [itex]x[/itex]; you have [itex]\ddot x[/itex] in terms of [itex]x[/itex] and [itex]\dot x[/itex]. You can't evaluate[itex]\int \dot x^2 \,dt[/itex] unless you already know [itex]x(t)[/itex] and [itex]\dot x(t)[/itex]. What you can say is that [tex]
\frac{d}{dt} \left( \frac12 \dot x^2 + \int f(x)\,dx \right) = -k\dot x^2 \leq 0,[/tex] ie. the system dissipates energy.
 
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  • #4
pasmith said:
By the chain rule: [tex]\int \dot x^2\,dt = \int \dot x \frac{dx}{dt}\,dt = \int \dot x \,dx.[/tex] But this only helps you if you know [itex]\dot x[/itex] in terms of [itex]x[/itex], ie. your system satisfies a first order ODE. But you originally asked this question in the context of a second-order nonlinear ODE [tex]
\ddot x = -f(x) - k\dot x.[/tex] In that case you do not have [itex]\dot x[/itex] in terms of [itex]x[/itex]; you have [itex]\ddot x[/itex] in terms of [itex]x[/itex] and [itex]\dot x[/itex]. You can't evaluate[itex]\int \dot x^2 \,dt[/itex] unless you already know [itex]x(t)[/itex] and [itex]\dot x(t)[/itex]. What you can say is that [tex]
\frac{d}{dt} \left( \frac12 \dot x^2 + \int f(x)\,dx \right) = -k\dot x^2 \leq 0,[/tex] ie. the system dissipates energy.
Thanks, I thought about the same process but was unsure of its veracity, guess I‘ll try to work in my system with the idea of energy dissipation term. Thank you very much for your answer.
 
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1. What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total accumulation of a variable over a given interval.

2. What is velocity squared?

Velocity squared is the square of an object's velocity, which is the rate of change of its position over time. It is calculated by multiplying the velocity by itself.

3. Why do we need to integrate velocity squared?

Integrating velocity squared allows us to find the total distance traveled by an object over a given time interval. This is useful in many scientific and engineering applications, such as calculating work or determining the displacement of a moving object.

4. How do you integrate velocity squared?

To integrate velocity squared, you can use the power rule for integration, which states that when integrating a function raised to a power, you add one to the power and divide by the new power. In this case, you would add one to the power of two and divide by the new power, which is three.

5. Are there any special cases for integrating velocity squared?

Yes, if the velocity squared function has a variable as the exponent, then you would use the chain rule to integrate. In this case, you would multiply the function by the derivative of the exponent before integrating.

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