What is the Integral of a Time-Independent Force in Velocity Calculation?

[MuIotaTau]

Homework Statement

In the process of solving a physics problem, I've run across an integration that I'm not sure I've ever had to perform before. It's of the form $$\int \frac{1}{a - (\frac{dy}{dx})(x)} dx$$ where $a$ is a constant.

Homework Equations

Table integral?

The Attempt at a Solution

I honestly have no clue where to even begin with this. I do not have a functional form for $y$, it's simply a quantity that changes with respect to $x$. Any hints on where to begin?

 

[Ray Vickson]

Homework Statement

In the process of solving a physics problem, I've run across an integration that I'm not sure I've ever had to perform before. It's of the form $$\int \frac{1}{a - (\frac{dy}{dx})(x)} dx$$ where $a$ is a constant.

Homework Equations

Table integral?

The Attempt at a Solution

I honestly have no clue where to even begin with this. I do not have a functional form for $y$, it's simply a quantity that changes with respect to $x$. Any hints on where to begin?

Your notation is confusing: is the denominator supposed to be $1 - a x y'(x)$ or $1-a y'(x)$, where $y'(x) = dy(x)/dx.$ That is, is the (x) an argument of f', or does it multuply f'(x)?

As far as I can see you can write more-or-less ANY integrand h(x) in that form: just put
$$h(x) = \frac{1}{a-y'(x)} \Longrightarrow y'(x) = \frac{h(x)-1}{a h(x)}$$ if your integrand is 1/(1-ay'), or put
$$h(x) = \frac{1}{a - x y'(x)} \Longrightarrow y'(x) = \frac{1-h(x)}{a x h(x)}$$
if your integrand is 1/(1-a*x*y').

So, to summarize: your integrand can be anything at all, so there is no way to do the integral.

 

[MuIotaTau]

Your notation is confusing: is the denominator supposed to be $1 - a x y'(x)$ or $1-a y'(x)$, where $y'(x) = dy(x)/dx.$ That is, is the (x) an argument of f', or does it multuply f'(x)?

As far as I can see you can write more-or-less ANY integrand h(x) in that form: just put
$$h(x) = \frac{1}{a-y'(x)} \Longrightarrow y'(x) = \frac{h(x)-1}{a h(x)}$$ if your integrand is 1/(1-ay'), or put
$$h(x) = \frac{1}{a - x y'(x)} \Longrightarrow y'(x) = \frac{1-h(x)}{a x h(x)}$$
if your integrand is 1/(1-a*x*y').

So, to summarize: your integrand can be anything at all, so there is no way to do the integral.

I'm sorry, $x$ is being multiplied in the denominator. Well, um, that's certainly disappointing. I guess it's back to the drawing board. Thank you for your help!

 

[Dick]

I'm sorry, $x$ is being multiplied in the denominator. Well, um, that's certainly disappointing. I guess it's back to the drawing board. Thank you for your help!

If it's a physics problem you probably know something about the relation between y' and x. What kind of problem is it?

 

[MuIotaTau]

If it's a physics problem you probably know something about the relation between y' and x. What kind of problem is it?

Momentum transport. In this case, $x$ is time, and $y'(x)$ is the change in mass with respect to time. So really the integral is $$\int_0^t \bigg(\frac{1}{M + m - (t')(\frac{dm}{dt'})}\bigg) dt'$$ where $t'$ is a dummy variable. $M$ and $m$ are the masses of a truck and of an original mass of sand, respectively, so they're constants, and $(t)(\frac{dm}{dt})$ is the amount of sand that falls out the truck after a time $t$.

 

[Dick]

Momentum transport. In this case, $x$ is time, and $y'(x)$ is the change in mass with respect to time. So really the integral is $$\int_0^t \bigg(\frac{1}{M + m - (t')(\frac{dm}{dt'})}\bigg) dt'$$ where $t'$ is a dummy variable. $M$ and $m$ are the masses of a truck and of an original mass of sand, respectively, so they're constants, and $(t)(\frac{dm}{dt})$ is the amount of sand that falls out the truck after a time $t$.

Since you are multiplying t' times dm/dt', I think it's pretty likely you are assuming the rate of mass change dm/dt' is constant. Are you?

 

[MuIotaTau]

Since you are multiplying t' times dm/dt', I think it's pretty likely you are assuming the rate of mass change dm/dt' is constant. Are you?

Oh no, you're right, I certainly am! So I should be able to manipulate that into something of the form $$\int \frac{1}{1 + x} dx$$ right? Which is just $ln(1 + x)$?

 

[Dick]

Oh no, you're right, I certainly am! So I should be able to manipulate that into something of the form $$\int \frac{1}{1 + x} dx$$ right? Which is just $ln(1 + x)$?

Sure, so the integral isn't a real problem. Now I'd worry that the integral is computing something like the average value of 1/mass over a portion of the trip, which doesn't sound very useful for momentum problems, but then I don't know exactly what you are doing.

 

[MuIotaTau]

Sure, so the integral isn't a real problem. Now I'd worry that the integral is computing something like the average value of 1/mass over a portion of the trip, which doesn't sound very useful for momentum problems, but then I don't know exactly what you are doing.

I'm actually calculating the impulse and dividing by the mass to get velocity, but the force in this particular case happens to be time independent, so I just pulled it out of the integral.

 

[Dick]

I'm actually calculating the impulse and dividing by the mass to get velocity, but the force in this particular case happens to be time independent, so I just pulled it out of the integral.

Ok, carry on.