How do I find v(t) from v(x)?
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BigBeachBanana
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You'd need x(t) the function that describes how position changes with respect to time.
Dr.Peterson
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Since v = dx/dt, this is a differential equation, dx/dt = 4212x/(x+0.5), with initial condition x(0) = 0. You would solve for x(t), and differentiate that to find v(t).
How much do you know about solving differential equations? This one is easy in one sense, and difficult in another.
No experience. Only calculus education I’ve had has been calc 1 and 2, just simple derivatives and integrals. And it’s been years at that. This problem is just from personal interest.
I’ve already tried to go about this the following way
V=dx/dt dt=dx/V dt=(x+0.5)/4212x dx
Then tried integrating to get:
(t^2)/2 + C=x/4212 + (x^2)/16848 + C
Canceled both Cs since initial time and position is 0, then rearranged to
x^2 + 4x - 8424t^2=0 to use the quadratic equation.
x= [-4 +/- sqrt(16+33712t^2)] / 2
Im pretty sure this gives me the correct function of x(t) but because of my limited calculus knowledge I don’t know for sure.
Dr.Peterson
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Unfortunately, your integration is wrong (though your approach is valid, and is the one kind of differential equation students commonly learn in a basic calculus course).
But the correct integration results in something you can't solve for x.
BigBeachBanana
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[imath]v(t) = \dfrac{dx}{dt}[/imath] and [imath]v(x) = v(x(t)) [/imath]
Your first step is correct.
But then you have jumped to a step that if someone from the Chess (Math) Community sees it would think that you have done the calculations in your head!
Why don't you write each step carefully before solving it? The next step should be:
[imath]\displaystyle \int dt = \int \frac{x + 0.5}{4212x} \ dx[/imath]
Then, set the integrals' limits from the initial condition.
[imath]\displaystyle \int_{0}^{t} dt = \int_{0}^{x} \frac{x + 0.5}{4212x} \ dx[/imath]
Because I am a gentleman, I will solve the left side for you and you solve the right side.
[imath]\displaystyle t = \int_{0}^{x} \frac{x + 0.5}{4212x} \ dx[/imath]
Solving the integral is the easy part. As an expert once said, in any problem involving calculus, the calculus part is the easiest part. What comes next is a little difficult.
After solving the integral on the right side, you will get a second degree polynomial function. To get [imath]x(t)[/imath], you will have to complete the square and solve for [imath]x[/imath]. While solving just be careful for the signs of the square root.
When [imath]x(t)[/imath] is ready, you will go back to the original equation, [imath]v(x)[/imath], and you will replace every [imath]x[/imath] with this new [imath]x(t)[/imath] that you have just found and you will get [imath]v(t)[/imath].
Bonus Questions. Why did we get two signs for the square root, [imath]\pm\sqrt{}[/imath]?
Can we just take the positive one, [imath]+\sqrt{}[/imath]? Yes? No?
Can we just take the negative one, [imath]-\sqrt{}[/imath]? Yes? No?
Or
We have to take both signs, [imath]\pm\sqrt{}[/imath].
OP, you have to ignore my reply.Your first step is correct.
But then you have jumped to a step that if someone from the Chess (Math) Community sees it would think that you have done the calculations in your head!
Why don't you write each step carefully before solving it? The next step should be:
[imath]\displaystyle \int dt = \int \frac{x + 0.5}{4212x} \ dx[/imath]
Then, set the integrals' limits from the initial condition.
[imath]\displaystyle \int_{0}^{t} dt = \int_{0}^{x} \frac{x + 0.5}{4212x} \ dx[/imath]
Because I am a gentleman, I will solve the left side for you and you solve the right side.
[imath]\displaystyle t = \int_{0}^{x} \frac{x + 0.5}{4212x} \ dx[/imath]
Solving the integral is the easy part. As an expert once said, in any problem involving calculus, the calculus part is the easiest part. What comes next is a little difficult.
After solving the integral on the right side, you will get a second degree polynomial function. To get [imath]x(t)[/imath], you will have to complete the square and solve for [imath]x[/imath]. While solving just be careful for the signs of the square root.
When [imath]x(t)[/imath] is ready, you will go back to the original equation, [imath]v(x)[/imath], and you will replace every [imath]x[/imath] with this new [imath]x(t)[/imath] that you have just found and you will get [imath]v(t)[/imath].
Bonus Questions. Why did we get two signs for the square root, [imath]\pm\sqrt{}[/imath]?
Can we just take the positive one, [imath]+\sqrt{}[/imath]? Yes? No?
Can we just take the negative one, [imath]-\sqrt{}[/imath]? Yes? No?
Or
We have to take both signs, [imath]\pm\sqrt{}[/imath].
It is based on:
[imath]\displaystyle \int dt = \int \frac{x + 0.5}{4212} \ dx[/imath]
This is the last time to solve a problem without wearing my glasses.
I will try to correct my mistake!
Corrected Version
Why don't you write each step carefully before solving it? The next step should be:
[imath]\displaystyle \int dt = \int \frac{x + 0.5}{4212x} \ dx[/imath]
Then, set the integrals' limits from the initial condition.
[imath]\displaystyle \int_{0}^{t} dt = \int_{0}^{x} \frac{x + 0.5}{4212x} \ dx[/imath]
Because I am a gentleman, I will solve the left side for you and you solve the right side.
[imath]\displaystyle t = \int_{0}^{x} \frac{x + 0.5}{4212x} \ dx[/imath]
Solving the integral is the easy part. As an expert once said, in any problem involving calculus, the calculus part is the easiest part. What comes next is a little difficult.
New.
The solution to the integral will involve logarithmic function which means that the position cannot start at [imath]\displaystyle x = 0[/imath] unless you will assume that [imath]\displaystyle x = 0.0000000001[/imath].
There is a way with which the integral above can be solved in a blink of an eye.
hint: Apply Linearity
Let us assume that you have done everything correctly in solving the integral. Then, you will face the big problem that, you will not be able to isolate [imath]\displaystyle x[/imath] which means you will not have [imath]\displaystyle x(t)[/imath] to get [imath]\displaystyle v(t)[/imath].
But there is a solution to this kind of problems. Let us assume you want to find [imath]\displaystyle v(t = 5)[/imath]. You will use the solution that you have found by solving the integrals and numerical analysis. The numerical analysis will help you to approximate the value of [imath]\displaystyle x(t = 5)[/imath]. After you get this value, you will plug it in the original function of [imath]\displaystyle v(x)[/imath] and you will get [imath]\displaystyle v(t=5)[/imath] which is the velocity when time is [imath]\displaystyle 5[/imath] seconds.
Your first step is correct.
But then you have jumped to a step that if someone from the Chess (Math) Community sees it would think that you have done the calculations in your head!
Why don't you write each step carefully before solving it? The next step should be:
[imath]\displaystyle \int dt = \int \frac{x + 0.5}{4212x} \ dx[/imath]
Then, set the integrals' limits from the initial condition.
[imath]\displaystyle \int_{0}^{t} dt = \int_{0}^{x} \frac{x + 0.5}{4212x} \ dx[/imath]
Because I am a gentleman, I will solve the left side for you and you solve the right side.
[imath]\displaystyle t = \int_{0}^{x} \frac{x + 0.5}{4212x} \ dx[/imath]
Solving the integral is the easy part. As an expert once said, in any problem involving calculus, the calculus part is the easiest part. What comes next is a little difficult.
New.
The solution to the integral will involve logarithmic function which means that the position cannot start at [imath]\displaystyle x = 0[/imath] unless you will assume that [imath]\displaystyle x = 0.0000000001[/imath].
There is a way with which the integral above can be solved in a blink of an eye.
hint: Apply Linearity
Let us assume that you have done everything correctly in solving the integral. Then, you will face the big problem that, you will not be able to isolate [imath]\displaystyle x[/imath] which means you will not have [imath]\displaystyle x(t)[/imath] to get [imath]\displaystyle v(t)[/imath].
But there is a solution to this kind of problems. Let us assume you want to find [imath]\displaystyle v(t = 5)[/imath]. You will use the solution that you have found by solving the integrals and numerical analysis. The numerical analysis will help you to approximate the value of [imath]\displaystyle x(t = 5)[/imath]. After you get this value, you will plug it in the original function of [imath]\displaystyle v(x)[/imath] and you will get [imath]\displaystyle v(t=5)[/imath] which is the velocity when time is [imath]\displaystyle 5[/imath] seconds.
Dr.Peterson
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I think you need to turn your focus to your first attempt at this question,
There you talked about the origin of the problem (or at least a very similar one), and I suggested that your equation may not represent what you really want to do. A correctly formulated problem is more likely to be solvable.
Please explain in detail what your equation means, and how you got it.
How do I find v(t) from v(d)?
How do I find the function for velocity with respect to time from an equation for velocity with respect to distance? The equation of interest is v = 200x / (x+0.5) I have very limited calculus knowledge (1 and 2) which is really only simple derivatives and integrals, and it’s been a few years...
www.freemathhelp.com
There you talked about the origin of the problem (or at least a very similar one), and I suggested that your equation may not represent what you really want to do. A correctly formulated problem is more likely to be solvable.
Please explain in detail what your equation means, and how you got it.
Sure, the equation v=4212x/(x+0.5) represents a velocity equation for a projectile being accelerated, where 4212 is the theoretical max velocity assuming an infinitely long length of acceleration, and 0.5 is the distance in which it takes to reach 1/2 Vmax.
The acceleration of the projectile is based on pressure, which of course would be inversely proportional to the distance traveled. I believe I was able to correctly find a(x) just messing around.
A=a(t) A=dv/dt A=dv/dt * dx/dx
A=dv/dx * dx/dt A=dv/dx * V
I then took the derivative of v(x) and multiplied it by v(x). The corresponding graph did represent perfectly (I assume) what the pressure/acceleration graph should look like.
Last edited:
How was this equation [v=4212x/(x+0.5)] derived? Did you derive the equation or the equation was given to you ? What is the domain of V(x)?
This equation is the Le Duc equation of interior ballistics. It is the space-velocity curve of a projectile whose acceleration is determined by a rapid increase in pressure, and then subsequently decreases due to the increase of volume from the projectiles increasing distance.
Because of this, the domain of the function is [0,+infinity), as only positive displacement makes sense for this application.
Again, 4212 (in ft/s) represents the theoretical Vmax assuming an infinitely long length of acceleration, and 0.5 represents the distance traveled in which 1/2 Vmax is achieved (in ft)
Dr.Peterson
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I tried looking into this, but haven't yet found a thorough recent source. It appears that this is empirically derived; and also that no one I've found tries to turn it into a formula for v(t). The latter is probably because of what we've said here, that it would require numerical methods.
But from what I've seen, since it deals with velocity in the muzzle of a gun, the domain ought to be just from 0 to the length of the gun.
Again, I haven't looked deeply enough to be trusted on any of this!
Here is a link to a more thorough breakdown.
Maybe someone who has a better understanding than me can deduce something from it. The closest I was able to find from this link was just the simple substitution x’(t) = r*x(t) / (x(t)+s).
(Where r and s would represent 4212 and 0.5, respectively, in my example).
It is frustrating how difficult this is turning out to be. Seeing as this equation represents a real world phenomenon in which x corresponds to a definite t, I would’ve thought there would’ve been a simple way to deduce t from x.
Your main goal is to find [imath]v(t)[/imath]. By definition, [imath]v(t) = x'(t)[/imath]. This means [imath]\displaystyle v(t) = v(x(t)) = \frac{4212x(t)}{x(t) + 0.5}[/imath].
Now, say, you want to find [imath]v(t = 5)[/imath]. Since [imath]x = x(t)[/imath], they have to tell you both the position and time. For example, if the position [imath]x = 10[/imath] m and time is important, they have to tell at what time the position is [imath]10[/imath] m. In other words, they cannot just say [imath]x = 10[/imath] m, they have to tell you [imath]x(5) = 10[/imath] m because [imath]x = x(t)[/imath] is the position at a specific moment. Therefore, [imath]\displaystyle v(t = 5) = v(x(5)) = v(10) = \frac{4212*10}{10 + 0.5}[/imath]. If they did not mention the time at all, it means time is not important.
Now, you will say what about the reverse? They want me to find the time when the velocity, say, [imath]v(x(t)) = 15 [/imath] m/s. They will not tell you to do that because the Le Duc equation is designed (derived) to find the velocity when the position (and time) is known, or to find the position when the velocity is known. Or sometimes, they will play around to ask you to find [imath]r[/imath] or [imath]s[/imath] when other variables are known.