How Can You Solve a Differential Equation Involving Exponential Drag Force?

[MichaelTam]

Homework Statement

MIT pretest.

Relevant Equations

𝐅⃗=−𝑏𝑒^(𝑐𝑣)𝐢̂ , find v(t), by using differential equation of F=ma

There ‘s an equation, I think it should like a “drag factor” because the -(b) now has a negative symbol, which means it should be affecting the particle going parallel of the x-axis to the right, I suppose F=ma need to be apply into this equation, but before this, I need to get it into a differential equation first, but I don’t know how to isolate the exponential element, what I know is to give natural log to them , but the next step I don’t know how to do.

A particle of mass 𝑚 moving parallel to the x-axis is acted on by a velocity dependent force directed against its motion. The force is given by:

𝐅⃗=−𝑏𝑒^(𝑐𝑣)𝐢̂

where 𝑏 is a positive constant (units N), 𝑐 ia also a positive constant (units s·m−1), and 𝑣 is the speed, the magnitude of the particle's velocity (units m·s−1).

If at 𝑡=0, the particle is moving with speed 𝑣0. Find the speed 𝑣(𝑡) as a function of time 𝑡. Express your answer in terms of some or all of the following: 𝑏, 𝑚, 𝑡, 𝑐, and v_0 for 𝑣0. Use ln() for the natural logarithm function and e^() for the exponential function if needed.
This problem I can’t reach the solution have two main reason, the first is , I get too few knowledge of exponential equation (mathematical problem) and I still feel confuse about the equation of how to derive it with an exponential element in the equation in order to get the differential equation and substitute the v(t).

 

[MichaelTam]

I am still finding the way to get the differential equation from 𝐅⃗=−𝑏𝑒^(𝑐𝑣)𝐢̂.
And what is the unit of c (s*m^-1)?(meaning)?And I know F is the only force that acting on the particle that slowing down its motion.
Can you explain this problem to me, because I am not sure the v is the F or the F is add to the v?Is it the v is the −𝑏𝑒^(𝑐𝑣)𝐢̂ according to the diagram?

 

[MichaelTam]

My understanding to this problem is , the particle is moving in a constant velocity, but at $v_0$ a force is applied to slow down the particle, and the force is name as F
which is −𝑏𝑒^(𝑐𝑣)𝐢̂, and you need to find v(t), I am not really sure the problem is asking in this way?
And if my understanding to this problem is correct ,the F=ma the total force acting on the particle should be $F=v{t} - b e^{ c v{t}}$?in the v{t} situation.
And the $v_0$ situation should be F=$v{0} - b e^{c v{0}}$
so it should be F=-b?

 

[kuruman]

Start with Newton's second law and write the acceleration as the derivative of the velocity.

 

[MichaelTam]

Thanks

 

[MichaelTam]

With the help of kuruman (Homework helper) I can find that:
F=ma
$F= m { dv/dt }$ where the t is the time.
$F=-b e^{c v}$
$-b e^{c v}= m { dv/dt }$
$dv={-b e^{c v}}/m$ times $dt$,but I don’t know how to remove or isolate the exponential, can you teach me to do?because the In() will move to the $dv/dt$ if I remove the exponential.
I got $c v = In{ m dv/{-b dt}}$
I cannot find how to type in the natural log?

 

[kuruman]

You have $-b e^{c v}= m { dv/dt }.$ You need to separate variables. What algebraic steps do you need to take to have things with $v$ only on one side and things with $t$ only on the other side? Hint: To get rid of something on one side, divide both sides by that something. What could that be?

 

[haruspex]

Duplicate of https://www.physicsforums.com/threads/velocity-dependent-force.992948/#post-6385373

 

[PeterDonis]

Duplicate of https://www.physicsforums.com/threads/velocity-dependent-force.992948/#post-6385373

I have moved the relevant posts from the other thread to this one. The other thread is now closed.

 

[MichaelTam]

I got $dv= \frac {-be^{cv}} m dt$
but I don’t know how to put the $dv$ and $v$ together because there have exponential in it, can you teach me how to get $dv +v=...$
because I need to isolate the v from the exponential, then I can do the integration effectively.I am worrying about after I integrate it, I can’t separate the v(t) and v(0) into two same variable group e.g (v(t)+v(t)).

 

[haruspex]

I got $dv= \frac {-be^{cv}} m dt$
but I don’t know how to put the $dv$ and $v$ together because there have exponential in it, can you teach me how to get $dv +v=...$
because I need to isolate the v from the exponential, then I can do the integration effectively.I am worrying about after I integrate it, I can’t separate the v(t) and v(0) into two same variable group e.g (v(t)+v(t)).

That's not what was meant by 'putting them together'
Just rearrange the equation so that all references to v are on one side and all references to t are on the other.
You do need to know how to integrate $\int e^x.dx$

 

[MichaelTam]

I don’t know I should write in which forms?

 

[MichaelTam]

It is the dv is respond to the c(v) or it should be a variable refer to my picture?
If my understanding is correct, it should be

 

[MichaelTam]

If the constant k1 and k2 are not need to add on the equation(because of the t=0), the answer is at the upper part without the k.However, there are still lot of step that I am confused.

 

[MichaelTam]

Hi can anyone help me please?Can anyone told me where I am wrong?

 

[haruspex]

If the constant k1 and k2 are not need to add on the equation(because of the t=0), the answer is at the upper part without the k.However, there are still lot of step that I am confused.

You certainly don’t need two constants, and substituting t=0 shows the constant is zero.

 

[MichaelTam]

so I can ignore all constant k result in the integration?

but constant of the integration of v(t) will not equal to zero.
If my understanding to this question is correct,
I got the answer of :
$e^{-c v_0} - e^{-c v_t}=\frac {-b c t} m$
And I got:
$\frac {ln{\frac {b c t} m + e^{-c v_0}}} c$

 

[haruspex]

It is the dv is respond to the c(v) or it should be a variable refer to my picture?
If my understanding is correct, it should be View attachment 268502

I think you have a sign error that crept in near the end.
You had $e^{-cv_0}-e^{-cv}=-\frac{bct}m$. Take it from there again.

 

[MichaelTam]

I reply #17 here
so I can ignore all constant k result in the integration?

but constant of the integration of v(t) will not equal to zero.
If my understanding to this question is correct,
I got the answer of :
$e^{-c v_0} - e^{-c v_t}=\frac {-b c t} m$
And I got:
v(t)=$\frac {ln{(\frac {b c t} m + e^{-c v_0}})} c$

 

[MichaelTam]

Is this correct now?

 

[haruspex]

I reply #17 here
so I can ignore all constant k result in the integration?

but constant of the integration of v(t) will not equal to zero.
If my understanding to this question is correct,
I got the answer of :
$e^{-c v_0} - e^{-c v_t}=\frac {-b c t} m$
And I got:
v(t)=$\frac {ln{(\frac {b c t} m + e^{-c v_0}})} c$

As I have explained to you twice now, you do not need a constant of integration if you use the range bounds on both sides. Look what happens if we include a constant and plug in t=0:
$e^{-c v_0} - e^{-c v_t}=\frac {-b c t} m+k$
$e^{-c v_0} - e^{-c v_0}=0+k$
$0=k$

 

[MichaelTam]

So it is $e^{-c v_0} = e^{- c v_t}$?

 

[haruspex]

So it is $e^{-c v_0} = e^{- c v_t}$?

What is "it" in your question?

 

[MichaelTam]

#21 ’s result?

 

[haruspex]

#21 ’s result?

Post #21 was just to show you that you do not need a constant of integration because you used the form $\int _0^t ... = \int _{v(0)}^{v(t) } ...$.
Your equations in post #19 are correct.

 

[MichaelTam]

Ok, I am just scared of the result...

 

[MichaelTam]

So my equation in post 19 is correct, my answer also correct?

 

[MichaelTam]

So, using the post equation in 19, I got 2, but using the 25post ,I get 1.

 

[MichaelTam]

It is the most simplified answer?

 

[MichaelTam]

Ops, I forget, only 2 is the correct way, but is it the step is wrong?

 

[haruspex]

So, using the post equation in 19, I got 2, but using the 25post ,I get 1.
View attachment 268528

I have no idea how you keep getting $e^{-c v_0} = e^{-c v_t}$. Where does that come from?
What I wrote in post #21 was, suppose we allow a constant of integration, k:
$e^{-c v_0} - e^{-c v_t}=\frac {-b c t} m+k$
If we evaluate that at t=0 we get:
$e^{-c v_0} - e^{-c v_0}=0+k$
Look carefully at the left hand side. Everything cancels there, leaving zero, so k=0.

 

[kuruman]

Your problem is that you do not understand the difference between a definite integral and an indefinite integral.
1. Definite integral
$$\int_{x_0}^x u~du=\frac{1}{2} \left. u^2 \right |_{x_0}^x=\frac{1}{2}x^2-\frac{1}{2}x_0^2.$$You just evaluate the antiderivative at the upper and lower limits and subtract the latter from the former.

2. Indefinite integral
$$\int u~du=\frac{1}{2} u^2 +k.$$There are no upper and lower limits; constant $k$ is a placeholder for the lower limit. In this particular problem you have a definite integral, you know the lower limit, therefore the placeholder is not needed or, as has already been shown, it must be zero.

 

[MichaelTam]

So, is my answer in the equation two is the simplest answer?
Why my answer in the equation 2 is still incorrect?

 

[MichaelTam]

Ops, I find my self have some typing error, the answer in the equation two is correct, thank you kuruman and haruspex!

 

[MichaelTam]

Are there any way to learn more calculus to get more strategy?