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?

 

[haruspex]

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

It looks like you need to reread integration basics, paying attention to bounds and to definite and indefinite integrals.

 

[ThEmptyTree]

I got v(t) = -1/c ln( e^(-c v_0) + bc/m t ).

 

[haruspex]

I got v(t) = -1/c ln( e^(-c v_0) + bc/m t ).

Good.

 

[idekthanks]

Homework Statement:: MIT pretest.
Relevant Equations:: 𝐅⃗=−𝑏𝑒^(𝑐𝑣)𝐢̂ , find v(t), by using differential equation of F=maHello, would you mind sharing what course this is? I am familiar with edx but can’t find it

 

[haruspex]

@idekthanks , the thread is over a year old. @MichaelTam might not be visiting the forum any more.

 

[deepread]

@idekthanks Maybe this is a late reply, but I leave it for future readers.
The course contents are available here - https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/

 

[Delta2]

This problem becomes more complex if we alter the initial condition of $v_0$ to be not parallel to x-axis but to have some other direction. For example if $v_0$ is in the y-axis then the differential equation becomes $$m\frac{dv_x}{dt}=-be^{c\sqrt{v_0^2+v_x^2}}$$ and who can solve this