Solve this integral involving a quadratic and linear air resistance equation

[happyparticle]

Homework Statement

quadratic and linear air resistance

Relevant Equations

f = ma
f = -bv -cv²

Hi,
I'm trying to solve this integral and then isolate V, but I can't get the right answer. I don't know where is my errors. I probably muffed the integral.

$-bv -cv² = m\frac {dv}{dt}$

$\int_0^t dt = - m \int_{Vo}^v \frac {dv}{bv+cv^2}$

I get this after the integration

$t = -m[ln\frac{V}{Vo} - \frac cb ln(\frac {b+cV}{b+cVo})]$

Finally, I get

$v = \frac{e^{\frac{-bt}{cm}} VoB}{b +cVo- e^{\frac{-bt}{cm}}VoC}$

But the right answer is

$v = \frac{b +cVo}{b +cVo- e^{\frac{-bt}{m}}VoC} -1$

 

[TSny]

$v = \frac{e^{\frac{-bt}{cm}} VoB}{b +cVo- e^{\frac{-bt}{cm}}VoC}$

But the right answer is

$v = \frac{b +cVo}{b +cVo- e^{\frac{-bt}{m}}VoC} -1$

Could these be equivalent?

 

[happyparticle]

Could these be equivalent?

To be honest I was thinking about it, but I don't know how to verify.

 

[TSny]

Note that

$\large \frac{A}{B}-1 = \frac{A-B}{B}$

 

[happyparticle]

In this case B = ${b +cVo- e^{\frac{-bt}{m}}VoC}$

I don't think it's equivalent.

 

[TSny]

Sorry, I now see they are not quite the same. I agree with your answer.

Your answer is equivalent to

$\large \frac{b+cV_0}{b+cV_0 - e^{\frac{-bt}{m}}bV_0}-1$

which has a factor of $b$ rather than $c$ in the last terms of the denominator.

[Edit: Your answer is not quite equivalent to this. See other corrections below]

 

[TSny]

Rats, I just caught something else. Your argument of the exponential should not have $c$ in it.

 

[TSny]

$t = -m[ln\frac{V}{Vo} - \frac cb ln(\frac {b+cV}{b+cVo})]$

I think you're missing a factor of $\frac{1}{b}$ in front of the log in the first term on the right and the second term on the right should not have the factor of $c$ in front of the log.

You can check that the dimensions of the terms in your equation are not consistent.

I'm now getting "their answer".

 

[happyparticle]

I don't know where I made a mistake.

$\int_0^t dt = -m \int_{Vo}^V \frac{dv}{bv+cv²}$

$t = -m \int_{Vo}^V \frac{dv}{v(b+cv)}$
Then I did a integration with partial fractions. A = $\frac1b$ B = $\frac {-c}b$

$t = \frac{-m}b \int_{Vo}^V \frac 1v -\frac cb \int_{Vo}^V \frac{1}{b+cv}$

 

[TSny]

I don't know where I made a mistake.

$\int_0^t dt = -m \int_{Vo}^V \frac{dv}{bv+cv²}$

$t = -m \int_{Vo}^V \frac{dv}{v(b+cv)}$
Then I did a integration with partial fractions. A = $\frac1b$ B = $\frac {-c}b$

$t = \frac{-m}b \int_{Vo}^V \frac 1v -\frac cb \int_{Vo}^V \frac{1}{b+cv}$

OK, that looks good. Of course there should be $dv$ in the integrals and the second integral will also have a factor of $m$.

Be careful with the second integral. Integration will bring in a factor of $\large \frac1c$ which will cancel the $c$ in the fraction $\large \frac cb$.

 

[happyparticle]

I don't get $\frac 1c$

$t = \frac{-m}b[ln\frac {V}{Vo} - \frac cb ln (\frac{b+cV}{ b+cVo})]$

The only way I see to get $\frac 1c$

$- \frac cb \frac 1c \int_{Vo}^V \frac {1}{\frac bc + V}$

However, I don't know why I have to do it.

 

[TSny]

$\large \int \frac{dv}{b+cv}$

If you let $u = b+cv$, how is $du$ related to $dv$?

 

[happyparticle]

I'm not sure to understand.

$\int \frac{1}{b+cv} dv$

$ln |b +cv|$

That's what I did.

 

[haruspex]

I'm not sure to understand.

$\int \frac{1}{b+cv} dv$

$ln |b +cv|$

That's what I did.

If you differentiate $ln |b +cv|$, using the chain rule , do you get $\frac{1}{b+cv}$?

 

[happyparticle]

I found $\frac {cv}{b+cv}$
You are right.

How can I know I have to do multiply by $\frac 1c$

 

[haruspex]

I found $\frac {cv}{b+cv}$
You are right.

How can I know I have to do multiply by $\frac 1c$

One way is to check that the original is the derivative of the integral.
Another is a change of variable: $\int\frac v{b+cv}dv=\int\frac 1c\frac v{b+cv}d(cv)$.

 

[happyparticle]

Alright, thanks guys for the help

 

[happyparticle]

Me again, sorry.

I found what you guys said, but I'm still stuck.

$- \frac {tb}{m} = ln \frac {V}{Vo} - \frac {1}{b} ln \frac {\frac bc + V}{\frac bc + Vo}$

I can't do that

$\frac 1b ln(\frac {...}{...})$

 

[TSny]

I found what you guys said, but I'm still stuck.

$- \frac {tb}{m} = ln \frac {V}{Vo} - \frac {1}{b} ln \frac {\frac bc + V}{\frac bc + Vo}$

The first term on the right should have a factor of $\frac 1b$ just like the last term. The left side should not have the factor of $b$ yet. So you can factor out an overall factor of $\frac 1b$ on the right side and then combine the two log terms. The $b$ in the $\frac 1b$ factor can then be taken to the left side to give the $b$ in $\frac {tb}{m}$.

 

[happyparticle]

I know, but I can't find where I miss it.

You said you got the right answer. Is your A = 1/b and B = -c/b in your integration with partial fractions ?

 

[TSny]

I know, but I can't find where I miss it.

You said you got the right answer. Is your A = 1/b and B = -c/b in your integration with partial fractions ?

Yes, your A and B look good. Both A and B have a $b$ in the denominator. This $b$ ends up on the left side to give you the $b$ in $\frac {tb}{m}$

 

[happyparticle]

I'm not sure to understand.
I already have the $b$ in $- \frac {tb}{m}$ from $\frac mb$

Furthermore, I really don't see where I miss the
$\frac 1b$

 

[TSny]

Earlier you had

I don't know where I made a mistake.

$\int_0^t dt = -m \int_{Vo}^V \frac{dv}{bv+cv²}$

$t = -m \int_{Vo}^V \frac{dv}{v(b+cv)}$
Then I did a integration with partial fractions. A = $\frac1b$ B = $\frac {-c}b$

$t = \frac{-m}b \int_{Vo}^V \frac 1v -\frac cb \int_{Vo}^V \frac{1}{b+cv}$

Everything here looks good except the last line. There is a missing factor of $m$ in front of the last integral and there should be $dv$'s in the integrals. The negative sign in $B = -\frac cb$ means that the last integral ends up with a positive sign. Thus, see if you agree that you should get

$t = -\frac mb \int_{Vo}^V \frac {dv}v + \frac{mc}b \int_{Vo}^V \frac{dv}{b+cv}$

 

[TSny]

Finally, I get

$v = \frac{e^{\frac{-bt}{cm}} VoB}{b +cVo- e^{\frac{-bt}{cm}}VoC}$

But the right answer is

$v = \frac{b +cVo}{b +cVo- e^{\frac{-bt}{m}}VoC} -1$

After reviewing everything, I get an answer very close to your answer. The only difference is that your exponential factors have an argument of $\frac{-bt}{cm}$ while it should be $\frac{-bt}{m}$. You can check that $\frac{-bt}{m}$ is dimensionless.

Also, the "right answer" that you quoted seems to be missing an overall factor of $\frac bc$. As given, the answer cannot be correct since the left side has dimensions of velocity while the right side is dimensionless. It should be written as $$v = \frac bc \left(\frac{b +cVo}{b +cVo- e^{\frac{-bt}{m}}VoC} -1\right)$$.

Then, this would be equivalent to your way of writing the answer once your argument of the exponential term is corrected. Your way of writing the answer looks nicer.

I must apologize for not seeing this earlier.

 

[happyparticle]

Alright, no problem. You helped me a lot.

I found this answer.

$V = \frac {e^{-\frac{tb}{m}}Vo\frac bc}{-e^{-\frac{tb}{m}}Vo+\frac bc + Vo}$

 

[TSny]

I found this answer.

$V = \frac {e^{-\frac{tb}{m}}Vo\frac bc}{-e^{-\frac{tb}{m}}Vo+\frac bc + Vo}$

That looks correct.

 

[happyparticle]

Phew! Thanks a lot for your patience.

 

[Chestermiller]

The form of the solution that appeals to me is:
$$v=\frac{V_0e^{-\frac{bt}{m}}}{1+\frac{cV_0}{b}(1-e^{-\frac{bt}{m}})}$$where the numerator is the solution when c = 0, and the 2nd term in the denominator captures the added drag effect of c not being zero.

 

[happyparticle]

I can't see where the 1 in the denominator comes from.

Even with all the tries I never found a +/- 1.

 

[Chestermiller]

I can't see where the 1 in the denominator comes from.

Even with all the tries I never found a +/- 1.

Just multiply numerator and denominator of your answer in post #25 by c/b

 

[happyparticle]

Ah, sorry! I feel bad.