Solve this integral involving a quadratic and linear air resistance equation

In summary: V}##, but then you have to cancel the ##b## in the denominator of the first term on the right side. That will give you the overall factor of ##\frac 1b##.In summary, the conversation is about solving an integral and isolating a variable, but the individual is having trouble getting the correct answer and is seeking help from others. They discuss their approach and equations, pointing out mistakes and correcting them. Ultimately, they realize that they need to factor out an overall factor of 1/b on the right side and combine the logarithmic terms to cancel out the b in the denominator of the first term.
  • #1
happyparticle
410
20
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

##
 
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  • #2
EpselonZero said:
##
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?
 
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  • #3
TSny said:
Could these be equivalent?
To be honest I was thinking about it, but I don't know how to verify.
 
  • #4
Note that

##\large \frac{A}{B}-1 = \frac{A-B}{B}##
 
  • #5
In this case B = ##
{b +cVo- e^{\frac{-bt}{m}}VoC}
##

I don't think it's equivalent.
 
  • #6
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]
 
Last edited:
  • #7
Rats, I just caught something else. Your argument of the exponential should not have ##c## in it.
 
Last edited:
  • #8
EpselonZero said:
##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".
 
Last edited:
  • #9
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}##
 
  • #10
EpselonZero said:
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##.
 
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  • #11
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.
 
  • #12
##\large \int \frac{dv}{b+cv}##

If you let ##u = b+cv##, how is ##du## related to ##dv##?
 
  • #13
I'm not sure to understand.

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

## ln |b +cv|##

That's what I did.
 
  • #14
EpselonZero said:
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} ##?
 
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  • #15
I found ##\frac {cv}{b+cv}##
You are right.

How can I know I have to do multiply by ##\frac 1c##
 
  • #16
EpselonZero said:
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)##.
 
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  • #17
Alright, thanks guys for the help
 
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  • #18
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 {...}{...})##
 
  • #19
EpselonZero said:
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}##.
 
  • #20
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 ?
 
  • #21
EpselonZero said:
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}##
 
  • #22
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##
 
Last edited by a moderator:
  • #23
Earlier you had
EpselonZero said:
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}##
 
  • #24
EpselonZero said:
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.
 
  • #25
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}##
 
  • #26
EpselonZero said:
I found this answer.

## V = \frac {e^{-\frac{tb}{m}}Vo\frac bc}{-e^{-\frac{tb}{m}}Vo+\frac bc + Vo}##
That looks correct.
 
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  • #27
Phew! Thanks a lot for your patience.
 
  • #28
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.
 
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  • #29
I can't see where the 1 in the denominator comes from.

Even with all the tries I never found a +/- 1.
 
  • #30
EpselonZero said:
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
 
  • #31
Ah, sorry! I feel bad.
 

Related to Solve this integral involving a quadratic and linear air resistance equation

1. What is an integral involving a quadratic and linear air resistance equation?

An integral involving a quadratic and linear air resistance equation is a mathematical expression that represents the total resistance experienced by an object moving through a fluid, such as air. It takes into account both the quadratic and linear components of air resistance, which are dependent on the velocity of the object.

2. Why is it important to solve this type of integral?

Solving this type of integral is important because it allows us to accurately predict the behavior of objects moving through a fluid, such as air. This is crucial in fields such as engineering and physics, where understanding the effects of air resistance on an object's motion is essential.

3. What are the steps to solve an integral involving a quadratic and linear air resistance equation?

The steps to solve this type of integral involve first setting up the equation by identifying the variables and constants involved, then integrating the equation using techniques such as substitution or integration by parts. Finally, the integral can be solved using algebraic manipulation and the appropriate integration rules.

4. Can this type of integral be solved analytically or does it require numerical methods?

It depends on the specific equation and its complexity. In some cases, an integral involving a quadratic and linear air resistance equation can be solved analytically using integration techniques. However, in more complex cases, numerical methods such as approximation or simulation may be necessary to find a solution.

5. How does solving this integral help in real-world applications?

Solving this integral helps in real-world applications by allowing us to make accurate predictions about the behavior of objects moving through a fluid, such as airplanes or cars. This information is crucial for designing efficient and safe vehicles, as well as understanding the effects of air resistance on their motion.

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