Difficult integral - Falling mass

[Cincin]

Homework Statement

Solve the differential equation, dt/dv= 1/g (1/1-a^2*v^2) where a = (k/mg)^1/2 to yield v= 1/a(1-e^-2agt/1+e^-2agt).

Homework Equations

F=ma
Newton's 2nd Law
Integration Laws

The Attempt at a Solution

See image. I think I'm getting messed up on the integration laws.

 

[SteamKing]

Please post threads in the proper forum by subject.

Solving differential equations and calculating integrals are by nature calculus problems, not pre-calculus.

 

[Ray Vickson]

Homework Statement

Solve the differential equation, dt/dv= 1/g (1/1-a^2*v^2) where a = (k/mg)^1/2 to yield v= 1/a(1-e^-2agt/1+e^-2agt).

Homework Equations

F=ma
Newton's 2nd Law
Integration Laws

The Attempt at a Solution

See image. I think I'm getting messed up on the integration laws.

Your differential equation
\frac{dt}{dv} = \frac{1}{g} \left( \frac{1}{1} - a^2 v^2 \right) = \frac{1}{g} ( 1 - a^2 v^2)
(which is what you wrote) does not have the solution
v = \frac{1}{a} \left( 1 - e^{-2} \frac{agt}{1} + e^{-2} agt \right)
which is what you wrote. If these are not what you mean, you need to start using parentheses.

I did not look at the attached image; I look only at typed work, which is the preferred format in this Forum.

 

[Cincin]

Your differential equation
\frac{dt}{dv} = \frac{1}{g} \left( \frac{1}{1} - a^2 v^2 \right) = \frac{1}{g} ( 1 - a^2 v^2)
(which is what you wrote) does not have the solution
v = \frac{1}{a} \left( 1 - e^{-2} \frac{agt}{1} + e^{-2} agt \right)
which is what you wrote. If these are not what you mean, you need to start using parentheses.

I did not look at the attached image; I look only at typed work, which is the preferred format in this Forum.

Sorry for the misconception. I meant to write:
dt/dv = 1/g ((1)/(1-a^2v^2))
V = 1/a ((1-e^(-2agt))/(1+e^(-2agt)))

Hopefully this is clearer.

 

[Samy_A]

Sorry for the misconception. I meant to write:
dt/dv = 1/g ((1)/(1-a^2v^2))
V = 1/a ((1-e^(-2agt))/(1+e^(-2agt)))

Hopefully this is clearer.

Try to use LaTeX (https://www.physicsforums.com/help/latexhelp/ )
Do you mean:
$\displaystyle \frac{dt}{dv}= \frac{1}{g}\frac{1}{1-a²v²}$
$\displaystyle v=\frac{1}{a}\frac{1-e^{-2agt}}{1+e^{-2agt}}$?

 

[Cincin]

\frac{dt} {dv} = \frac{1} {g} ( \frac{1} {1 - a^2 v^2} )

v = \frac{1} {a} ( \frac{1 - e^(-2agt)} {1 + e^(-2agt)} )

The (-2agt) is to the power of e
e^(-2agt)
Sorry for the confusion guys! :(

 

[Cincin]

Try to use LaTeX (https://www.physicsforums.com/help/latexhelp/ )
Do you mean:
$\displaystyle \frac{dt}{dv}= \frac{1}{g}\frac{1}{1-a²v²}$
$\displaystyle v=\frac{1}{a}\frac{1-e^{-2agt}}{1+e^{-2agt}}$?

\frac{dt} {dv} = \frac{1} {g} ( \frac{1} {1 - a^2 v^2} )

v = \frac{1} {a} ( \frac{1 - e^{-2agt}} {1 + e^{-2agt}} )

This is what I mean to write.
Sorry for the confusion guys! :(

 

[Samy_A]

Great!

Now back to the exercise:

Something went wrong in the part I circled in red.
The derivative of $\log (1-a²v²)$ is $\frac{-2av}{1-a²v²}$.

EDIT: To fix the error above: the derivative of $\log (1-a²v²)$ is $\frac{-2a²v}{1-a²v²}$.

 

[Cincin]

Great!

Now back to the exercise:

View attachment 98245

Something went wrong in the part I circled in red.
The derivative of $\log (1-a²v²)$ is $\frac{-2av}{1-a²v²}$.

Yeh, I know there is probably an issue here in my work, but I wrote: $\ln (1 - a^2 v^2)$ when derived, this gives $\frac{1}{1-a^2 v^2}$ doesn't it?

 

[Samy_A]

Yeh, I know there is probably an issue here in my work, but I wrote: $\ln (1 - a^2 v^2)$ when derived, this gives $\frac{1}{1-a^2 v^2}$ doesn't it?

No, it gives $\frac{-2a²v}{1-a^2 v^2}$ (I forgot the a² in my previous post).

You have to apply the chain rule.
If $f(x)=\ln g(x)$, then $f'(x)=\frac{g'(x)}{g(x)}$.

 

[Cincin]

No, it gives $\frac{-2a²v}{1-a^2 v^2}$ (I forgot the a² in my previous post).

You have to apply the chain rule.
If $f(x)=\ln g(x)$, then $f'(x)=\frac{g'(x)}{g(x)}$.

Yeahh I'm really stuck on this one. I don't know how I can get my $g'(x) = 1$ whilst the bottom $= 1 - a^2 v^2$??

 

[Samy_A]

To calculate $\int \frac{1}{1-a²v²}dv$, look at this insight from @Mark44: https://www.physicsforums.com/insights/partial-fractions-decomposition/

You can decompose $\frac{1}{1-a²v²}$, and then you get two easy integrals.

 

[Cincin]

To calculate $\int \frac{1}{1-a²v²}dv$, look at this insight from @Mark44: https://www.physicsforums.com/insights/partial-fractions-decomposition/

You can decompose $\frac{1}{1-a²v²}$, and then you get two easy integrals.

So I took that advice on board and I ended up getting $t = \frac {-ln(av+1)+ln(av-1)} {-2ag}$

 

[Samy_A]

So now you can solve that to get v in function of t.
Remember that $\ln \frac{b}{c}=\ln b -\ln c$.

 

[K West]

Hi Cincin, Nice to see you are working on your assignment.