# Difficult integral - Falling mass

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1. Mar 30, 2016

### Cincin

1. The problem statement, all variables and given/known data
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).

2. Relevant equations
F=ma
Newton's 2nd Law
Integration Laws

3. The attempt at a solution
See image. I think I'm getting messed up on the integration laws.

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2. Mar 31, 2016

### SteamKing

Staff Emeritus
Please post threads in the proper forum by subject.

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

3. Mar 31, 2016

### Ray Vickson

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

4. Mar 31, 2016

### Cincin

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.

5. Mar 31, 2016

### Samy_A

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

6. Mar 31, 2016

### 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!! :(

7. Mar 31, 2016

### Cincin

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

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

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

8. Mar 31, 2016

### 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 $\frac{-2a²v}{1-a²v²}$.

Last edited: Mar 31, 2016
9. Mar 31, 2016

### Cincin

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?

10. Mar 31, 2016

### Samy_A

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 $f'(x)=\frac{g'(x)}{g(x)}$.

11. Mar 31, 2016

### Cincin

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

12. Mar 31, 2016

### Samy_A

13. Mar 31, 2016

### Cincin

14. Mar 31, 2016

### Samy_A

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

15. Apr 18, 2016

### K West

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