Time it takes for submerged object to rise to surface

[xodin]

Actually, I lied: you just have to refer to a table of standard derivatives to see that: $\frac{d}{dx}arctanh(x) = \frac{1}{1 - x^2}$. You can check this if you like. Your calc textbook should also have a proof of this in it. Now just rescale the variables appropriately and you get the result.

Ah, ok, I think I'm getting close now! Great stuff on the terminal velocity too, that part makes total sense, thanks again!

$\frac {1}{a} \int \frac {dv}{1-\alpha^{2} v^{2}} = t_{0} + t$

$\frac {tanh^{-1}\left(\alpha v\right)}{a} = t_{0} + t$

Ignoring $t_{0}$:

$tanh^{-1}\left(\alpha v\right) =a t$

$\alpha v = tanh\left(a t\right)$

$v = \frac {tanh\left(a t\right)}{\alpha}$

Somewhere I lost an alpha! :P

 

[VantagePoint72]

Let's just focus on:
$\int \frac {dv}{1-\alpha^{2} v^{2}}$

Set $u=\alpha v, du = \alpha dv$. Thus we have:
$\frac{1}{\alpha}\int \frac {du}{1-u^{2}} = \frac{1}{\alpha}arctanh(u)$

There's your missing alpha. Not just sub u back in and multiply the $\frac{1}{a}$ I ignored.

Incidentally, if you want to show that $arctanh(x) = \int \frac{dx}{1 - x^2}$, you can use trig substitution like I said—but with hyperbolic trig functions, not regular ones like you tried. Substitute $x = tanh(u)$ and away you go. Of course, if you knew to do this off the top of your head, you could have just solved the original differential equation by inspection without all this fiddling. Solving diff eq's is as much an art as a science. Such is life.

 

[xodin]

Let's just focus on:
$\int \frac {dv}{1-\alpha^{2} v^{2}}$

Set $u=\alpha v, du = \alpha dv$. Thus we have:
$\frac{1}{\alpha}\int \frac {du}{1-u^{2}} = \frac{1}{\alpha}arctanh(u)$

There's your missing alpha. Not just sub u back in and multiply the $\frac{1}{a}$ I ignored.

Incidentally, if you want to show that $arctanh(x) = \int \frac{dx}{1 - x^2}$, you can use trig substitution like I said—but with hyperbolic trig functions, not regular ones like you tried. Substitute $x = tanh(u)$ and away you go. Of course, if you knew to do this off the top of your head, you could have just solved the original differential equation by inspection without all this fiddling. Solving diff eq's is as much an art as a science. Such is life.

You are awesome, thank you once again. I was just coming back to look and see if I should have used u substitution there to get my missing alpha, and apparently should have. I should have known better on that one, but rushed it! I can't wait to take the course, but will be taking multivariable first for intro to mechanics followed by differential equations next semester.

I really appreciate both your's and voko's posts, but especially your added explanations! Enjoy the rest of your evening.

 

[VantagePoint72]

My pleasure, good luck in the course.

 

[voko]

$$\frac {1} {a} \int_0^v \frac {dv} {1 - \alpha^2 v^2} = \frac {1} {2a} \int_0^v \left[ \frac {1} {1 - \alpha v} + \frac {1} {1 + \alpha v}\right]dv = \frac {1} {2\alpha a} \left[ -ln (1 - \alpha v) + ln(1 + \alpha v)\right]_0^v = \frac {1} {2\alpha a} \ln \frac {1 + \alpha v} {1 - \alpha v}$$ Using Euler's formula for hyperbolic functions, we can see immediately that the latter is $\frac {1} {\alpha a} \tanh^{-1} \alpha v$, so $$\frac {1} {\alpha a} \tanh^{-1} \alpha v = \int_0^t dt = t$$

 

[voko]

The second integration is even easier: $$\frac {1} {\alpha} \int_0^t \tanh \alpha a t dt =\frac {1} {\alpha} \int_0^t \frac {\sinh \alpha a t} { \cosh \alpha a t }dt = \frac {1} {\alpha} \int_0^t \frac { (\frac {\cosh \alpha a t} {\alpha a})'} { \cosh \alpha a t }dt =$$$$= \frac {1} {\alpha^2 a} \left[\ln \cosh \alpha a t\right]_0^t = \frac {1} {\alpha^2 a} \ln \cosh \alpha a t = \int_{x_0}^{x} dx = x - x_0$$

 

[xodin]

Cool, thanks for sharing the way you worked them out! I think I prefer the u substitution and tanh^-1 identity over using logarithms and Euler's formula, but mostly just because I already have all the hyperbolic identities and their inverses memorized from differentiation. That's still an interesting way to see it done though, and I'm going to try to remember how to do it. I saw that last identity yesterday (Euler's formula) in my book when looking for hyperbolic identities, but didn't think it could be applied until I saw how you first expanded it and integrated into logarithms.

After the first integration was explained yesterday, the second integration became obvious, but still, thanks for walking me through the steps you took for both!