Integrating $\frac{dv}{dt}=g-kv^2$ from $v=0$ to $v_t$

[Locrian]

I'm missing something simple here:

$$\frac{d\mbox{v}}{d\mbox{t}} = g - k\mbox{v}^2$$

I'm integrating from v=0 to v=v_t (where v_t is a known constant independant of velocity and time) and from t=0 to t=t_f, where t_f is the variable I wish to solve for in the end.

I'd rather not embarass myself by giving out the answer I produced. The actual answer has a hyperbolic tangent and I honestly am not sure how to get that. Could someone nudge me in the right direction?

 

[arildno]

The inverse of the hyperbolic tangent, the Artanh function is given by:
$$Artanh(x)=log(\frac{1+x}{1-x})$$

 

[Galileo]

You can separate the variables and integrate:

$$\int_{0}^{v(t )}\frac{dv}{g-kv^2}=\int_0^t t'dt'$$

 

[Locrian]

Aha! And the derivative of that is:

$$\frac{1}{1-x^2}$$ Which should take care of the problem. If only it had seemed more obvious before I had to ask! Thank you for your time arildno.

 

[Locrian]

Thank you too for helping Galileo. I'm afraid this seems rather obvious in retrospect, which I thought it might be.

Have a good friday.

 

[arildno]

I'm sorry, I made a mistake:
We should have:
$$Artanh(x)=\frac{1}{2}log(\frac{1+x}{1-x})$$
This can be seen by solving for "y" from the following equation:
$$x=Tanh(y)\equiv\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}$$