Problem inverting a function

Question

I have this function:

$$v(t)=\sqrt{\frac F c} \tanh \left(\frac{\sqrt{Fc}}{m} t \right)$$

I can visually see that t=6.3 when v=27.8, so why don't I get t=6.3 upon putting v=27.8 in this supposedly inverse function?!?

$$t(v)=\sqrt{\frac{m^2}{Fc}}\tanh^{-1}\left(v\sqrt{\frac c F}\right)$$

Answer 1

We invert the function as we would any other function:

$$v=\sqrt{\frac{F}{c}}\tanh\left(\frac{\sqrt{Fc}}{m}t\right)$$

Dividing both sides by $\sqrt{\frac{F}{c}}$, we get:

$$\sqrt{\frac{c}{F}}\,v=\tanh\left(\frac{\sqrt{Fc}}{m}t\right)$$

Taking the inverse hyperbolic tangent of both sides:

$$\tanh^{-1}\left(\sqrt{\frac{c}{F}}\,v\right)=\frac{\sqrt{Fc}}{m}t$$

We then divide both sides by $\frac{\sqrt{Fc}}{m}$, to give:

$$t=\frac{m}{\sqrt{Fc}}\tanh^{-1}\left(\sqrt{\frac{c}{F}}\,v\right)$$

So your expression for $t(v)$ is correct, so it must be merely a numerical issue you are having.

Answer 2

Winplot has no definition for atanh()

Hence plotting atanh(x) results in plotting a*tanh(x) with obviously wrong results

By defining in WinPlot a new function $ath(x)=\frac 1 2 ln \frac{1+x}{1-x}$ I eventually got right results.