A quick simple derivatives question

[comiconor]

I've never really noticed whether this is true but if I know that:

dt/ds = k/(1-s/r)

for example, how does one find ds/dt? Is it the inverse, as you would expect, or is there some other method?

 

[Fredrik]

The formula for the derivative of the inverse of a differentiable function f looks like this:
$$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}.$$ So if $t=f(s)$, $s=f^{-1}(t)$ and
$$f'(s) =\frac{dt}{ds} =\frac{k}{1-\frac{s}{r}},$$ we have
$$\frac{ds}{dt} =(f^{-1})'(t) = \frac{1}{f'(f^{-1}(t))} = \frac{1}{f'(s)} = \frac{1-\frac s r}{k}.$$ So it turns out that the super-naive calculation
$$\frac{ds}{dt} =\frac{1}{\frac{dt}{ds}} = \frac{1}{\frac{k}{1-\frac{s}{r}}} =\frac{1-\frac s r}{k}$$ would have worked.

I haven't really thought about the exact circumstances in which this works. If you want to be safe, you should always rewrite things so that you can apply the formula at the start of this post.

 

[Matterwave]

This is just an application of the inverse function theorem:

http://en.wikipedia.org/wiki/Inverse_function_theorem

This article tells you when you can do this.