Does the opposite of the brachistochrone exist? [closed]

Suppose a particle only influenced by gravity is going from point a to point b on an x,y plane. Is there an equation in which the particle takes the longest amount of time, yet still makes it to point b? I know the equation for brachistochrone is the parametric for an inverted cycloid, yet I'm not sure of the opposite. For reference, I am a high school student wanting to try this for a research project. I really only need to know whether this has already been done or not, but attempted solutions are obviously appreciated.

• There are infinite curves with this property Oct 2, 2016 at 19:23
• What about a segment parallel to the ground, with a smooth curve joining the right endpoint of such a segment with the point on the ground we want to reach? The object never start sliding along such a curve, hence the arrival time is infinite. Oct 2, 2016 at 19:23
• I do not believe there is a solution to this problem because the set of curves is open and bounded above in time. Oct 2, 2016 at 19:25
• @lordoftheshadows: bounded by what? Please have a look at my previous comment. Oct 2, 2016 at 19:27
• @JackD'Aurizio You can construct a curve that takes an arbitrary length Oct 2, 2016 at 19:28

You can always find a curve which takes a longer time to reach point b. Thus, there isn't one unique answer, unlike the brachistochrone. However, if the particle is only influenced by gravity, it will necessarily follow the least action principle, i.e. the path of the brachistochrone.

There is no solution to this. You can construct a curve that takes an arbitrary amount of time to reach point B. Voila:

We will scale points $A$ and $B$ to be $A(0,a)$ and $B(b,0)$. We will para metrically define $f(x)$. $x=0$ $x>\epsilon$. $x<\epsilon$ we have the curve that goes from $(0,\epsilon)$ to $(b,0)$. As you decrease $\epsilon$ the length of time it takes to traverse the entire curve increases. $\epsilon$ cannot be zero otherwise the particle will never reach $B$ however the time strictly increases as you decrease $\epsilon$. You can find the exact amount of time as a function of $\epsilon$ if you wish to verify it yourself.

Apologies for the bad formatting, I'm not very familiar with mathjax.