The Sandpiper

Let’s solve this the way Ronak Agarwal has it, ending up with an expression using the complete elliptical integral of the second kind, so that numerical approximation can be done readily with a math calculator.

Let $R$ be the sandpiper’s speed, relative to the walker. Since this is scale invariant, we can set the distance $X=1$ . Let $x$ be the position of the walker on his path as a function of $\theta$ which is the angle between the walker’s path and the line between the walker and the sandpiper. Check the following graphic

For the incremental change $dx$ in the walker’s position, there is the corresponding incremental change $d\theta$ in the angle. The incremental travel of the sandpiper is $Rdx$ . Then the following is true

${ \left( Cos\left( \theta +d\theta \right) -Cos\left( \theta \right) +dx \right) }^{ 2 }+{ \left( Sin\left( \theta +d\theta \right) -Sin\left( \theta \right) \right) }^{ 2 }={ \left( Rdx \right) }^{ 2 }$

This can be rearranged and simplified to

${ \left( -Sin\left( \theta \right) +\dfrac { dx }{ d\theta } \right) }^{ 2 }+1-{ Sin }^{ 2 }\left( \theta \right) ={ \left( R\dfrac { dx }{ d\theta } \right) }^{ 2 }$

Solving for $dx$ and integrating, the condition being that when the walker reaches where the sandpiper was feeding, the sandpiper is perpendicular $\left( \dfrac { \pi }{ 2 } \right)$ from the walker, we have

$\displaystyle \int _{ 0 }^{ 1 }{ dx=1=\frac { 1 }{ { R }^{ 2 }-1 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( -Sin\left( \theta \right) +\sqrt { { R }^{ 2 }-1+{ Sin }^{ 2 }\left( \theta \right) } \right) } } d\theta$

the right side which, if it equals to $1$ , eventually can be simplified to

$\dfrac { 1 }{ R } \displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { 1-\dfrac { 1 }{ { R }^{ 2 } } { Cos }^{ 2 }\left( \theta \right) } } d\theta$

and the final step is to note that this definite integral, from $\theta=0$ to $\dfrac{\pi}{2}$ , for reasons of symmetry, will yield the same result as

$\dfrac { 1 }{ R }\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { 1-\dfrac { 1 }{ { R }^{ 2 } } { Sin }^{ 2 }\left( \theta \right) } } d\theta$

and there we have Ronak’s expression using the complete elliptical integral of the second kind, that is, we find $R$ such that

$\dfrac { 1 }{ R } E\left( \dfrac { 1 }{ { R }^{ 2 } } \right) =1$

resulting in the approximation

$R=1.30566196258124609424078019268446…$

Let $x$ be the reciprocal of the speed of the sandpiper than $x$ satisfies :

$xE({x}^{2})=1$

Where $\displaystyle E(m)=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { 1-m{ (sin }^{ 2 }\theta ) } d\theta }$

that is complete elliptical integral of second kind with parameter $m={k}^{2}$

I used wolfram alpha to get the numerical value.

Will post proof of this thing later.