Optimization in mechanics

Great question Ronak! I actually knew that the curve will be a cycloid. I've seen this in many science museums where they say straight line path is not always the fastest. The proof by Fertmat's principle was also familiar to me, although I couldn't do it now, I remember my science camp teacher telling me about this. For proof that this curve is a cycloid, refer here and here

Once we have the parametric equation:

$x=a(t-sin(t))+c_1, y=a(1-cos(t))+c_2$

We now have to find the member of this family that passes through both $(1,1)$ and $(0,0)$ . At the initial point, the curve has to be perpendicular to horizontal(refer wiki proof). so now, for $t=0$ , the curve must be at $(1,1)$ . Hence $c_1=c_2=1$ . Now, in order to pass through $(0,0)$ , $a$ will be such that:

$a(t_0-sin(t_0))=a(1-cos(t_0))=-1$ .

Solving for $t$ and $a$ , we get:

$t_0=2.412$ and $a=-0.5729$

Now observe that velocity at $(x,y)$ is given by: $v=\sqrt { 2g(1-y) }$

Distance moved along curve when velocity is $v$ is: $\sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } dx$

Hence the total time taken will be given by:

$T=\int _{ 1 }^{ 0 }{ \frac { \sqrt { 1+{ \left( \frac { dy }{ dx } \right) }^{ 2 } } }{ \sqrt { 2g(1-y) } } dx }$

We know $\frac { dy }{ dx }$ and $x$ in terms of $t$ . Hence we will write everything in terms of $t$ and substitute $x=a(t-sin(t))$ in the integral to get:

$T = \sqrt { \frac { |a| }{ g } } \int _{ 0 }^{ t_0 }{ \frac { \sqrt { 1-cos(t) } }{ \sqrt { 1-cos(t) } }dt }$

$\Rightarrow T=\quad t_0\sqrt { \frac { |a| }{ g } }$

Now substitute known values to get answer as $\boxed{577}$ .

Proof that curve is cycloid

the time integral is given by

$\int _{ 0 }^{ 1 }{ \sqrt { \frac { 1+\left( y' \right) ^{ 2 } }{ 2gy } } } dx\quad =\quad T$

(note i have shifted origin and flipped figure horizontally in mind for convinience, it wont affect answer)

so we have our Lagrange as

$\sqrt { \frac { 1+\left( y' \right) ^{ 2 } }{ 2gy } }$

which is independent of x (explicitly)

and we know that , when

$\frac { \partial L }{ \partial x } =0\\ \\ L-y'\frac { \partial L }{ \partial y' } =k\\$

so we have here a simple relation (which you can check for yourself)

$y(1+(y')^{ 2 })=c$

whose solution is a cycloid (motion of particle attached to rim of wheel)

Now, look at Raghav's solution

Nice problem, however, this is the first problem everyone solves after Learning the lagrange method,

Whats interesting however is that Isaac Newton solved this without calculus with in a day

Here is another interesting problem where cycloid comes into play

https://brilliant.org/problems/geometry-kinematics-challenge/?group=UZg8UXKeTEJn&ref_id=650641