August 2024

1 Introduction ..... 1
2 Generalized integral time formula for any curve and descent for a straight line ..... 2
3 Time taken for the descent by various conic sections ..... 3
3.1 Parabola ..... 4
3.2 Circle ..... 4
3.3 Ellipse ..... 5
3.4 Hyperbola ..... 7
4 Conclusion ..... 8

The brachistochrone problem, proposed by Johann Bernoulli in 1696, aims to find, for points $A$$A$AA$A$ and $B$$B$BB$B$ with $A$$A$AA$A$ located above $B$$B$BB$B$, a curve such that descent of an object due to gravity along that curve takes the shortest possible time. The problem considers only gravitational force on the object and ignores other forces such as friction. It was solved by a number of famous mathematicians such as Johann Bernoulli himself, Jakob Bernoulli, and Isaac Newton, all of them proposing a different approach [4]. Till date, numerous researchers have tried to expand upon the problem by including external factors such as non-uniform gravitational field [1], friction [2] and some other challenges, examining the problem further. These methods have been quite effective but require heavy machinery such as the calculus of variations and the Euler-Lagrange differential equation. Additionally, all of the papers are predominantly focused on deriving the equation of the curve with quickest descent and are not focused on other curves. In this paper, we focus on a more elementary approach and study descent along the conic sections from a point $A$$A$AA$A$ with coordinates $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$ to a point $B$$B$BB$B$ with coordinates $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$. In Section 2, we utilise a general approach for calculating the time it takes to descend along a curve defined by a function $y=f(x)$$y=f(x)$y=f(x)y=f(x)$y=f(x)$. In the following section, 3 we give a detailed account of the time period in cases of parabola, circle, ellipse, and hyperbola respectively. Using an online tool [5] to find the value of integrals, we compare the findings of time periods of different curves. The curves when arranged in decreasing order of their time periods are as follows : line > parabola of form $y=f(x)=$$y=f(x)=$y=f(x)=y=f(x)=$y=f(x)=$ hyperbola with $y$$y$yy$y$ axis as major axis $>$$>$>>$>$ ellipse with $y$$y$yy$y$ axis as major axis $>$$>$>>$>$ circle $>$$>$>>$>$ ellipse with $x$$x$xx$x$ axis as major axis $=$$=$==$=$ hyperbola
with $x$$x$xx$x$ axis as major axis $>$$>$>>$>$ parabola of form $x=f(y)$$x=f(y)$x=f(y)x=f(y)$x=f(y)$. Applying these findings in real life, we find that water parks generally have curved slides to provide a faster and more enjoyable experience to the person. Additionally, we understand that the shortest curve is not always the fastest one.
The methodology mainly comprises of independent calculations for calculating the time. Additionally, a small literature review was done to gather historical background on the problem.

Let us take an object of mass $m$$m$mm$m$ sliding from point $A$$A$AA$A$ to point $B$$B$BB$B$. Throughout the paper, we will consider the initial position position $(A)$$(A)$(A)(A)$(A)$ to be $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$ and the final position $(B)$$(B)$(B)(B)$(B)$ to be $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$. In order to find the time taken, we will first use the law of conservation of energy. The energy at the initial point is just the gravitational potential energy $mgh$$mgh$mghm g h$mgh$ where $h$$h$hh$h$ is the height. We have $h=1$$h=1$h=1h=1$h=1$ since our initial point is $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$. The energy at any other point $P$$P$PP$P$ on the curve will be $\frac{1}{2}m{v}^2+mgy$$\frac{1}{2}m{v}^2+mgy$(1)/(2)mv^(2)+mgy\frac{1}{2} m v^{2}+m g y$\frac{1}{2}m{v}^2+mgy$ where $v$$v$vv$v$ is the velocity of the object at $P$$P$PP$P$ and $h$$h$hh$h$ is the $y$$y$yy$y$-coordinate. By conservation of energy,

We have found the velocity at any point. By definition, velocity is the derivative of displacement with respect to time i.e $v=\frac{ds}{dt}$$v=\frac{ds}{dt}$v=(ds)/(dt)v=\frac{d s}{d t}$v=\frac{ds}{dt}$. Rearranging gives $dt=\frac{ds}{v}$$dt=\frac{ds}{v}$dt=(ds)/(v)d t=\frac{d s}{v}$dt=\frac{ds}{v}$ and integrating we obtain

Note that $ds$$ds$dsd s$ds$ is the small change in displacement in a small period of time $dt$$dt$dtd t$dt$. Our aim is to express $ds$$ds$dsd s$ds$ in terms of $dx$$dx$dxd x$dx$. In the below graph, the blue line represents the curve $y=x^2$$y=x^2$y=x^(2)y=x^{2}$y=x^2$ and the purple line refers to the tangent to the curve at $x=1$$x=1$x=1x=1$x=1$. The angle made by the tangent line with the positive $x$$x$xx$x$ axis is denoted by $\theta$$\theta$theta\theta$\theta$. Hence, the slope of the tangent line is equal to $\tan \theta$$\tan \theta$tan theta\tan \theta$\tan \theta$. Through the right angled triangle formed with $ds$$ds$dsd s$ds$ as the hypotenuse and $dx$$dx$dxd x$dx$ as the base, $ds=\frac{dx}{\cos \theta }$$ds=\frac{dx}{\cos \theta }$ds=(dx)/(cos theta)d s=\frac{d x}{\cos \theta}$ds=\frac{dx}{\cos \theta }$.

Now, the derivative of the curve at $x=1$$x=1$x=1x=1$x=1$ is the slope of the line i.e. $\tan \theta$$\tan \theta$tan theta\tan \theta$\tan \theta$. Now from equation ${\sec }^2\theta =$${\sec }^2\theta =$sec^(2)theta=\sec ^{2} \theta=${\sec }^2\theta =$ ${\tan }^2\theta +1$${\tan }^2\theta +1$tan^(2)theta+1\tan ^{2} \theta+1${\tan }^2\theta +1$ we have

As $\theta \le \frac{\pi }{2},\cos \theta$$\theta \le \frac{\pi }{2},\cos \theta$theta <= (pi) (2),cos theta< asciimath>\theta \leq \frac{\pi}{2}, \cos \theta$\theta \le \frac{\pi }{2},\cos \theta$ is positive. Now

So, $ds=\sqrt{({\left(\frac{dy}{dx}\right)}^2+1)}dx$$ds=\sqrt{\left({\left(\frac{dy}{dx}\right)}^2+1\right)}dx$ds=sqrt((((dy)/(dx))^(2)+1))dxd s=\sqrt{\left(\left(\frac{d y}{d x}\right)^{2}+1\right)} d x$ds=\sqrt{({\left(\frac{dy}{dx}\right)}^2+1)}dx$ and

This is the general time formula, which is applicable to any curve $y=f(x)$$y=f(x)$y=f(x)y=f(x)$y=f(x)$.
Consider the case of a line passing through the points $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$ and $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$. It is given by $y=1-x$$y=1-x$y=1-xy=1-x$y=1-x$ so that $\frac{dy}{dx}=-1$$\frac{dy}{dx}=-1$(dy)/(dx)=-1\frac{d y}{d x}=-1$\frac{dy}{dx}=-1$. Hence, the time taken by an object to roll down the line $y=x-1$$y=x-1$y=x-1y=x-1$y=x-1$ from $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$ to $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$ is equal to

(Taking the value of g as 9.8 ).

Figure 2.1: Graph of $y=x^2$$y=x^2$y=x^(2)y=x^{2}$y=x^2$ along with its tangent $y=2x-1$$y=2x-1$y=2x-1y=2 x-1$y=2x-1$ at $x=1$$x=1$x=1x=1$x=1$

In this section, we discuss the time taken by an object of any mass to travel from $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$ to $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$ through various conic sections listed below.

Figure 3.1: A graph summarising the curves, which are labelled in the order of increasing descent time.

The code for the above graph is accessible at https://github.com/Aarush2012/Brachistochrone-curvegraph.git.

Consider a parabola of the form $\mathrm{y}=a{x}^2+bx+c$$\mathrm{y}=a{x}^2+bx+c$y=ax^(2)+bx+c\mathrm{y}=a x^{2}+b x+c$\mathrm{y}=a{x}^2+bx+c$ passing through the points $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$ and $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$. Let us substitute the coordinates and find the equation of such a parabola. We get:

The general equation for a parabola passing these points is

We can now apply the general formula (2.1). First, we differentiate: $\frac{dy}{dx}=2ax-a-1$$\frac{dy}{dx}=2ax-a-1$(dy)/(dx)=2ax-a-1\frac{d y}{d x}=2 a x-a-1$\frac{dy}{dx}=2ax-a-1$. Substituting, we get

Note that the time $t$$t$tt$t$ is dependent on the value of the parameter $a$$a$aa$a$, and we can try to find the minimum of $t(a)$$t(a)$t(a)t(a)$t(a)$ to minimize the descent time.

We calculate the value of the above integral by plotting it into Desmos [5]. As $a$$a$aa$a$ is a variable, we add a slider to change the value of $a$$a$aa$a$. The observations are that the value of the integral is the least when $a=1$$a=1$a=1a=1$a=1$ with time taken for descent equal to $0.595227\dots$$0.595227\dots$0.595227 dots0.595227 \ldots$0.595227\dots$ seconds. At $a=0.9$$a=0.9$a=0.9a=0.9$a=0.9$, the value comes out be 0.5958 and at 1.1 , it is 0.59543 . From $a=0$$a=0$a=0a=0$a=0$ to $a=1$$a=1$a=1a=1$a=1$, the time taken decreases and after 1 , it keeps on increasing. The curve is represented by $y=x^2-2x+1=(x-1{)}^2$$y=x^2-2x+1=(x-1{)}^2$y=x^(2)-2x+1=(x-1)^(2)y=x^{2}-2 x+1=(x-1)^{2}$y=x^2-2x+1=(x-1{)}^2$
Now, let us discuss the time taken by a parabola of the form $x=a{y}^2+by+c$$x=a{y}^2+by+c$x=ay^(2)+by+cx=a y^{2}+b y+c$x=a{y}^2+by+c$. Substituting the coordinates in the equation, $c=1$$c=1$c=1c=1$c=1$ and $a+b+1=0$$a+b+1=0$a+b+1=0a+b+1=0$a+b+1=0$ so that $b=-1-a$$b=-1-a$b=-1-ab=-1-a$b=-1-a$. The parabola is of the form $x=a{y}^2-(a+1)y+1$$x=a{y}^2-(a+1)y+1$x=ay^(2)-(a+1)y+1x=a y^{2}-(a+1) y+1$x=a{y}^2-(a+1)y+1$. Differentiating the equation with respect to $y,\frac{dx}{dy}=2ay-a-1$$y,\frac{dx}{dy}=2ay-a-1$y,(dx)/(dy)=2ay-a-1y, \frac{d x}{d y}=2 a y-a-1$y,\frac{dx}{dy}=2ay-a-1$. Our second step is to express $y$$y$yy$y$ in terms of $x$$x$xx$x$. For this, we will express $x$$x$xx$x$ in terms of a perfectly squared equation, which will help us isolate $y$$y$yy$y$. We have

. Therefore

so that

Clearly, the graph of the descent curve cannot pass through the second quadrant. In addition, it has to pass through the points $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$ and $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$. This implies that in the equation above we need to choose the minus sign and restict $a$$a$aa$a$ to the interval $a\in [0,1]$$a\in [0,1]$a in[0,1]a \in[0,1]$a\in [0,1]$. So we have

Now, we substitute the values of $\frac{dy}{dx}$$\frac{dy}{dx}$(dy)/(dx)\frac{d y}{d x}$\frac{dy}{dx}$ and $y$$y$yy$y$ in the equation (2.1), we obtain an integral depending on $a$$a$aa$a$. The least value of this integral is $0.58377\dots$$0.58377\dots$0.58377 dots0.58377 \ldots$0.58377\dots$. which occurs at $\mathrm{a}=0.91$$\mathrm{a}=0.91$a=0.91\mathrm{a}=0.91$\mathrm{a}=0.91$.

Consider a circle of $(x-a{)}^2+(y-b{)}^2=r^2$$(x-a{)}^2+(y-b{)}^2=r^2$(x-a)^(2)+(y-b)^(2)=r^(2)(x-a)^{2}+(y-b)^{2}=r^{2}$(x-a{)}^2+(y-b{)}^2=r^2$ passing through the points $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$ and $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$. Here, $a$$a$aa$a$ stands for the $x$$x$xx$x$-coordinate of the centre and $b$$b$bb$b$ refers to the $y$$y$yy$y$ coordinate of the centre; $r$$r$rr$r$ is the radius of the circle. Substituting the points in the equation we get

Equating the left sides, we have

so that $a=b$$a=b$a=ba=b$a=b$. Now the first equation gives $r^2=2a^2-2a+1$$r^2=2a^2-2a+1$r^(2)=2a^(2)-2a+1r^{2}=2 a^{2}-2 a+1$r^2=2a^2-2a+1$ and we have

We can rewrite it as follows:

Differentiating with respect to $x$$x$xx$x$, we get

so that

In the integral, all the $y$$y$yy$y$ terms should be represented in terms of $x$$x$xx$x$ so that we can integrate with respect to $x$$x$xx$x$. We express $y$$y$yy$y$ as

We use the negative sign because it represents the lower part of the circle, which is used to traverse from $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$ to $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$. As for the parabola case, we substitute $\frac{dy}{dx}$$\frac{dy}{dx}$(dy)/(dx)\frac{d y}{d x}$\frac{dy}{dx}$ and $y$$y$yy$y$ in equation (2.1). The least value of the integral is $0.58512\dots$$0.58512\dots$0.58512 dots0.58512 \ldots$0.58512\dots$ which appears when $a=1.3$$a=1.3$a=1.3a=1.3$a=1.3$. An important point to consider here is that $a\ge 1$$a\ge 1$a >= 1a \geq 1$a\ge 1$. This is due to the fact that when $a<1$$a<1$a < 1a<1< latex>$a<1$ the circle no longer uses its lower part to pass through those points. Instead, the graph will be represented by $\{y=\sqrt{2a^2-2a+1-(x-a{)}^2}+a$$\left\{y=\sqrt{2a^2-2a+1-(x-a{)}^2}+a\right.${y=sqrt(2a^(2)-2a+1-(x-a)^(2))+a:}\left\{y=\sqrt{2 a^{2}-2 a+1-(x-a)^{2}}+a\right.$\{y=\sqrt{2a^2-2a+1-(x-a{)}^2}+a$, which is the upper part of the circle. Figure 3.2 graphically demonstrates the above idea.

Figure 3.2: Curves represented graphically demonstrating why the negative sign was chosen for positive values of $a$$a$aa$a$

Now, for the case $a<1$$a<1$a < 1a<1< latex>$a<1$, we find the value of the integral by using the corresponding $y$$y$yy$y$ value 3.2 . The integral's value keeps on decreasing as x becomes more negative until it reaches a limiting value of $0.6388\dots$$0.6388\dots$0.6388 dots0.6388 \ldots$0.6388\dots$ seconds. This is same as the least time taken by a line as when $x\to -\mathrm{\infty }$$x\to -\mathrm{\infty }$x rarr-oox \rightarrow-\infty$x\to -\mathrm{\infty }$, the part of the circle joining $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$ and $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$ becomes increasingly flatter, eventually taking the shape of a line.

Consider an ellipse with x -axis as its major axis and $\mathrm{x},\mathrm{y}$$\mathrm{x},\mathrm{y}$x,y\mathrm{x}, \mathrm{y}$\mathrm{x},\mathrm{y}$ coordinate of its center as $x_0$$x_0$x_(0)x_{0}$x_0$, 1 . i.e. $\frac{{(x-x_0)}^2}{a^2}+\frac{(y-1{)}^2}{b^2}=1$$\frac{{\left(x-x_0\right)}^2}{a^2}+\frac{(y-1{)}^2}{b^2}=1$((x-x_(0))^(2))/(a^(2))+((y-1)^(2))/(b^(2))=1\frac{\left(x-x_{0}\right)^{2}}{a^{2}}+\frac{(y-1)^{2}}{b^{2}}=1$\frac{{(x-x_0)}^2}{a^2}+\frac{(y-1{)}^2}{b^2}=1$ passing through the points $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$ and $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$. Substituting the coordinates in the equation, we get

. Inserting the value of $x_0$$x_0$x_(0)x_{0}$x_0$ as $+a$$+a$+a+a$+a$ in the first equation, $\frac{(1-a{)}^2}{a^2}+\frac{1}{b^2}=1$$\frac{(1-a{)}^2}{a^2}+\frac{1}{b^2}=1$((1-a)^(2))/(a^(2))+(1)/(b^(2))=1\frac{(1-a)^{2}}{a^{2}}+\frac{1}{b^{2}}=1$\frac{(1-a{)}^2}{a^2}+\frac{1}{b^2}=1$ So, $\frac{1}{b^2}=1-\frac{(1-a{)}^2}{a^2}=\frac{-1+2a}{a^2}$$\frac{1}{b^2}=1-\frac{(1-a{)}^2}{a^2}=\frac{-1+2a}{a^2}$(1)/(b^(2))=1-((1-a)^(2))/(a^(2))=(-1+2a)/(a^(2))\frac{1}{b^{2}}=1-\frac{(1-a)^{2}}{a^{2}}=\frac{-1+2 a}{a^{2}}$\frac{1}{b^2}=1-\frac{(1-a{)}^2}{a^2}=\frac{-1+2a}{a^2}$ $\therefore b^2=\frac{a^2}{2a-1}$$\therefore b^2=\frac{a^2}{2a-1}$:.b^(2)=(a^(2))/(2a-1)\therefore b^{2}=\frac{a^{2}}{2 a-1}$\therefore b^2=\frac{a^2}{2a-1}$. So , the equation of the ellipse in this case is

We take the value of $x$$x$xx$x$ as $+a$$+a$+a+a$+a$ as the ellipse is defined only for positive $a$$a$aa$a$ values. . Differentiating the above equation with respect to $x,2(x-a)+2(2a-1)(y-1)\frac{dy}{dx}=0$$x,2(x-a)+2(2a-1)(y-1)\frac{dy}{dx}=0$x,2(x-a)+2(2a-1)(y-1)(dy)/(dx)=0x, 2(x-a)+2(2 a-1)(y-1) \frac{d y}{d x}=0$x,2(x-a)+2(2a-1)(y-1)\frac{dy}{dx}=0$

. The next step will be to isolate y. $(y-1{)}^2=\frac{a^2-(x-a{)}^2}{2a-1}$$(y-1{)}^2=\frac{a^2-(x-a{)}^2}{2a-1}$(y-1)^(2)=(a^(2)-(x-a)^(2))/(2a-1)(y-1)^{2}=\frac{a^{2}-(x-a)^{2}}{2 a-1}$(y-1{)}^2=\frac{a^2-(x-a{)}^2}{2a-1}$.

. Due to the same reason in the curves discussed, we choose the negative sign. Plugging the respective values of $\frac{dy}{dx}$$\frac{dy}{dx}$(dy)/(dx)\frac{d y}{d x}$\frac{dy}{dx}$ and $y$$y$yy$y$ into the time period equation 2.1 , we find that value of this integral varies inconsistently for values of $a$$a$aa$a$. However, gradually, with a huge increase in the value of a, there is a decreasing trend in the value of this integral. Hence, as $a\to \mathrm{\infty }$$a\to \mathrm{\infty }$a rarr ooa \rightarrow \infty$a\to \mathrm{\infty }$, the value of the integral reaches a limiting value. Let's go back to the original equation we derived 3.1. Expanding it, we obtain $x^2-2ax+a^2+(2a-1)(y-1{)}^2=a^2$$x^2-2ax+a^2+(2a-1)(y-1{)}^2=a^2$x^(2)-2ax+a^(2)+(2a-1)(y-1)^(2)=a^(2)x^{2}-2 a x+a^{2}+(2 a-1)(y-1)^{2}=a^{2}$x^2-2ax+a^2+(2a-1)(y-1{)}^2=a^2$. As our aim is to examine the curve near $x=0$$x=0$x=0x=0$x=0$ and $x=1,x^2$$x=1,x^2$x=1,x^(2)x=1, x^{2}$x=1,x^2$ can be neglected. $\therefore 2ax=(2a-1)(y-1{)}^2$$\therefore 2ax=(2a-1)(y-1{)}^2$:.2ax=(2a-1)(y-1)^(2)\therefore 2 a x=(2 a-1)(y-1)^{2}$\therefore 2ax=(2a-1)(y-1{)}^2$. Now, $2a=2a-1$$2a=2a-1$2a=2a-12 a=2 a-1$2a=2a-1$ as we are finding the equation in which $a\to \mathrm{\infty }$$a\to \mathrm{\infty }$a rarr ooa \rightarrow \infty$a\to \mathrm{\infty }$

We expressed an ellipse of type 3.1 with a significantly large value of $a$$a$aa$a$ in the form of a parabola. Expressing the equation 3.3 in terms of $x$$x$xx$x$,

$\cdot \frac{dy}{dx}=-\frac{1}{2\sqrt{x}}$$\cdot \frac{dy}{dx}=-\frac{1}{2\sqrt{x}}$*(dy)/(dx)=-(1)/(2sqrtx)\cdot \frac{d y}{d x}=-\frac{1}{2 \sqrt{x}}$\cdot \frac{dy}{dx}=-\frac{1}{2\sqrt{x}}$. Hence, the least time taken by an object to reach from $(1,0)$$(1,0)$(1,0)(1,0)$(1,0)$ to $(0,1)$$(0,1)$(0,1)(0,1)$(0,1)$ through $3.3=$$3.3=$3.3=3.3=$3.3=$
${\int }_0^1\sqrt{\frac{1+\frac{1}{4x}}{2\ast 9.8(1+\sqrt{x}-1)}}dx=0.58436\dots$${\int }_0^1 \sqrt{\frac{1+\frac{1}{4x}}{2\ast 9.8(1+\sqrt{x}-1)}}dx=0.58436\dots$int_(0)^(1)sqrt((1+(1)/(4x))/(2**9.8(1+sqrtx-1)))dx=0.58436 dots\int_{0}^{1} \sqrt{\frac{1+\frac{1}{4 x}}{2 * 9.8(1+\sqrt{x}-1)}} d x=0.58436 \ldots${\int }_0^1\sqrt{\frac{1+\frac{1}{4x}}{2\ast 9.8(1+\sqrt{x}-1)}}dx=0.58436\dots$. seconds.
Let us take the case of an ellipse with y axis as its major axis and center as $(1,y_0)$$\left(1,y_0\right)$(1,y_(0))\left(1, y_{0}\right)$(1,y_0)$. The equation of such an ellipse would be $\frac{{(y-y_0)}^2}{a^2}+\frac{(x-1{)}^2}{b^2}=1$$\frac{{\left(y-y_0\right)}^2}{a^2}+\frac{(x-1{)}^2}{b^2}=1$((y-y_(0))^(2))/(a^(2))+((x-1)^(2))/(b^(2))=1\frac{\left(y-y_{0}\right)^{2}}{a^{2}}+\frac{(x-1)^{2}}{b^{2}}=1$\frac{{(y-y_0)}^2}{a^2}+\frac{(x-1{)}^2}{b^2}=1$. Applying the former strategy,