Mathematics AA SL

Sample Internal Assessment

Table of content

Rationale

Aim

Background information

Introduction

Process of calculation

Conclusion

Bibliography

11 mins Read

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I was too young when i was first taken to a waterpark. I can't exactly recall my first experience but it was certainly quite good. I am sure of this because I would visit there quite often.

There was one waterpark in my town and it was not much far from my house. That was the most common place I would visit during my holidays. My parents would accompany me.

It was a thrilling place. I loved water. There were many rides too and I loved all rides but somehow would avoid the two water rides every time.

As I grew older, I started visiting the park with my friends. That was a different kind of fun all together. We would enjoy very much every time we visited, but one thing never changed and that was the fear of the water rides.

I did not fear all rides. I quite enjoyed the ones on the land and air. I feared only the water rides. Specifically, there were two such rides and of which was the scariest. It was the water slide. There was a long tunnel through which one was pushed out of it into the pool in a good projectile motion.

It was not the projection that frightened me. The thought that I might land out of the pool and not into it, scared me. I tried to overcome the fear but every time I would climb the stairs, I would give up.

Once my uncle and aunt visited us. We went to the water park in a weekend. I told him about my fear. He tried to convince me that it was safe by going down the water slide a multiple time, but all in vain.

He was a physics professor. On returning home in the evening, he talked to me about my fear.

He introduced me to projectile and asked me to study about it. While studying, I came to know that these slides are designed to ensure that people dive straight into the pool and not out of it. I realised that it was pointless to be afraid of the slide.

In this IA, i have tried and calculated the minimum range of pool required to be constructed right after a steep slide in a water park, provided that it offers maximum thrilling experience by creating maximum terminal velocity of projectile.

The main motive of this IA, is to study and analyze the Brachistochrone problem. Consequently, study on projectile or a two-dimensional motion is also another motive of this IA. Maximum range of any projectile equation should be found in association with the Brachistochrone problem.

In the proceedings of this IA, several concepts of Physics and Mathematics are extensively necessary. Variation calculus and differential calculus will be required to solve the derivations of this IA. Most importantly, one special form of differential equation called Euler Lagrange Equation is essential for this IA. It is a second order partial differential equation whose solution determines the solution for the given function being stationary. According to the Euler Lagrange’s equation, if **I** is defined as:

I = \(\displaystyle\int\) f (x, y, y) dx

Since, \(\frac{∂y}{∂x}=y\)

Then the function **I** has a stationary value if the following equation is satisfied:

\(\frac{∂f}{∂y}-\frac{d\Big(\frac{∂f}{∂y}\Big)}{dx}=0\)

There is one more special case in Euler Lagrange’s equation. If and only if, above mentioned function f(x,y,y) is not explicitly dependent on *x*, then, the equation can be modified as the following:

\(f -\) y\(\frac{∂f}{∂y}\) = c

where *c* is a constant. This special function is called Beltrami’s Identity.

Pythagorean equation is also important in this IA, which is denoted as:

a^{2 }+ b^{2 }= c^{2}

Where, a, b, and c are three sides of a right-angled triangle.

Regarding Physics, Law of Conservation of Energy will be required in this IA. Law of Conservation of Energy states that, “Energy can neither be created and nor be destroyed. It can only be converted from one form to another.”

Thus, if a body falls from point A to B, then the total energy of the body at point A and that of B will remain constant. Some amount of potential energy will get converted into kinetic energy until it reaches B, and by this way, Law of Conservation of Energy is maintained. Additionally, multiple and submultiple angles are also necessary. The formula that will be required in this IA is mentioned below:

2 \(\frac{\theta}{2}\) = 1 - cos cos θ

The parametric equation of a cycloid is necessary of this IA. Cycloid is a curve which is traced by a point on a circle when the circle moves along a straight line. The parametric equation of cycloid are as follows:

x = θ - sin sin θ

y = 1 - cos cos θ

In this IA, I will analyze the projectile motion of the body the reason being, the relationship between displacement and time will be necessary. The displacement time relationship in any dimension is given by:

s = ut + \(\frac{1}{2}\)at ^{2}

Where, *s* is the displacement done by the body, *u* is the initial velocity of the body, *a* is acceleration of the body, and *t* is time.

In this IA, I will find the range of pool required just after a water slide for a safe dive into the pool. In that case, the terminal velocity and the angle at which the body is leaving from the slide are very essential parameters. In modern water parks, slides are way taller and steep than that of present in the previous days. Nowadays, the water park authority’s prime requisite is the enhance the thrilling experience considering the safety of the individuals to an utmost extent. As we all know, more the speed of descent through the slide, more thrilling the experience becomes for the individual. As there is no initial velocity of the body in slides, the only force that the body gets is gravitational force which increases the speed of the body during the descent. Consequently, maximum speed can be calculated by calculating the time required to cover the entire path of the slide. Minimum the time be, more will be the velocity of descent. In order to find that, the shape of the slide is very essential to determine.

The initial and final positions of any water slide is fixed. Let, us consider, A is the starting point of the slide and B is the ending point of the slide. In this diagram, we have considered three different types of pathways which can be followed to cover up the distance but the type of pathway which will offer maximum velocity is required to be found.

Thus, the shape of the pathway that should be followed to have minimum time period required to cover the distance between the two points provided no additional force is applied to the body except the gravitational force should be analyzed in this IA.

In continuation with that, the range of the projectile will also be derived.

**Case 1: Determination of shape of Pathway that will offer minimum time or maximum speed:**

Let us consider the highest point of the slide in the co-ordinate system be (0,h) and the end point of the slide be (l, a) and the path to be followed be y = f(x). Therefore, total time taken T can be expressed as:

\(T={\displaystyle\int\limits^{B}_{A}}\) dt…………(1)

Moreover, we can write,

\(dt =\frac{ds}{v(x,y)}\)…………(2)

Here, *dt* is a small instant of time, *ds* is the distance covered in *dt* instant of time and *v* is a function of velocity which depends on both *x* and *y*.

From the Pythagorean Equation, we can write,

\(ds=\sqrt{dx^{2}+dy^{2}}\)

=> \(ds=\sqrt{dx^{2} (1+\frac{dy^{2}}{dx^{2}}})\)

=> \(ds=dx\sqrt{1+\frac{dy^{2}}{dx^{2}}}\)…………(3)

This is because, *ds* is a very small component of distance traversed from the initial position and *dx* is the small transition in X-axis and *dy* is the small transition in Y-Axis

Now, from the law of conservation of energy, we can write,

Potential Energy _{atA }= Kinetic Energy _{at any point }+ Potential Energy _{at any point}

∴mgh = \(\frac{1}{2}\) mv^{2} + mgy

=> mgh - mgy = \(\frac{1}{2}\)mv^{2}

=> v = \(\sqrt{2g(h-y)}\)…………(4)

From equation 1, 2, 3, and 4, we can write,

\(T={\displaystyle\int\limits^{B}_{A}}dt\)

=> T = \({\displaystyle\int\limits^{B}_{A}} \frac{\sqrt{1\ +\ \frac{dy^2}{dx^2}dx}}{\sqrt{2g(h\ -\ y)}}\)

=> T = \({\displaystyle\int\limits^{B}_{A}} \frac{\sqrt{1\ +\dot{y^2}\ dx}}{\sqrt{2g(h\ -\ y)}}\)

For minimum value of T, the function of T must be stationary but the function of which is inside the integral is not explicitly dependent on *x*. Thus, we can use Beltrami’s Identity for solve the function.

\( \frac{\sqrt{1\ +\dot{y^2}\ }}{\sqrt{2g(h\ -\ y)}}-y\frac{\cdot\ \partial\frac{\sqrt{1+\dot{y}^{2}}}{\sqrt{2g(h-y)}}}{\partial \dot{y}} = c\)

∴\(\frac{ \partial\frac{\sqrt{1+\dot{y}^2}}{\sqrt{2g(h-y)}}}{\partial \dot{y}} = \frac{\dot{y}}{\sqrt{1+\dot{y}^2}\sqrt{2g(h-y)}}\)

Therefore, putting the above value in the equation we get,

\( \frac{\sqrt{1\ +\dot{y^2}\ }}{\sqrt{2g(h\ -\ y)}}-\)\(\dot{y}\frac{\frac{{{}^{\dot{y}}}}{\sqrt{1+\dot{y}^2}\sqrt{2g(h-y)}}}{\partial \dot{y}} = c\)

\(=> \frac{\sqrt{1\ +\dot{y^2}\ }}{\sqrt{2g(h\ -\ y)}}-\frac{\dot{y}^{2}}{\sqrt{1+\dot{y}^2}\sqrt{2g(h-y)}}=c\)

=> \(\sqrt{1+\dot{y^2}}\sqrt{2g(h-y)}[\frac{\sqrt{1+\dot{y^2}}}{\sqrt{2g(h-y)}}-\frac{\dot{y^2}}{\sqrt{1+\dot{y^2}}\sqrt{2g(h-y)}}]\)

\(=c\sqrt{1+\dot{y}^2}\sqrt{2g(h-y)}\)

\(=> 1 + \dot{y}^2-\dot{y}^2=c\sqrt{1+\dot{y^2}}\sqrt{2g(h-y)}\)

\(=> 1=c\sqrt{1+\dot{y^2}}\sqrt{2g(h-y)}\)

Squaring both sides, we get,

2g(h - y) \(\biggl(1+\dot{y}^2\biggl)c=1\)

=> (h - y) \(\biggl(1+\dot{y}^2\biggl)=\frac{1}{2gc}\)

=> (h - y) \(\biggl(1+\dot{y}^2\biggl)=c_1\)

where, c_{1 }= \(\frac{1}{2gc}\)

\(=> h + \dot{hy}^2- y - \dot{yy}^{2} = c_{1}\)

\(=> \dot{y}^{2}=\frac{c_{1}-(h-y)}{h-y}\)

\(=>\dot{y}=\sqrt{\frac{c_{1}-(h-y)}{h-y}}\)

\(=>\dot{y}=\sqrt{\frac{c_{1}-(h-y)}{h-y}}\)

\(=>\frac{∂y}{∂x}=\sqrt{\frac{c_{1}-(h-y)}{h-y}}\)

\(=>∂x=\sqrt{\frac{{}h-y}{c_1-(h-y)}}∂y…………(5)\)

Let, y = h - c_{1 }\(\frac{\theta}{2}\)

∴\(\frac{dy}{dθ}=-\,c_{1}\) sin sin \(\frac{\theta}{2}\) cos cos \(\frac{\theta}{2}\)

=> dy = - c_{1} sin sin \(\frac{\theta}{2}\) cos cos \(\frac{\theta}{2}\)dθ

From equation (5), we can write,

\(∂x=\sqrt\frac{h-h+c_1\frac{\theta}{2}{}{}}{{c_1-(h-h+c_1\frac{\theta}{2}})}[-\,c_1\) sin sin \(\frac{\theta}{2}\) cos cos \(\frac{\theta}{2}\)dθ]

=> \(∂x=\sqrt{\frac{\frac{\theta}2}{1\ -\ \frac{\theta}2}}[-\)- c_{1} sin sin \(\frac{\theta}{2}\) cos cos \(\frac{\theta}{2}\)dθ]

=> \(∂x=\sqrt{\frac{\frac{\theta}2}{\\ \frac{\theta}2}}[-\) - c_{1} sin sin \(\frac{\theta}{2}\) cos cos \(\frac{\theta}{2}\)dθ]

=> \(∂x=\frac{sin\ sin\frac{\theta}{2}}{cos\ cos\frac{\theta}{2}}[-\) c_{1} sin sin \(\frac{\theta}{2}\) cos cos \(\frac{\theta}{2}\)dθ]

\(=> ∂x= \frac{\theta}{2}\)dθ

Integrating both sides,

\(\displaystyle \int∂x=\displaystyle \int\frac{\theta}{2}\)dθ

=> x = - c_{1} \(\displaystyle \int\frac{1-cos\,cos\,\theta}{2}\)dθ

=> x = k_{1 }(θ - sin sin θ) + k_{2}…………(6)

since, k_{1} = - \(\frac{c_1}{2}\)

Putting the value of c_{1} in the equation of y, we get,

y = h + 2k_{1}\(\frac{\theta}{2}\)

=> y = h + 2k_{1} \(\frac{1-\,cos\,cos\,\theta}{2}\)

=> y = h + k_{1 }(1 - cos θ) …………(7)

Equation 6 and 7 are the solution of the nature of the pathway which is actually the parametric equation of a cycloid. So, we can say that, for Time to be minimum, or speed to be maximum, the shape of the pathway must be cycloid in nature.

**Case 2: Determination of maximum range of pool:**

For the velocity to be maximum, it is evident that, the base of the slide (cycloid shaped) should make an angle equal to 0 to have maximum velocity. If the angle of the slide at the base is more than zero degrees, then, the pathway should look like Figure 2. As marked in the figure, the body will move in upward direction (positive Y axis) resulting in a decrease in the speed of descent as acceleration of the body will act in downward direction.

The terminal velocity of the projectile will be:

Potential Energy _{at A }= Kinetic Energy _{at B }+ Potential Energy_{ at B}

Potential Energy _{at A }= Kinetic Energy _{at B }+ Potential Energy _{at B}

=> mgh = \(\frac{1}{2}\,mv^2+mga\)

v = \(\sqrt{2g(h-a)}\)

Now, we have the find the maximum distance the body will cover (R as per Figure 3) after getting away from the slide. Let us consider the motion in two dimensions separately. The body will cover a certain distance in X direction as well as in Y direction.

**Parameters in X direction:**

Acceleration = 0 m/sec^{2}

Displacement = R metre

Initial Velocity = \(\sqrt{2g(h-a)}m/sec\)

Note: The initial velocity will act completely in X direction because the horizontal angle at the base of slide is 0.

**Parameters in Y direction:**

Acceleration = g = 9.81^{8} m/sec^{2}

Displacement = a metre

Initial Velocity = 0 m/ sec

Therefore, Let the time taken to cover the distance to get into the pool after getting away from the slide be T sec.

From the Displacement – Time relationship, we can write,

S_{y} = u_{y} t + \(\frac{1}{2}\,gt^2\)

=> a = \(\frac{1}{2}\,gt^2\)

=> T = \(\sqrt{2ag}\)

If the time taken to cover the height between the slide and the pool is \(\sqrt{2ag}\) sec, then the body will move in X direction for the same time period also.

From the Displacement – Time relationship, we can write,

S_{x} = u_{x} t + \(\frac{1}{2}\,gt^2\)

=> R = \(\sqrt{2g(h-a)}T\)

=> R =\(\sqrt{2g(h-a)}\sqrt{2ag}\)

=> R = \(2g\sqrt{a(h-a)}\)

In this IA, we have studied that to cover a certain distance in a plane (not in one direction) under the action of gravitational force only, the trajectory should not be a straight line. Straight line will provide the least distance between the two points but the velocity will be less unless and until the pathway is curved. The trajectory which will offer maximum velocity resulting the object cover the distance in minimum time is a locus of a Cycloid. In case of cycloid, the body initially falls downward gaining a considerable amount of speed due to the acceleration due to gravity and uses that speed to cover the remaining distance in the plane quickly. As a result, the shape of the slide in the water park should be cycloid to offer maximum descent speed of the individuals. Moreover, as the shape of the slide is cycloid, the base of the slide must be equal to zero degree to attain maximum terminal speed increasing the thrill of the ride. If the angle becomes greater than zero degree with the horizon, then the body will tend to slide upward in the slide resulting in a decrease in speed. Further-more, the range of distance a person make cover in air after getting slided, is:

R = 2g \(\sqrt{a(h-a)}\) metres

Where the g is the acceleration due to gravity, a is the height of the base of the slide from the pool and h is the topmost height of the slide.

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