1 Introduction 2. 2 Functions Introduction to Functions Common Notation... 3

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2 Contents 1 Introduction Functions.1 Introduction to Functions Common Notation Modelling Functions 4.1 Angular Radius and Solar Eclipses Centre of Mass Exponential Models Angles of Burning Matches Falling Through the Earth Epilogue 1 1

3 Chapter 1 Introduction It can be said that the entirety of physics is just the modelling of different phenomena as mathematical functions, so it is safe to say that being able to model functions is a very important skill to learn. Not to mention it is a skill that many elite universities look for in their physics and engineering students. To be a good physicist, or even a physicist at all, it is necessary to have the ability to express physical ideas using mathematics. Hence it is a topic that comes up very often in interviews and entrance exams. Schools only teach you how to use these mathematical functions, plugging in values and/or carrying out various algebraic operations on them. Almost never would you learn how to make these functions. You might as well never even encounter this task of modelling functions until your very first university entrance exam or interview. This is the exact motivation for me writing this book, to prevent the previously stated situation from happening and to acquaint students with this very fundamental skill. In this book my main method of teaching will be through solving different example problems.

4 Chapter Functions.1 Introduction to Functions Let s start off by defining what a function really is. This will not be a very rigorous definition; rather one to introduce you to the concept as smoothly as possible. A function s purpose is to link two numbers together. Inputting a number x into a function will give you a number f(x). Never will you get two different numbers from inputting the same x, as this goes against the definition of a function. It may help you to think of a function as a rule or better yet a machine. A streamlined machine that gives you a certain output after you input information into it. A function does not need to work for every number out there. Each number for which a function is defined for is said to lie in the function s domain.. Common Notation Now let s say your teacher wants to ask you to write down an equation relating the distance travelled by an object with the time passed since the object started to move, he or she may say write down the distance travelled by this object as a function of time. Assuming this object has a constant velocity, v, you would write D(t) = vt. In this problem it is quite obvious what the domain is, but at times it can be helpful to state it explicitly next to your functions like this D(t) = vt, t 0.

5 Chapter Modelling Functions More often than not you will need to utilise calculus to model a phenomenon, as it is usually easier to think about the rate of change of a phenomenon than the actual phenomenon. The reason for this is that differentiation is fundamentally a simplifying process, it removes any pesky constants that may be difficult to comprehend and decreases the powers of variables. So generally, the first order differential equation of a function is easier to write down than the function itself. Now that the significance of calculus in modelling functions has been stated, let us start off with a very simple example. One that doesn t even require calculus. If you are familiar with functions and their notation you may find it time saving to skip this first problem as it is very routine; though the way I present it may be routine to you.1 Angular Radius and Solar Eclipses Problem 1: Write the angular radius of a star as seen from a planet orbiting it as a function of the distance, x, from the surface of the planet to the surface of the star. Where the radius of the star is R. Solution 1: sin( θ) = R x+r = θ = arcsin( R x+r ) So the answer is θ(x) = arcsin( R x+r ). Now we can consider a few other things. A really insightful way to explore a function is to take appropriate limits of the variable. In physics, it is often very helpful to utilise limits while trying to understand different phenomena as extremes are easy to understand. This is also a spectacular way to boost intuition. 4

6 CHAPTER. MODELLING FUNCTIONS 5 lim x 0 θ arcsin( R R ) = θ π lim x θ arcsin(0) = θ 0 Meaning if the planet is right next to the star, an observer that is fireproof will find 180 degrees of their vision being occupied by the sun. And if the planet is really far away from the sun it will start to look like a speck in the sky, taking up almost no portion of the observer s field of vision. Yes, in this case the result was quite boring, nothing out of the ordinary, but at times it can be very tricky to understand a function and limits really do help.. Centre of Mass Problem : There is a measuring cylinder of length L, with a base of negligible mass and thickness. Water is poured into the measuring cylinder to a height H. Define the distance of the centre of mass from the base as a function of H. The thickness of the glass of the measuring cylinder is negligible. Solution : It is often of great help to draw the graph of a phenomenon you are looking express mathematically. Doing this gives you a good feel for the problem. So let s start off by thinking about some points that we are sure about: 1) When H = 0 the centre of mass should be at L. )The centre of mass should be back at L when H=L. )The graph should decrease then start to increase at one point, depending on the radius of the cylinder and its weight. Now let s plot an estimate graph. [Turn to next page]

7 CHAPTER. MODELLING FUNCTIONS 6 The next step is to start thinking of a formula to determine centre of mass, of any object, in one dimension. We know that for a uniform object of height x, the centre of mass lies at a height of x. (Usually the word uniform in this context means that the object s mass is evenly distributed per unit volume or area.) Now by combining two objects, of height x and y, we form a new object. At what height will this new object s centre of mass lie? It will depend on the centres of mass of each of the objects separately. We cannot simply add both of the centres of mass, x and y, and divide by two to get the average. This would be assuming that each of the objects centre of mass matters equally, this may not be true as the two objects may be of different masses. This means we need to form a sort of weighting system, which gives due significance to each of the centres of mass. This needs to be done with respect to the fraction of total mass each of the objects occupies. So now we may simply add the centres of mass, x and y, while assigning to them a coefficient which allows for weighting. Overall centre of mass= x. Mass x + y. Mass y Mass T Mass T Above is an equation of great specificity. The next step is called generalisation, the greatest source of pleasure for any mathematician. Y = (x 1m 1 +x m +...+x n m n ) m T Where Y is the centre of mass of an object comprising of n objects with centres of mass x 1, x,..., x n and of masses m 1, m,..., m n and where m T = m 1, m,..., m n. This formula can also be used in problems with more than one dimension, but I will leave the task of figuring how to do this to you. Now back to Problem ; to use our freshly formulated formula we will first need to express

8 CHAPTER. MODELLING FUNCTIONS 7 the masses of the cylinder and water. M W = p.v V = πr H = M W = pπr H And, Centre of mass of water of height H = H Centre of mass of measuring cylinder of height L = L So, f(h) = M W H +M C L f(h) = pπr H f(h) = pπr H M t +M C L M t +M C L pπr H+M C Where M T is the total mass, M C is the mass of the measuring cylinder, M W is the mass of the water, p is the density of water, and r is the radius of the measuring cylinder.. Exponential Models Problem : There are two tanks A and B; A is evacuated and the other is at a very high pressure. Tank A is also of far less volume than the pressurised tank. A is then connected to B at time t=0. Express the pressure of A as a function of time. Solution : This is a classic problem, very similar to the other classic problem which asks whether tea will cool to a greater extent with cold milk added before or after letting the tea sit out. We will answer the tea problem after we are done with Problem. Again, the best way to approach this problem is to first and foremost draw a graph.

9 CHAPTER. MODELLING FUNCTIONS 8 Just by first glance we can see that the rate of change of our function, f(t), decreases as it approaches the value of the pressure inside tank B, P B. Using this piece of information or just intuition alone you may deduce that the rate of change of f(t) is directly proportional to the difference between f(t) and P B. For ease of notation we will refer to f(t) as y for now. Also note that the purpose of mentioning that the pressure is very high and that tank A is much smaller than B is so that we may accurately approximate P B as a constant throughout. dy dt = k(p B y) 1 dy = kdt P B y 1 P B y dy = kdt ln (P B y) = kt + c

10 CHAPTER. MODELLING FUNCTIONS 9 P B y = e (kt+c) y = P B 1 e (kt+c) f(t) = P B 1, 0 t e (kt+c) Marvelous! That was fun, wasn t it? We end up with a concise, and powerfully elegant equation. I recommend you play around with the significant limits of t and see what values of f(t) you end up with. Now about that tea problem. Let s begin by properly defining the problem There are two cups that are full of tea, let us call them cup A and cup B. Cup A is left to cool as is and cup B is introduced to some cold milk before it is left to cool along with cup A. After the cooling process is over, cold milk is poured into cup A. Which teacup ends up being colder at the end? To fully utilise the time during which the cups are left to cool, the cup would have to have maximum difference in temperature from the environment, because this process works under the same principle as the tanks exchanging pressure. The rate of change of temperature is directly proportional to the difference between the fourth powers of the temperatures of the environment and the tea. The cup which is hotter during the cooling phase is cup A and so it will end up being the coldest after this whole exercise. Problem 4: A rock of infinite density is falling, under gravity, through a fluid which has a property such that it exerts a frictional force on the rock directly proportional to its velocity, v. Gravitational acceleration is g and the rock is of mass m. Find the velocity of the rock as a function of time. Solution 4: The purpose of stating that the rock is of infinite density is to clarify that there is no need to take buoyancy into account. friction v friction = kv force = gm kv ma = gm kv m dv dt = gm kv dv gm kv = dt m ln(mg kv) k = t m + c mg kv = 1 e k( m t +c ) v(t) = 1 k (mg 1 e k( m t +c)) Now comes the fun part, finding what speed the rock approaches as t becomes very large. Due to the task s triviality, I will leave it as an exercise to the reader.

11 CHAPTER. MODELLING FUNCTIONS 10.4 Angles of Burning Matches Problem 5: Define the time taken to burn a match of length L, as a function of its angle θ to the vertical. Solution 5: Sometimes it is easier to start off a problem by taking what is supposed to be the variable as a constant. So you re able to think about how the other values, physically and mathematically, relate to each other. Let s think about the rate of change of length L, or in other words the rate of burning of the match. A straight match, with an angle θ = 0 to the vertical, will have a constant rate of burning, c. Something like this: dl = c. dt Now if you tilt the match by an angle θ it will start to add onto that c. We need a function that increases from 0 to a maximum value, as the angle θ goes from 0 to π. The only function that would do such a thing would be sine but that increases from 0 to a maximum as θ goes from instead of sinθ. Our equation is now: 0 to π, so we will tweak it slightly by making it sin θ dl = ksin θ + c. dt Next we integrate, dl = ksin θ + c L = t(ksin θ + c). Now by rearranging, L t = (ksin θ +c). L t(θ) = (ksin θ +c). Come to think of it, this does seem slightly obvious now that it is in front of us. dl was the rate dt of change of length L with respect to time, which can be interpreted as velocity; and so ksin θ +c is our velocity, L is our distance, and t is our time. We know that distance divided by velocity gives time, so we can come to the our conclusion this way as well. This is an example of an instance where two, completely different, phenomena have a symmetry to them allowing the techniques of the already explored phenomenon to be applied onto the other less explored phenomenon. People have a perception that all physics is inspired through mathematical concepts, this although is true for most of the time, it is not a hard fact. Take for example when mathematicians at Trinity College s Mathematical Society were tackling the problem of squared squares, which fell into a very obscure subcategory of mathematics; they used the mathematics of circuits, an already very developed field, to solve this problem. At times it is very useful to find symmetries, relating to more well known and established fields of mathematics, in obscure problems. This way you can use the already developed powerful tools instead of trying to develop new ones.

12 CHAPTER. MODELLING FUNCTIONS 11.5 Falling Through the Earth Problem 6: A straight hole is drilled through the diameter of the Earth. You can assume the Earth to be a homogeneous perfect sphere. A particle is released from one end of the hole at t=0. Express the acceleration of the particle falling through the hole as a function of its distance from the centre, r, and in terms of the radius, R of the earth and gravitational acceleration on the surface g. Then write r as a function of time, hence also arrive upon velocity and acceleration as functions of time. Solution 6: As you fall through the Earth you will have to look at the earth as two separate objects, one being the inner sphere and one being the outer sphere. This makes the problem seem overly complicated, but if you think about it for a second you can eliminate one source of trouble; that is the mass of the outer sphere. Since all the acceleration provided by it will cancel out. Let s look into the problem and figure out why this is so. So let s start off by examining the most simple instance, one where you are exactly at the centre of the earth. Now all of the mass of the earth above you will cancel out the mass below you, hence giving a resultant acceleration of zero. Now move on to a circumstance where you are closer to one side of the Earth. Now from the closer side you have lesser mass but you are obviously closer to it, and about the far away side, you are further from it but it also has more mass. So again resultant acceleration is zero. You can now come to the conclusion that the net acceleration produced is always zero. a = GM r M = 4πr p, where p is density = a = 4πr G( p ) r When r is the radius of the earth, g = Rearranging to p = g 4GRπ 4πr G( p ) 4πR G( p ) R. Inputting p into a = r We get the answer a(r) = rg R Now moving on to the second part. We know that the distance from the centre is a looping value as a result of the first part of the problem. We also unknowingly proved that the motion of the particle is simple harmonic. As a r, and since acceleration is always in the opposite direction of the displacement from the centre of the Earth, we may use SHM equations to present r, velocity and acceleration as functions of time. w R = R Rg, where w is angular velocity w = g R r(t) = Rcos( g R t) v(t) = g R Rsin( g R t) a(t) = g R Rcos( g R t)

13 Chapter 4 Epilogue That is all that I had for you for now; I contemplated adding more problems and content to this poor excuse of a book but considering just these 1 pages took me several hours to write I will stop here, at least for now. I might post an updated version soon, though. On an ending note, if you want some guidance on what to do after reading this and how to continue becoming a better physicist, my best advise is to do physics, think physics, and live physics. 1