Finding the length of a section of a curve

This morning I was set a new problem, which I feel I will be working on for a fair while - what is the volume of a gas-filled pillowcase shape (two rectangles stuck together around the edges, and inflated to the maximum possible volume)? I already have my suspicions as to how the 3-D surface of a half-pillowcase shape could be modelled as a function - a 2-D function needs to be found which, for a given length of the curve (the length of one of the rectangles making up the pillowcase), maximises the area under it. Using a bit of intuition, this function should rise and fall very quickly at its ends, for example y = |√(x+5)|, and have a relatively high height inbetweentimes. This leads me to common naturally-occurring functions such as hyperbolae, catenaries or even quadratics! 

The first key bit of information needed is a method for calculating the length of a section of a curve. I have done this today, generalising the method up to an integral about halfway down; from there I took the quadratic example forward, since it turns out pretty neatly:

Calculus in a simple resistance problem

The problem goes as follows: a resistor, resistance R, is connected in series with a battery, emf ε and internal resistance r; find the maximum power dissipation in the resistor R, and the resistance that results in this output.

Although it is quite clear how this can be solved with algebra, some thought about the theory behind Ohm's Law and Kirchoff's Second Law is required to set the answer in some context: since the power dissipation in the resistor is equal to I²R, one would think that raising the current is more influential on the output than raising the resistance. However, since the current drawn from the battery is intrinsically linked to the total resistance of the circuit, there is a delicate balance to be struck.
The nature of this balance is difficult to ascertain without calculation though, so without further ado:

The result is that the power dissipated across the resistor is a maximum (assumed from the context of the function whose stationary point is found, since only one positive-R stationary point exists) when R = r. This nicely helps to explain why a short-circuited battery is likely to heat up, set fire or explode - the short-circuit has a very low resistance, so overall a massive current is drawn which causes the wire to heat up. The wire will heat up the most when the resistance of the wire matches the internal resistance of the battery. The overall implication is that, if one was designing an electric resistance heater to emit the maximum heat output for any given emf, the length of wire coiled up should, via the resistivity equation, reflect the internal resistance of the power supply.

However, a separate issue is maximising the efficiency of the circuit in terms of energy - for environmental and financial reasons it is always of high priority to do this in power-hungry domestic appliances, and the heater is no exception. Since the efficiency of the circuit is the ratio of power dissipation in the battery to the total power dissipation of the circuit, maximum efficiency would be achieved when the resistance of R is infinitely high. The issue with this is, just like trying to connect a 12V power pack across a piece of plastic, you cannot expect any power to be generated at all - this is because the high resistance results in a tiny current being drawn from the battery and in any case, the power output would be highly suboptimal according to the result I derived above. The moral of the story for the heater is that the most important factor here is to cut the internal resistance of the power supply as much as possible.

If the heater is being connected to the mains power, this opens an entirely new can of worms, namely the definition of the National Grid's internal resistance. This is a combination of the cumulative resistance of electrical cabling across the country, inefficiencies in combustion of fuels in power stations, friction in turbine halls, and delayed responses in adjusting power distribution across the country following surges and drops in energy demand, amongst other more subtle factors. Nevertheless more simple domestic factors to consider are reducing usage of extension leads where possible, and keeping the heater cable well insulated so that its resistance doesn't itself rise with time!

A curiosity - the reciprocal triangle

I was absent-mindedly solving simple calculus problems on Brilliant.org this afternoon when I stumbled across an interesting one about the reciprocal function. The question asked the user to prove that the area enclosed by the coordinate axes and the tangent to the curve at any point on the curve is constant, and to find this area. This is a basic proof of this unusual property:

The logical extension of this observation is to see what happens when the function is manipulated. First, imagine a reciprocal function of order n:

This sets nicely into context how unique the basic reciprocal function is - the k in the numerator will only be cancelled when n = 1.

Next, consider a function with a linear stretch factor a in the y-direction (which, due to the symmetry of the function, is the same as a stretch factor ¹/_a in the x-direction):

This is of mild interest too - the original problem took a = 1, such that the area of the triangle was 2.

Finally, consider a function translated by a units in the positive x-direction, and b units in the positive y-direction:

Clearly, as the sketch demonstrates, the translation of the curve disrupts the symmetry of the curve about the x and y axes, such that there is no longer a triangle of constant area. Trivially, it could easily be proven that there is a region of constant area, with the area under the tangent within the limits x > a and y > b, since these asymptotes effectively become the new x and y axes within the translated reference frame of the new curve. Furthermore it can be seen that the constant-area is reinstated when a = b = 0, such that k cancels out in the numerator and denominator to leave the previous area of 2.

The hanging cable - the intrepid part 2

I mentioned at the end of my post yesterday (see "The Hanging Cable") that I might be able to optimise the maximum height of the cable by adding a taper. The aim of today, aside from the fairly mundane past papers I had scheduled, was to model this as best I could. The result of the monster integral, which I have verified using Wolfram Alpha, is a complicated function in H_max which I don't believe can be solved by conventional algebraic means. Nevertheless, the working is fairly satisfying, if I have made it clear at all, and there are several interesting qualitative conclusions that can be drawn from the model.

It is important to note that I have not repeated the first-principles derivation of this method below. For this, see here.

The function at the bottom is equated to zero to indicate how the required root is to be found. I have entered it into Desmos graphing calculator here, where it is easy to read off the x-intercept as the value of H_max for the material. The sample data on it is for 2800 maraging steel; when a taper angle of 1° ( ≈ 0.0175^c ) is selected, a clearly too large but a good arbitrary starting figure, the graph of y = f(H_max) looks like this:

This shows that the theoretical maximum length of cable, fitting this specification, is 7288km, a pretty awesome distance. However, for a little perspective, let's consider how fat the cable will be at the very top, where it is hanging from its superlatively streadfast loop:

0.5d = Htanθ = 7.288·10⁶ · tan(0.0175^c) = 127553.0213 ≈ 128km (3sf)

If this doesn't seem unwieldy and inappropriate enough, next consider the volume and mass of this so-called 'cable':

V = ¹/₃πH_max³tan²θ = ¹/₃π · (7.288·10⁶)³ · tan²(0.0175^c) = 1.241705149·10¹⁷ ≈ 1.24·10¹⁷m³ (3sf)
m = ρV = 1.24·10¹⁷ · 8100 = 1.005781171·10²¹ ≈ 1.01·10²¹kg (3sf)

This cable therefore makes up 0.0114% of the Earth's volume, enough to excavate the grand canyon nearly 30000 times (using the common estimate of 5.45 trillion cubic yards, where 1 yard = 0.9144 metres), and 0.0169% of the Earth's mass! Now, for the worst part, consider the estimated cost of this object, with the assumption that all the forges in the world could produce enough maraging steel between them. It is difficult to find a precise figure for the price of any particular variety, but in general the world steel price is about $60 per tonne, or 6 cents per kilogram:

Cost = 0.06·m ≈ $(6.06·10¹⁹)

This cost, just over 60 quintillion US dollars or 4.21 quintillion GBP, would be enough to pay off the £1.56 trillion UK deficit nearly 27 million times!

In essence what I am trying to illustrate is that a 1 degree taper is a truly ridiculous idea for extending the snapping length of the cable, even though it does a very good job of doing so. Let's see what happens as the angle is changed...

Since I am using graphical means to solve f(H_max) = 0, I can see no way to easily find the equation for a graph of H_max against taper angle. However some empirical experimentation with Desmos shows that the smaller the taper angle, the greater the value of H_max. This is rather counterintuitive - the smaller the taper angle, the closer this model should come to the model established in the previous article.

Nevertheless I reckon I have found a possible flaw: since the cable starts with a radius of zero, having such a small taper means it is pretty much non-existent for the first few kilometres. I see this as analogous to the critical assembly of a system - just like ants can support weights disproportionate to their own mass due to their tiny size, having such a microscopic cable allows very disproportionate behaviours to occur in comparison with the macroscopic world. This idea doesn't entirely explain away the fact that the previous model completely decoupled the value of H_max from the diameter of the cable, but I'm working on that bit! Perhaps the cable could instead be modelled as a frustum, with a significant radius at the bottom?

The hanging cable

A interesting problem was put forward by my physics teacher on Friday, relating to material properties - what is the longest possible constant-diameter cable, from a material with a given yield stress and density, that can be hung vertically in the Earth's gravitational field (with its end just touching the ground) without snapping?

In essence the problem is very simple. The point of maximum stress is right at the top of the cable, where it is hanging from some kind of steadfast loop capable of supporting the material in its entirety; at this point the stress is equal to the weight of the cable divided by the cross-sectional area. However when the materials used are strong enough to last kilometres into the air, the changing value of g starts to become a factor in the weight force experienced by the maximum stress point. I did some scribbling today, and came up with a little proof of a nice formula that takes into account the changes in g, with the use of a definite integral as the infinite sum of a series in its limit:

The most important insight to be gleaned from the final formula for the maximum height of cable is that the cross-sectional area of the cable is irrelevant to the height achieved, as long as it remains constant. It depends only on the radius of the Earth (constant), the universal gravitational constant (constant), the mass of the Earth (constant), the density of the material (constant for a given material) and the breaking stress of the material (constant for a given material) - note that it is assumed yield stress = breaking stress, since it is unnecessarily complicated to consider the plastic properties of the cable between yield and fracture.

An example of the use of the formula is to consider a material - steel is a popular choice for high-stress cabling. Wikipedia tells me the breaking stress of a certain type, 2800 maraging steel, is 2617MPa and the density is 8100g per cubic metre. The formula returns a H_max value of 33072.8304159 ≈ 33.0km. If the uniformity of the Earth's gravitational field was assumed, the H_max would be significantly different: 

Cross-sectional area of cable = 0.25πd²
Volume of cable = 0.25Hπd²
Mass of cable = 0.25Hρπd²
Tension at top of cable = 0.25Hρgπd²
Stress at top of cable = Hρg
H_max = σ_frac/ρg = 32934.3954896 ≈ 32.9km

There is an absolute difference of 138.43493 ≈ 138m here, which converts to a surprisingly sizeable percentage error of 0.419%!

Now it has been established how important the non-uniformity of the Earth's gravitational field is in such a matter, one must now wonder how the shape of the cable could maximise the value of H_max - my hypothesis is that having a slight linear taper on the cable such that it is fatter at the top than at the bottom, with the cross-sectional area scale factor per metre climbed matching the scale-factor for decrease in gravitational field strength per metre climbed, would be optimal. This is because having less mass at the bottom is essential for marginal gains, where the gravitational pull is strongest, and a taper to this specification should account logically for the inverse-square nature of Newtonian gravity.

Steel was a fairly old-school example for testing the value of H_max: a huge focus of the material science field is the applications of carbon nanotubes, tailor-made materials based on graphene and its chemical derivatives. Wikipedia informs me that one particular type, Armchair Single-Walled NanoTubes, has a breaking stress of 126.2GPa, nearly 50 times the strength of 2800 maraging steel. Ignoring the considerable strain produced by such a breaking stress (0.231), an oversight it seems, the H_max values can be calculated with the calculus and simplistic methods, given an approximate density value for Armchair SWNT of 1660kg per cubic metre (from here):

Calculus H_max = -35976055.5613
Simplistic H_max = 7749653.04644

This is pretty stunning as a result. The nanotubes are so strong that the calculus formula breaks down to form a negative result - this implies the cable can stretch infinitely out into space (assuming it is not affected by the gravitational fields of other bodies, which isn't technically true of course), effectively escaping the effects of the Earth's pull, without fracture. I suppose the combination of the cable being nearly 5 times less dense and 50 times stronger means there is an effective relative increase in potential height of 250 times. 

Hyperbolae 2 - introducing the third dimension

This is a summary of today's musings. In essence I have done two fairly basic things that marginally extend the 2-dimensional hyperbola: I have taken the infinitesimal-lamina principle behind solid revolution and used it to model how the function will rotate around the x-axis. Then I have returned to the basic calculus of revolution to derive an expression for the volume of a lobe, given values of the stretch-coefficients a and b, and the limit of integration c. Enjoy!

Investigating hyperbola-type implicit functions 1 - asymptotes

"Which way round was it again?"
During one of my FP1 past-papers, of which I have been trawling through the entire lot, it struck me that there was one part of the course that was very much a matter of "here's the result, learn it". Hyperbolas are only touched-upon in the AS further maths specification, and for me that makes the given theory more difficult to engage with. Hence, I found in one question that I could not remember the generalisation for the asymptotic gradient of a hyperbola. Well, with two options in mind, I thought asking for help somewhat non-proactive so I decided to see if I could work it out with a little bit of limit notation.

 Having achieved this, feeling extremely pleased with my elevated understanding of the function despite having not blasted through the 3 or 4 AS level questions I would have otherwise spent the time with, I began to consider what would happen if the same general form of hyperbola was used with higher-order indices. My findings are on the A4 scans below.

The second page deals with more trivial cases, where x and y are raised to different powers - there produce fairly uninteresting polynomial-type curves, which could effortlessly be rearranged to express the first-quadrant branches explicitly. Nevertheless an implicit-differentiation method for proving the limits of the gradients is still fairly stimulating, and the result is pretty nice, if quite obvious.