Consider the pendulum with pivot at \((0,r)\), where \(r\) is the length of the pendulum. The position of the end of the pendulum is given by \((x,y) = (r \sin \theta,rr \cos \theta )\). The law of conservation of energy states that if there is no friction or lose of energy, then the potential energy plus the kinetic energy of a system must be conserved. The potential energy \(P\) of a pendulum is proportional to how high it is (\(y\)), the mass of the pendulum (\(m\)), and the acceleration of gravity (\(g\)): \(P=mgy\). The kinetic energy \(K\) of a pendulum is one half of the mass times velocity squared: \(K = m v^2 /2\), where \(v\) is the velocity. The total energy \[E = P + K = mg y + m v^2/2.\] The velocity of the end of the pendulum, \[v = (\dot{x},\dot{y}) = (r \dot{\theta} \cos \theta,r \dot{\theta} \sin \theta )\] \[v^2 = v \cdot v = r^2 \dot{\theta}^2 \cos^2 \theta +r^2 \dot{\theta}^2 \sin^2 \theta = r^2 \dot{\theta}^2\] When we substitute, we find the energy \[E = m g (rr \cos \theta) + m r^2 \dot{\theta}^2/2 .\] Since the energy is conserved over time, the total energy never changes, and it's time derivative must vanish. Differentiating, \[\dot{E} = 0 = m g \dot{\theta} r \sin \theta + m r^2 \dot{\theta} \ddot{\theta}\] Factoring the righthand side, \[0 = m r \dot{\theta} \left( g \sin \theta + r \ddot{\theta} \right)\] Since the angular velocity is not zero, we must have \[\ddot{\theta}= \frac{g}{r} \sin \theta .\]
This derivation requires (1) faith in the conservation of ``energy'' and (2) an understanding of what potential energy and kinetic energy are.
Newton's second law of motion says the acceleration at any time is equal to the force per unit mass applied to it (\(\vec{a} = \vec{F}/m\)), or \[m \frac{d^2\vec{p}}{dt^2} = \sum_i F_i.\] Here, all the variables \(\vec{p}\), \(\vec{a}\), and \(\vec{F}\) are vectors.
Suppose we apply this to the pendulum. Looking at the diagram above, there appear to be two forces acting on the pendulum. There is a force from gravity, \(F_g\), pulling the pendulum straight down, and there is a force from the suspending rod \(F_r\) that always pulls toward the pivot. We do not know how large \(F_r\) is, but we know it should be strong enough to keep the end of the pendulum at constant distance \(r\) from the pivot of the pendulum. Call this unknown force \(k\). The total force is represented by \[\begin{gather*} \vec{F} = \vec{F}_g + \vec{F}_r = \begin{pmatrix} 0 \\mg \end{pmatrix} + k \begin{pmatrix} \sin \theta \\ \cos \theta \end{pmatrix} \end{gather*}\] To make sure the length of the pendulum stays constant, we would need \(\vec{F}\) to be perpendicular to the pendulum. Mathematically, two vectors are perpendicular if and only if their dotproducts are \(0\), so we must have \[\vec{F} \cdot (\sin \theta, \cos \theta ) = 0,\] \[k \sin^2 \theta + k \cos^2 \theta  mg \cos \theta = 0,\] \[k = mg \cos \theta.\] Plugging this back into our formulation of the forces \(\vec{F}\), \[\begin{gather*} \vec{F} = m g \begin{pmatrix} \cos \theta \sin \theta \\ 1  \cos \theta \cos \theta \end{pmatrix} = m g \begin{pmatrix} \cos \theta \sin \theta \\ \sin^2 \theta \end{pmatrix} \end{gather*}\]
The force is represented in polar coordinates, so we should also express the acceleration in polar coordinates. In polar coordinates, \[\begin{gather*} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r \sin \theta \\ r  r \cos \theta \end{pmatrix} \end{gather*}\] Differentiating with respect to time and observing that the radius \(r\) is constant, \[\begin{gather*} \begin{pmatrix} \ddot{x} \\ \ddot{y} \end{pmatrix} = \begin{pmatrix}  \dot{\theta}^2 r \sin \theta + \ddot{\theta} r \cos \theta \\ \dot{\theta}^2 r \cos \theta + \ddot{\theta} r \sin \theta \end{pmatrix} \end{gather*}\]
Substituting into Newton's second law now, the pendulum's motion should be governed by the equation \(F/m = a\) is given by \[\begin{gather*} g \begin{pmatrix} \cos \theta \sin \theta \\ \sin^2 \theta \end{pmatrix} = \begin{pmatrix}  \dot{\theta}^2 r \sin \theta + \ddot{\theta} r \cos \theta \\ \dot{\theta}^2 r \cos \theta + \ddot{\theta} r \sin \theta \end{pmatrix} \\ \frac{g}{r} \begin{pmatrix} \sin \theta \\ \sin \theta \end{pmatrix} = \begin{pmatrix} \ddot{\theta}  \dot{\theta}^2 \frac{\sin \theta}{\cos \theta} \\ \ddot{\theta} + \dot{\theta}^2 \frac{\cos \theta}{\sin \theta} \end{pmatrix} \end{gather*}\] Now, however, we have reached a problem. The same kind of problem that natural scientists encountered before 1659.
Mathematically, we are trying to solve for one unknown function \(\theta(t)\), the angle of the pendulum over time. However, we have two different equations to solve  usually, two equations can not be solved simultaneously for one unknown. We might be able to resolve this if we could show both equations were really the same, but in this case they are not. Subtracting one from the other, we find that we must have \[\dot{\theta}^2 \left( \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \right) = 0.\] The only solutions are pendulums that are at a constant angle (any constant angle) and don't move (\(\dot{\theta} = 0\)). Obviously, this does not agree with reality  our equations are predicting pendulums can not move, but clearly they actually move rather regularly. This stumped people for a while, but the the problem was eventually resolved by Christian Huygens.
Physically, what did our derivation yield? Well, the force exhurted on the pendulum must provide all the accelerations needed. There appear to be two components in the acceleration: one is the angular acceleration, with \(\ddot{\theta}\). The other depends on the angular velocity, \(\dot{\theta}\). But our model of the forces does not depend in any way on the speed of the pendulum. Which is the correct answer? Should there be a component of force that depends on the speed of the pendulum rather than gravity?
Thought Experiments
It takes only a second of spinning a pendulum to convince us that it's harder to hold on to the pendulum the faster it moves.
If we take the pendulum into space, there would be no significant gravity (\(g=0\)), so the left hand side of our formula would disappear. Would the pendulum rod exert a force on the pendulum? Yes. There must still have to be force that keeps the pendulum moving in a circle rather than a straight line.
Thus, the forces SHOULD depend on the speed of the pendulum also. Christian Hugyens formalized this idea around 1660 by creating the idea of centrifugal force  the force exerted on an object necessary to keep it in a circular orbit when no other forces are being applied to the system. This force \(\vec{F} _ {cf}\) was proportional to the math and the square of the velocity, but inversely proportional to the radial distance : \(\vec{F} _ {cf} = m v^2 / r\). In vectorform for our pendulum equation, the centrifugal force \[\vec{F} _ {cf} = m r \dot{\theta}^2 \begin{pmatrix} \sin \theta \\ \cos \theta \end{pmatrix}\] The total force on the pendulum is thus \[\begin{gather*} \vec{F} = \vec{F}_g + \vec{F}_r + \vec{F}_{cf}= \begin{pmatrix} 0 \\mg \end{pmatrix} + m g \cos \theta \begin{pmatrix} \sin \theta \\ \cos \theta \end{pmatrix} + m r \dot{\theta}^2 \begin{pmatrix} \sin \theta \\ \cos \theta \end{pmatrix} \end{gather*}\] Substituting into Newton's law, \[m \begin{pmatrix}  \dot{\theta}^2 r \sin \theta + \ddot{\theta} r \cos \theta \\ \dot{\theta}^2 r \cos \theta + \ddot{\theta} r \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\mg \end{pmatrix} + m g \cos \theta \begin{pmatrix} \sin \theta \\ \cos \theta \end{pmatrix} + m r \dot{\theta}^2 \begin{pmatrix} \sin \theta \\ \cos \theta \end{pmatrix} \\ \begin{pmatrix} \ddot{\theta} r \cos \theta \\ \ddot{\theta} r \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\g \end{pmatrix} + g \cos \theta \begin{pmatrix} \sin \theta \\ \cos \theta \end{pmatrix}\] After simplifying either of thes two equations, we get our standard result \[\ddot{\theta} =  \frac{g}{r} \sin \theta.\]
Galileo proposed that the shape of a hanging curve was a parabola, at least for small deflections. This is approximately true, but not general. The classic solution arises from a problem proposed by Jakob Bernoulli from 1691. Robert Hooke figured out that if you flip a catenary over, it will be the ``perfect'' shape for an archway, as the forces from the weight of the stone itself flow along the curve of the arch.
Consider a cable hanging between two supports that is fully at rest. At each point along the cable, there is no acceleration, so the forces of gravity and tension in the cable must be perfectly balanced. Let \(y(x)\) be the height of the cable at each location \(x\), and that the cable has a uniform thickness and density \(\rho\) everywhere. At each point \(x\), there is a little piece of cable of length \(2 \mathrm{d} x\), \(\mathrm{d} x\) to the left and \(\mathrm{d} x\) to the right. The forces are tension to the left and right, and gravity pulling down on that part of the cable. The forces from the tension in the cable must point approximately along the cable. Near the point \(x\), \[y(x+\mathrm{d} x) = y(x) + \frac{\mathrm{d} y}{\mathrm{d} x} \mathrm{d} x + \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \frac{1}{2} \mathrm{d} x^2 + \ldots \quad \text{and} \quad y(x\mathrm{d} x) = y(x)  \frac{\mathrm{d} y}{\mathrm{d} x} \mathrm{d} x + \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \frac{1}{2} \mathrm{d} x^2 + \ldots\] The dominant parts of the tension forces are then \[F_{left} = k_L(x)\begin{pmatrix} 1 \\ \frac{\mathrm{d} y}{\mathrm{d} x} \end{pmatrix}\] and \[F_{right} = k_R(x) \begin{pmatrix} 1 \\ \frac{\mathrm{d} y}{\mathrm{d} x} \end{pmatrix}\] where \(k_L(x)\) and \(k_R(x)\) are the unknown magnitudes of the forces. Assuming there is a sizable piece of cable to both the left and right of \(x\), though, we expect \(k_L(x) \gg 0\) and \(k_R(x) \gg 0\).
Now, let us look at gravity. Assume we are looking at a very short pieces of the cable that is nearly straight, then the force from gravity will be \[F_{gravity}=\rho g 2 \mathrm{d} s \begin{pmatrix} 0 \\ 1 \end{pmatrix}\] where \(2 \mathrm{d} s\) is the length of the piece of cable between \(x\mathrm{d} x\) and \(x+\mathrm{d} x\). Using the arclength derivation from calculus, we can show that \(\mathrm{d} s = \mathrm{d} x \sqrt{1+y'^2}\).
Since the length of rope is very short, the force of gravity is very small relative to the two tension forces. Thus, the main part of the force balance is \[k_L(x) \begin{pmatrix} 1 \\ \frac{\mathrm{d} y}{\mathrm{d} x} \end{pmatrix} + k_R(x) \begin{pmatrix} 1 \\ \frac{\mathrm{d} y}{\mathrm{d} x} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\] This implies that we must have \(k_L(x) = k_R(x)\). Further, this must be the same everywhere between the two ends of the cable, so this constant should not depend on \(x\). Thus, we can can just call it \(k\), with the understanding that \(\partial k/\partial x = 0\).
We know that gravity must be important because without gravity, the cable could hang in arbitrary shapes or flap in the wind. What we have done so far assumes the cable is straight, but in reality it must be curved. \[F_{left} = k \begin{pmatrix} 1 \\ y'(xdx) \end{pmatrix} = k \begin{pmatrix} 1 \\ \frac{\mathrm{d} y}{\mathrm{d} x} + \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \mathrm{d} x \end{pmatrix},\] \[ F_{right} = k \begin{pmatrix} 1 \\ y'(x+dx) \end{pmatrix} = k \begin{pmatrix} 1 \\ \frac{\mathrm{d} y}{\mathrm{d} x} + \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \mathrm{d} x \end{pmatrix}\] Now, if we add all the forces, we get \[F_{left} + F_{right} + F_{gravity} = k \begin{pmatrix} 1 \\ \frac{\mathrm{d} y}{\mathrm{d} x} + \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \mathrm{d} x \end{pmatrix} + k \begin{pmatrix} 1 \\ \frac{\mathrm{d} y}{\mathrm{d} x} + \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \mathrm{d} x \end{pmatrix} + \rho g 2 \mathrm{d} s \begin{pmatrix} 0 \\ 1 \end{pmatrix}\] \[= \begin{pmatrix} 0 \\ 2 k \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \mathrm{d} x  \rho g 2 \mathrm{d} x \sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2} \end{pmatrix}\] Setting this last component to \(0\), dividing out the \(\mathrm{d} x\), and rearranging the terms, we find \[\begin{gather*} \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \frac{\rho g}{k} \sqrt{ 1 + \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right)^2 } \\ \left( \frac{k}{\rho g} \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \right)^2 = 1 + \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right)^2 \end{gather*}\] Note that we have not determined \(k\). It is an unknown constant still. In fact, it depends on the length of the cable and how far the cable anchors are separated.
Despite being a nonlinear equation, we are fortunate that this equation may be integrated directly. Let \(H = 2 \rho g/k\). \[\begin{gather} \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = H \sqrt{ 1 + \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right)^2 } \\ u = y', \quad u' = y'' \\ \int \frac{ \mathrm{d} u }{ \sqrt{ 1 + u^2 } } = H \int \mathrm{d} x \\ u = \sinh \theta, \quad \mathrm{d} u = \cosh \theta \mathrm{d} \theta, \quad \cosh^2\theta  \sinh^2 \theta = 1 \\ y' = u = \sinh \theta = \sinh( g \rho (x + C) ) \\ y(x) = \frac{1}{H} \cosh( H (x + C) ) + K \end{gather}\] We can also verify the solution using sympy1 2 3 4 5 6 7 

Note that neither constant \(C\) or \(K\) effects the shape of the catenary, only the position. If we have boundary conditions \(y(w/2)=y(w/2)=0\), then solving for the constants we find \(C=0\) and \(K = \cosh(Hw/2)/H\). \[y(x) = \frac{ \cosh( H x )  \cosh(Hw/2) }{H}\] The shape is controlled completely by \(H\). To find \(H\), we need to make sure the hanging cable as the same length as it did when we started. Thus, \[s = \int_{w/2}^{w/2} \sqrt{ 1 + \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right)^2 } \mathrm{d} x\] \[s = \int_{w/2}^{w/2} \sqrt{ 1 + \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right)^2 } \mathrm{d} x\] After doing this integral, we should find \[s H = \sinh\left( \frac{w H}{2} \right).\] This can be solved for \(H\) graphically and using 1d root finding as long as \(s \geq w\), although not in a simple closed form. \[H(w, w) = 0\] \[H(k s, k w) = \frac{1}{k} H(s,w)\]
\[y(x) = \frac{ \cosh( H(s,w) x )  \cosh(H(s,w) w/2) }{H(s,w)}\]
Note that this independent of gravity \(g\) and density \(\rho\). After calculating \(H\), we can determine the tension in the wire \(k = \frac{2 \rho g}{H}\).
to be completed
Suppose that instead of having a uniform density, we have a chain whose density varies. Can we weight the chain so that it hangs in a circular arc rather than a catenary?
If we move the poles closer by half, how do the minimum and tension change?