A slack rope secured at both ends hangs under its own weight. We consider a differential line segment of length ds, anywhere along the rope.  The weight of the rope below the differential line segment adds to the tension acting on the segment, and thus, unlike the horizontal rope, the tension is not constant throughout the length of the rope.
 Kite String Theory
 Our interest is in the flight characteristics of a kite, and we need to know the mathematical curve which describes the shape of the string while the kite is flying.  We derive this curve in two ways: 1) we consider the torque, yes, torque, acting on a string and 2) we consider the tension in the string.  Our string is idealized as a flexible one-dimensional material of uniform linear mass density which does not stretch or compress under tension.  Our string must be hanging smoothly in a uniform gravitational field under its own weight.  The curve we derive, the well-understood catenary, will describe any rope, thread, wire, chain, cable, string, etc., with said characteristics.  The kite acts to secure one end of the string.  The kite flyer secures the other.   We start with a differential line segment of the string beginning at point A and following some smooth curve to another point, B, as shown in the figure below.  The string tension acting on the endpoints of the segment are vectors which are tangent to the curve at their respective points.  The vectors are not co-linear since the kite string is curved.   The magnitude of the tension at B is greater than the magnitude of the tension at A.
 Even in the case that ds is of finite length, the horizontal components of the tension vectors at A and B are equal.  Let ρ be the linear mass density of the string.  The differential mass of the string element, then, is dm = ρ ds and the magnitude of the force due to gravity is dF = dm g = ρ g ds, where g is the (constant) acceleration due to gravity near the earth’s surface.    For ds ≠ 0, there is no change in the tension at point A.  Since the force of gravity points downward, the change in tension at B occurs only in the y-direction, leaving the x-component unchanged.  Thus, the horizontal forces are still equal but opposite, despite that points A and B are not coincident.  This last condition is independent of the length of the line segment ds, which means that the horizontal component of the tension is everywhere the same.
 Let us look at the horizontal components of the tenson vectors; we aim to show that the horizontal components are equal in magnitude.  In the limit as ds → 0 (as B → A for example) we see that the vector TB → -TA since B now coincides with A.  Because the string is hanging still, i. e., there is no acceleration of the string, the net force on the string in the x-direction is zero.  This means that the horizontal components of the two tension vectors point in opposite directions but are of equal magnitude.
 Since ds is small, we drop the A and B superscripts and write differential expressions for the tension and the tangent angle.  Write the tension vector as -T at point A, and the net tension at point B as T + dTy. See the figure, above, where we show that vector dTy is in the y-direction. If -T at point A makes an angle -θ with respect to the -x-axis, then we may write the tension at point B as the vector T + dTy at an angle θ + dθ with respect to the +x-axis.
 Catenary Kite
 EOS
 If an unstretchable rope is anchored at one end to an “immovable” object such as a wall, and then pulled taut so that the rope describes a straight, exactly horizontal line, then the pulling force, called tension, is carried by the rope to the wall and the rope pulls the wall with the same tension.  The rope is static and the tension is constant throughout the length of the rope.  A differential segment of the rope of length, dℓ, has a mass, dm, on which the force of gravity acts vertically downward.  Because the rope is not accelerated downward, there must be molecular forces in the material acting upward on the line segment, making the net vertical force zero.  The tensional force, T, acts on both ends of the differential line segment.
 Physics
 Now we return our attention to the tension in the kite string.  Writing Newton’s second law with zero acceleration in the y-direction, or, Consider now the tangent of the angle, Solve for the y-component of the tension: Now, let us take the derivative with respect to x of both sides, recalling that Tx  is a constant, We may rewrite this as Now make the substitution for dTy
 Our kite string follows some as-yet-to-be-determined smooth mathematical curve with end points (x1, y1) and (x2, y2), where we require x2 ≠ x1 and y2 > y1.  That is, the string shall not follow a vertical straight line.  The earth’s gravitational force exerts a torque on the string.  Even though the string is flexible and the string would fall straight down rather than rotate about a fixed axis if separated from the kite, nevertheless, we can calculate the static torque, τ, acting on the string.  Since the kite flyer supports one end of the string and the kite supports the other, there is no angular acceleration of the string (for a stationary flying kite). If we like, we could imagine a thin rod such as a stiff copper wire which can be bent into any shape and retain that shape — no kite necessary — as a physical manifestation of our mathematical curve.  Further, we would imagine the rod supported at the lower end only, with a clamped hinge.  The gravitational force exerts a torque on the rod and the clamp must act against that torque in such a manner that the hinge not turn.  Loosen the clamp and the rod rotates with an angular acceleration. We pick a rotational axis parallel to the z-axis, perhaps through the point (x1, y1).  Let r represent the position vector from this rotation axis to the line element ds along the string; we also have the vector differential force dF due to gravity acting on the string.  Then, and which points in the – y-direction.  The differential torque is the cross product and the magnitude of the total torque can be found by the line integral The differential element of length can be written in terms or either dx or dy, where we use the prime notation to indicate the derivative.  The same result obtains whether we integrate along the x- or the y-axis; we chose the dx form.  The integral gives us the magnitude of the torque The torque vector is which points in the – z-direction.
 Variations on a String
 We cannot perform this integration because we do not yet have the function y(x), which we need in order to calculate y’(x). A particular y(x) can be found by solving the Euler-Lagrange equation.  Euler’s equation was derived from the calculus of variations, and when solved gives us a function which corresponds to an extremum of the integral.  There are several forms of Euler’s equation, and we use this one since it includes derivatives with respect to x and y’ The integrand in our expression for torque is the Lagrangian, L(x, y’).  A Lagrangian, generally, can be a function of x, x’, y, y’ and even x’’ or y’’, but for our purpose we drop the explicit mention of all but x and y’ since those do not appear in our integrand.  We mentioned that the form of ds which we use makes no difference to the torque calculation; it does, however, make a difference — in a way — as regards Euler’s equation.  If we had chosen the other form of ds (the one which includes x’) we would need to derive (or look up) another form of the equation.  The solution to Euler’s equation would be the same, however.  Euler’s equation allows for the use of generalized coordinates in the Lagrangian.  Also, when writing the Lagrangian we drop the constants that appear in front of the integral since they would “divide out” (a technical mathematical term) and thus play no role in the solution of Euler’s equation.  Expanding Euler’s equation, we have This can be simplified without performing the derivative operations.  We are left with If the derivative of a (non-zero) function is zero, then that function must be a (non-zero) constant.  This means that we can write Now we take the partial derivative of the Lagrangian, and solve for y’, or, We integrate both sides to find y(x) which satisfies Euler’s equation. We take note that this is an inverse function.  We also remind ourselves that solutions to Euler’s equation produce functions, y(x), which represent an extremum of the integral.  In this case the extremum is a minimum, which gives a minimum torque.  This means that no other curve between the same endpoints and with the same boundary conditions has a lower torque.  Let us graph the function to see what it looks like.
 At a glance this does not look like the curve of a kite string, but we press on.  To garner an idea of just how much torque is acting on the inverse catenary, we consider the cross product r x dF = r dF sin(φ) = r dF sin(β), where φ is the angle between the vectors and β = 180° - φ. Place the rotational axis at x = a.  For small r, the angle β is small; in fact, as r → 0, β → 0, and sin(β) → 0.  This means that the contribution (per unit length of string) to the total torque is small for the string near the rotational axis.  For large r, that is, as r → ∞, β → 90°, and sin(β) → 1.  This means that the contribution (again, per unit length) to the torque is large far from the rotational axis. The constant a determines which particular shape the string takes.  For the shape shown in the diagram, we can easily see that a (relatively) large part of this curve is far from the rotational axis and thus the string has, let us say, a fair amount of torque.  Nevertheless, this curve represents the minimum torque compared to any other curve with the same boundary conditions and endpoints. This solution to Euler’s equation is an inverse-catenary.  Of course its inverse is the catenary, which is open upward, and is the curve that describes the kite string.  To make the conversion, all we need do is reflect the curve across the line y = x. In other words, let y → x, and let x → y. We like to reorient the curve with mathematical rigor, just for completeness.  To do so, let us invoke the inverse function theorem.  Use of this theorem demands that several conditions be met, viz, that on and near the x-values of interest, y must be differentiable, that the differential is not zero, that the inverse of y is defined and be differentiable.  If these conditions are met (which we can check later if necessary, but a simple observation of the graph of the curve will answer) then we use an elementary expression of the theorem: the derivative of the inverse is equal to the inverse of the derivative, Our solution to Euler’s equation gave us so, The derivative of y is Define a new name for the inverse of y(x), which is the curve we are after, From the inverse function theorem, and the chain rule: which, rearranged, becomes Now we want to integrate both sides of that last equation, but we need y’ in terms of y. Thus, Integrating both sides and applying appropriate boundary conditions leads to This is the catenary which describes the kite string.