Similar Formulas and Intelligent Design

When you study science or geometry, you come across several formulas in the form of a = kx and b = ½kx².  For example, the velocity of an object starting at rest with a constant velocity can be determined by multiplying acceleration and time, or v = at, and the distance of the same object can be found by taking half of its acceleration multiplied by the time squared, or d = ½at².  For another example, the momentum of an object can be determined by multiplying its mass times its velocity, or p = mv, and the kinetic energy of an object can be found by taking half of its mass times its velocity squared, or E = ½mv².

Sometimes the equations need to be manipulated to be in the form of a = kx and b = ½kx².  For example, the circumference of a circle is C = 2πr and the area of a circle is A = πr², but if we substitute τ = 2π, then C = τr and A = ½τr².  For another example, the area of a triangle is A = ½bh, but if height is defined as a ratio of the base, then h = kb and A = ½kb².

Other equations in the form of a = kx and b = ½kx² include angular velocity and displacement, angular momentum and rotational energy, force and potential energy of a spring, electric flux and energy density, and electric charge and electric energy:

a = kx	b = ½kx2
Velocity v = at	Distance d = ½at2
Angular Velocity ω = αt	Angular displacement θ = ½αt2
Momentum p = mv	Kinetic Energy E = ½mv2
Angular Momentum L = Iω	Rotational Energy K = ½Iω2
Force of a Spring F = kx	Potential Energy of a Spring E = ½kx2
Electric Flux Density D = εE	Electric Energy Density Q = ½εE2
Electric Charge Q = CV	Electric Charge Energy E = ½CV2
Circumference of  a Circle C = 2πr = τr	Area of a Circle A = πr2 = ½τr2
Height of a Triangle h = kb	Area of a Triangle A = ½bh = ½kb2

In fact, if we lift the requirement that k must be a constant, we can include even more formulas to the form of b = ½kx².  Another pair of similar yet unrelated formulas is gravitational force (F = ^Gm1m2/_r^2) and electric force (F = ^kq1q2/_d^2).  Gravitational force can be rearranged to ½Gm₁m₂ = b = ½Fr² and electric force can be rearranged to ½kq₁q₂ = b = ½Fd².  Even Einstein’s famous theory of relativity E = mc² can be manipulated to ½E = ½mc².

Those familiar with calculus may recognize that the derivative of b = ½kx² is a = kx, which means a is a rate of b with respect to x.  For example, since d = ½at² and v = at, velocity is a rate of distance with respect to time.  A good way to visualize this is to use the two geometry formulas of the circle and triangle.  Since A = ½τr² and C = τr for a circle, the circumference is a rate of its area with respect to its radius, which means that the area of a circle is the sum of all the circumferences it contains. 

Area of a Circle = Sum of the Circumferences

Since A = ½kb² and h = kb for a triangle, the height is a rate of its area with respect to its base, which means that the area of a triangle is the sum of all the heights it contains.

Area of a Triangle = Sum of the Heights

Although calculus can be used to explain the relationship between b = ½kx² and a = kx, it cannot explain the high frequency of which these formulas appear in nature.  If there were some relationship between all these formulas we might be tempted to attribute the high frequency to a coincidence, but these equations appear in completely unrelated and different fields of math and science, from velocity to circles, from springs to electric fields, and from momentum to triangles.  So many different formulas in the form of b = ½kx² and a = kx cannot be a result of random chance, but rather must be the result of an Intelligent Designer of an orderly universe.

Areas of Conics

Most of us know area formulas for basic geometrical objects, such as circles, squares, rectangles, and triangles.  (The area of a circle is A = πr², the area of a square is A = s², the area of a rectangle is A = lw, and the area of a triangle is A = ½bh.)  Others of us may even remember learning area formulas of other geometrical objects in high school geometry, such as parallelograms, trapezoids, and equilateral triangles.  (The area of a parallelogram is A = bh, the area of a trapezoid is ½(b₁ + b₂)h, and the area of an equilateral triangle is ¼√3s².)

Other than the circle, the area of conics are neglected in most geometry textbooks, even though the Cartesian equation for each conic are included in most algebra textbooks.  The area of a parabola is A = ²/₃bh (where b and h are the base and height of the rectangle that contains it).

The area of an ellipse is A = πab (where a and b are the major and minor axes of the ellipse). 

The area of a hyperbola is A = ab(^h+a/_a√(( ^h+a/_a)² – 1) – ln|^h+a/_a + √(( ^h+a/_a)² – 1)|) (where a and b are the axes of the hyperbola and h is the height).
 

The area formulas for the parabola and ellipse are relatively simple, the equation for the hyperbola is not.

To prove the area of a parabola, we can use integration.  The Cartesian equation for a parabola with its vertex at the origin is y = ax², which means another point on the parabola would be (b, ab²). 

The area of a rectangle with opposite vertices on the origin and (b, ab²) is A = b∙ab² = ab³.  The area of the parabola inside this rectangle is A = ∫₀^b ab² – ax² dx = ab²x – ¹/₃ax³] ₀^b = ab²b – ¹/₃ab³ = ab³ – ¹/₃ab³ = ²/₃ab³.  (The left half of the parabola would be the same by symmetry.)  Therefore, the area of the parabola is ²/₃ of the area of the rectangle that contains it, or A = ²/₃bh.

To prove the area of an ellipse, we can compare its Cartesian equation to the Cartesian equation of a circle.  The Cartesian equation for an ellipse centered at the origin where a and b are the major and minor axes is ^{x^2}/_a^2 + ^{y^2}/_b^2 = 1, which rearranged is y = ^b/_a√(a² – x²), and the Cartesian equation for a circle centered at the origin where a is a radius is x² + y² = a², which rearranged is y = √(a² – x²), so the difference in heights is a factor of ^b/_a. 

    

So if the area of a circle is A = πr², or in this case A = πa², the area of an ellipse must be A = ^b/_aπa² = πab.  (Technically speaking, the area of an ellipse is 4∫₀^{a b}/_a√(a² – x²)dx = ^b/_a 4∫₀^a √(a² – x²)dx = ^b/_aπa² = πab.)

To prove the area of the hyperbola, we can use integration once again.  The Cartesian equation for a horizontal hyperbola centered at the origin where a and b are the major and minor axes is ^{x^2}/_a^2 – ^{y^2}/_b^2 = 1, which rearranged is y = ^b/_a√(x² – a²). 

The area is then A = 2∫_a^{h+a b}/_a√(x² – a²)dx.  Using a right triangle with legs a and √(x² – a²) and hypotenuse x, tan q = ^{√(x^2 – a^2)}/_a or √(x² – a²) = a tan q, and sec q = ^x/_a or x = a sec q, which means dx = a sec q tan q dq. 

Using substitution, A = 2∫_sec-1(a)^{sec-1(h+a) b}/_a a tan q a sec q tan q dq = 2ab∫₀^{sec-1((h+a)/a)} tan² q sec q dq = 2ab∫₀^{sec-1((h+a)/a)} (sec² q – 1)sec q dq = 2ab∫₀^{sec-1((h+a)/a)} sec³ q – sec q dq.  Integrating results in 2ab[½ sec q tan q + ½ ln|sec q + tan q| – ln|sec q + tan q|] ₀^{sec-1((h+a)/a)}  = ab[sec q tan q – ln|sec q + tan q|] ₀^{sec-1((h+a)/a)}.  Since sec(sec^-1(x)) = x and tan(sec^-1(x)) = √(x² – 1), A = ab(^h+a/_a√(( ^h+a/_a)² – 1) – ln|^h+a/_a + √(( ^h+a/_a)² – 1)|).

Although the area of a hyperbola is not an easy formula, both area formulas for a parabola and an ellipse are relatively simple.  The area of an ellipse can also be used to show the special case of a circle where a = b = r (A = πab = π∙r∙r = πr²), and the area formulas of conics nicely correspond with the Cartesian equations of conics learned in algebra.  It makes one wonder why the area formulas for the parabola and ellipse are typically not included in a high school geometry curriculum or textbooks.