Trig, Polar Coordinates & Holditch's Formula

With hindsight, Robinson's solution to the old problem of rigorous use of infinitesimal numbers in calculus boils down to Keisler's Funcrtion Extension Axiom that we discuss in Section 2.  Extending the (Dedekind) "real" numbers to include infinitesimals algebraically is not at all difficult, but calculus depends on small approximations with transcendental functions like sine, cosine, and the natural logarithm.  Following are some intuitive approximations with non-algebraic graphs.

Sine and cosine in radian measure give the x and y location of a point on the unit circle measured a distance θ along the circle as shown next.

[Graphics:../HTMLFiles/Lect1_158.gif]

Now make a small change in the angle and magnify by 1/δθ to make the change appear unit size.

[Graphics:../HTMLFiles/Lect1_160.gif]

Since magnification does not change lines, the radial segments from the origin to the tiny increment of the circle meet the circle at right angles and appear under magnification to be parallel lines.  Smoothness of the circle means that under powerful magnification, the circular increment appears to be a line.  The difference in values of the sine is the long vertical leg of the increment "triangle" above on the right.  The apparant hypotenuse with length δθ is the circular increment.

Since the radial lines meet the circle at right angles the large triangle on the unit circle at the left with long leg Cos[θ] and hypotenuse 1 is similar to the increment triangle, giving

Cos[θ]/1≈δSin/δθ

We write approximate similarity because the increment "triangle" actually has one circular side that is ≈-straight.  In any case, this is a convincing argument that dSin/dθ = Cos[θ].  A  similar geometric argument on the increment triangle shows that dCos/dθ = -Sin[θ].

The Polar Area Differential

The derivation of sine and cosine is related to the area differential in polar coordinates.  If we take an angle increment of δθ and magnify a view of the circular arc on a circle of radius r, the length of the circular increment is r · δθ, by similarity.

[Graphics:../HTMLFiles/Lect1_170.gif]

A magnified view of circles of radii r and r + δr between the rays at angles θ and θ + δθ appears to be a rectangle with sides of lengths δr and r · δθ.  If this were a true rectangle, its area would be r · δθ · δr , but it is only an approximate rectangle.  Technically, we can show that the area of this region is r · δθ · δr plus a term that is small compared with this infinitesimal,

δA = r δθ δr + ε · δθδr

Keisler's Infinite Sum Theorem assures us that we can neglect this size error and integrate with respect to r dθ dr.

Holditch's Formula

The area swept out by a tangent of length R as it traverses an arbitrary convex curve in the plane is A = π R^2.  

[Graphics:../HTMLFiles/Lect1_183.gif]

We can see this interesting result by using a variation of polar coordinates and the infinitesimal polar area increment above.  Since the curve is convex, each tangent meets the curve at a unique angle, φ, and each point in the region swept out by the tangents is a distance ρ along that tangent.

[Graphics:../HTMLFiles/Lect1_186.gif]

We look at an infinitesimal increment of the region in ρ-φ-coordinates, first holding the the φ-base point on the curve and changing φ.  Microscopically this produces an increment like the polar increment:

[Graphics:../HTMLFiles/Lect1_191.gif]

Next, looking at the base point of the tangent on the curve, moving to the correct φ + δφ-base point, moves along the curve.  Microscopically this looks like translation along the tangent line (by smoothness):

[Graphics:../HTMLFiles/Lect1_193.gif]

Including this near-translation in the infinitesimal area increment produces a parallelogram:

[Graphics:../HTMLFiles/Lect1_194.gif]

of height ρ · δφ and base δρ, or area δA = ρ · δφ · δρ:

[Graphics:../HTMLFiles/Lect1_198.gif]

Integrating gives the total area of the region

∫_0^R∫_0^(2π) ρφρ = π R^2


Created by Mathematica  (September 22, 2004)