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Planimeters and Isoperimetric Inequalities. Robert L. Foote Wabash College. Isoperimetric Problem Given a fixed length for the boundary of a region in the plane, what shape has the largest area? Answer: A circle, attributed to Dido, Queen of Carthage, Greek and Roman mythology,
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Planimeters and Isoperimetric Inequalities Robert L. Foote Wabash College
Isoperimetric Problem Given a fixed length for the boundary of a region in the plane, what shape has the largest area? Answer: A circle, attributed to Dido, Queen of Carthage, Greek and Roman mythology, appears in Virgil’s Aeneid, ca. 25 BC Dido’s problem. To enclose the largest possible area by the sea using an ox hide. Solution: Cut hide into thin strips and form a semi-circle. Not proven until early 1900s! How are area and perimeter related for a circle?
Isoperimetric Inequality For every region, The inequality Is “sharp.” First proved by Hurwitz (1902) using Fourier series. This solves the isoperimetric problem Furthermore, if and only if the boundary is a circle. • The inequality implies that circles are solutions • The sharp part implies that circles are the only solutions
Polar and Linear Planimeters Jacob Amsler, 1854 Polar Planimeter • The wheel rolls and slides – it measures the component of its motion perpendicular to the tracer arm. • The area is proportional to the net roll of the wheel (not obvious!) Linear Planimeter
Line segment sweeping out a signed area Positive Negative
A Formula for Signed Area • Moving segment has four degrees of freedom • Motion of midpoint (2 degrees) • Rotation about midpoint • Change length dσm N N p p dm dm m m q q Component of dm in direction of N Roll (signed distance) of a wheel at m Planimeter: Moving segment sweeping out area with a wheel attached
The Area Difference Theorem If the endpoints of a moving segment each go around a region CCW, the signed area swept out by the segment is Intuitive reason
Proof of the Area Difference Theorem … To show: the signed area here is Recall (if you’ve seen Green’s Theorem)
N Proof of the Area Difference Theorem … p dp dm m q dq Integrates to Integrates to 0
Planimeter: A moving segment of fixed length with a wheel attached Location of wheel Roll of wheel determined by λ
Now three degrees of freedom • Translation forward/backward, measured by dσλ • Rotation about wheel, measured by dθ • Translation sideways, doesn’t contribute to dA No rotation: Rotation about wheel: Consider Consider
How a Planimeter Works! Integrate to get area: Ω • Area Difference Theorem: • Right endpoint goes around ∂Ω: AR = AΩ • Left endpoint goes around no area: AL = 0 • No net rotation of segment: • Area is proportional to roll of wheel • Net roll of wheel doesn’t depend on • location!
The Prytz “hatchet” Planimeter 1875, HolgerPrytz, Danish mathematician and cavalry officer Economical alternative to Amsler’splanimeters Behaves like a bicycle Front wheel: tracer point Rear wheel: chisel edge
Error It can measure area! σ is the net disp. of the chisel (red arc)
To Prove: Isoperimetric Inequality Isoperimetric Defect if and only if the boundary is a circle We’ll do better:
Put wheel at right endpoint (λ = 1) so it rolls along ∂Ω. New: Have planimeter make one CCW rotation. Trace differently … As before, AR = AΩ and AL = 0. This and size of region put geometric constraints on ℓ. ℓ = half-width ℓ = inradius ℓ = circumradius
Goal: Currently have Really, just complete the square in ℓ and rearrange!
Key Observation N dq is a component of ds q
This implies ? When do we have Need a more geometrically meaningful expression for Solve and substitute …
Ω Suppose But Ω is contained in the circle. The radius of the circumscribing circle determines ℓ. Then So ∂Ω is the circle. “Sharp” part of the Isoperimetric Inequality
Isoperimetric Inequality if and only if the boundary is a circle Bonnesentype of Isoperimetric Inequality • B has geometric significance • B = 0 iff the boundary is a circle Bonnesen found several of these, 1920's Osserman, Amer. Math. Monthly, Jan 1979
Isoperimetric Inequality in Spherical and Hyperbolic Geometries K is the Gaussian curvature of the geometry; for a sphere. Get equality if and only if the region is circular.
Segments sweep out area differently than in Euclidean geometry … q p q p … but similar proofs work for planimeters and the isoperimetric inequality.
Amsler's Spherical Polar Planimeter, 1884 Never manufactured. One prototype built. • Many other types of isoperimetric inequalities: • higher dimensions, other geometries, in physics …
Thanks! Check out the planimeters on display A Plethora of Planimeters! What’s for sale on eBay?