do dogs know the trammel of archimedes?

Do Dogs Know the Trammel of Archimedes?Author(s): Mark SchwartzReviewed work(s):Source: The College Mathematics Journal, Vol. 42, No. 4 (September 2011), pp. 299-308Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/10.4169/college.math.j.42.4.299 .Accessed: 24/09/2012 02:20

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Do Dogs Know the Trammel of Archimedes?Mark Schwartz

Mark Schwartz ([email protected]) is a professor ofmathematics at Ohio Wesleyan University. His interest inplanar curves and geometry came as a result of studyingoptimization problems. The present article illustrates onesuch connection. Outside of mathematics, his recreationalinterests include running, swimming, and spectator sports.He especially values family time and opportunities fortravel, good meals, and good conversation.

Elvis, Tim Pennings’ Welsh Corgi, first introduced in this JOURNAL in [4], actinginstinctively, presumably, appears to run/swim an optimal path from the beach to a ballthrown into the water. In other words, Elvis solves the well-known refraction problem,finding the least-time path from a point A in one medium to a point B in another. Onemight question Elvis’ knowledge of calculus (and in [1], where the optimal path isfound by purely algebraic methods, it is suggested that Elvis doesn’t need calculus).In this paper, we make another contribution to this series by solving the refractionproblem yet another way.

Despite how well-known the refraction problem is, it continues to reveal fasci-nating insights. Unlike [1] or [4] or any solution I know, this paper gives a geomet-ric/mechanical solution involving envelopes and the trammel of Archimedes. In fact,a machine constructed from cardboard (see Figure 1) solves the problem.

Solution by calculusTo get started, we consider the standard solution of the problem. Let S be a point inthe upper half plane, which is filled with medium M1, and let U be a point in the lowerhalf plane, which is filled with medium M2. Suppose Elvis locomotes at speed r1 inM1 and r2 in M2. The problem is to find the point T on the x-axis such that the timefor Elvis to travel from S to T to U is minimal. (See Figure 2, where, for definiteness,S is the point (0, s) with s > 0, and U is the point (u, v) with v < 0.)

We seek to minimize the function

f (t) =

√t2 + s2

r1+

√(u − t)2 + v2

r2,

where (t, 0) is a point on the x-axis. Differentiating f with respect to t , equating to 0and rearranging terms, we get

t

r1

√t2 + s2

=u − t

r2

√(u − t)2 + v2

. (1)

http://dx.doi.org/10.4169/college.math.j.42.4.299

VOL. 42, NO. 4, SEPTEMBER 2011 THE COLLEGE MATHEMATICS JOURNAL 299

Figure 1. A Rolling Circle Refracto-Solver.

S (0, s)

T (t, 0)

U (u, v)

Figure 2. The refraction problem.

Let α and β be the angle of incidence and angle of refraction, respectively, as in Figure2. Then equation (1) gives sinα

r1=

sinβr2

, or sinαsinβ =

r1r2

. This is Snell’s law, where thecommon ratio is the index of refraction. Finding the exact value of t minimizing f isdifficult, as it involves solving a quartic equation. Regardless, the algebraic formulafor t is of little interest geometrically and gives no particular insight to the nature ofthe solution.

300 „ THE MATHEMATICAL ASSOCIATION OF AMERICA

Another solution—an envelopeTo motivate the geometric approach, consider the inverse problem. Suppose T = (t, 0)is fixed. We determine the locus of points U = (u, v) for which T is the minimizingpoint of f . Working from (1) we get the equation

r2

r1tv +

√√√√s2 + t2

[1−

(r2

r1

)2](u − t) = 0.

To simplify notation, let r = r2r1

, the reciprocal of the index of refraction, and replaceu and v by x and y, respectively. Then we have

r ty +√

s2 + t2(1− r 2)(x − t) = 0. (2)

For fixed t , this is a line passing through T forming the angle of refraction β, in accor-dance with Snell’s law. Before proceeding to the next step, we distinguish two cases.In (2), we must have s2

+ t2(1− r 2) ≥ 0, so certain restrictions on t may be in effect.

Case I: r > 1 (M1 is the slower medium.) We must have

t2≤−s2

1− r 2=

s2

r 2 − 1, and so |t | ≤

s√

r 2 − 1.

Case II: r < 1 (M2 is the slower medium.) We must have t2≥−s2

1−r2 , which is alwaystrue. Thus there is no restriction on t .

In the following we assume r > 1. With this assumption, in a departure from the usual,Elvis starts in the slower medium (swimming) so his refracted path bends away fromthe vertical as he enters the faster medium (exit from the water to begin running). Laterwe show how to solve the more natural run-to-swim problem.

Our solution of the refraction problem is initiated by varying t and plotting theresulting family of lines (2). Figure 3 shows the result. The interesting pattern seenhere suggests the presence of an envelope of a familiar curve.

Figure 3. Family of lines.

Recall that under certain mild conditions, a family of curves will have an envelopewhich is generally a very visible feature of the plot of the family. Often (but not always)the envelope is a curve tangent at each point to a (different) member of the family. The


references [2] and [6] have particularly complete descriptions of envelopes. We give adefinition here which indicates how to calculate them.

Let a family of smooth curves be given by F(t, x, y) = 0 where F is a smooth real-valued function. The envelope E of the family is given by the simultaneous solution ofthe equations F(t, x, y) = 0 and ∂

∂t F(t, x, y) = 0. That is,

E = {(x, y) : there exists t such that F(t, x, y) =∂

∂tF(t, x, y) = 0}.

In the present case, the family of lines is given by the equation F(t, x, y) = 0 where

F(t, x, y) = r ty +√

s2 + t2(1− r 2)(x − t).

Finding the envelope requires the simultaneous solution of the equations F = ∂

∂t F =0. That is,

r ty +√

s2 + t2(1− r 2)(x − t) = 0,

r y −√

s2 + t2(1− r 2)+t (1− r 2)(x − t)√

s2 + t2(1− r 2)= 0.

The system is linear in x and y so we can solve to get

x =t3(r 2− 1)

s2,

y =

[s2+ t2(1− r 2)

]3/2

rs2.

Eliminating t gives the Cartesian equation xs√

r2−1

2/3

+

(ysr

)2/3

= 1. (3)

The graph of this envelope is the 4-cusped curve of Figure 4. It is the evolute of acertain ellipse, that is, it is the envelope of the family of normal lines to the ellipse.Consequently, (2) is precisely the family of normals to an ellipse.

EvolutesIf γ is any smooth, regular (γ ′ 6= 0), unit speed curve in the plane, with nonzero cur-vature, the family of normal lines to γ has an envelope, called the evolute of γ . Thefamily of normal lines to γ is given by F(t, x) = (x − γ ) · T = 0, where x is a pointin the plane, and T is the unit tangent vector. Computing the envelope, we solve si-multaneously

(x − γ ) · T = 0,

−1+ (x − γ ) · T ′ = 0.


Figure 4. Evolute of ellipse.

The first equation says x − γ = λN , for some scalar λ. Substituting into the secondequation and using the Serret-Frenet formula T ′ = κN , we get λN · κN = 1, and soλ = 1/κ . Hence, the evolute has parametric equation x = γ + 1

κN . (Recall, this is the

locus of the center of curvature of γ , an alternative definition of the evolute.)Now consider an ellipse with Cartesian equation

(xa

)2+ (

yb )

2= 1. Parametrically,

this is given by γ (t) = 〈a cos t, b sin t〉. From the formula just derived, the evolute ofthe ellipse is

γ +1

κN = 〈a cos t, b sin t〉 +

(a2 sin2 t + b2 cos2 t)3/2

ab

−〈b cos t, a sin t〉

(a2 sin2 t + b2 cos2 t)1/2

=

⟨a2− b2

acos3 t,

b2− a2

bsin3 t

⟩.

The Cartesian equation of this curve is ( xA )

2/3+ (

yB )

2/3= 1 where Aa = −Bb =

a2− b2. This is the same form as equation (3). Thus, the evolute is a star-like (astroid-

like) curve, also known as a Lame curve, with four cusps as shown in Figure 4. Solvingfor a and b in terms of A and B, we get a = AB2

B2−A2 , b = A2 BA2−B2 . This allows us to find

the ellipse from its evolute. In the present case, the 4-cusped curve in Figure 4 is theevolute of the ellipse with a = −s

√r 2 − 1, b = rs, and so the underlying ellipse has

equation (x

s√

r 2 − 1

)2

+

( y

rs

)2= 1.

The significance for us is that the family (2) is the family of normal lines to thisellipse. Our constructive method for solving the refraction problem uses this keyfact.

The Trammel of ArchimedesThe Trammel of Archimedes is a system consisting of a rod of fixed length, whoseends move on two perpendicular lines, e.g., the x and y-axes (see Figure 5).


P

Figure 5. Trammel of Archimedes.

As the ends of the rod move along the axes, a point P on the rod traces out an ellipse.We quickly prove this. Suppose the rod has length k and let P be a point located atsigned distance d from the center of the rod. If the initial position of the rod is flatalong the positive x-axis then P is located at ( k

2 + d, 0) to start. Let θ be the interiorangle of the trammel as in Figure 5. Then parametric equations for the path of P are

〈k cos θ, 0〉 +〈−k cos θ, k sin θ〉

k

(k

2− d

)=

⟨(k

2+ d

)cos θ,

(k

2− d

)sin θ

⟩.

Thus, x = ( k2 + d) cos θ , y = ( k

2 − d) sin θ , parametric equations for an ellipse.An equivalent method for generating an ellipse is to roll a circle inside or outside a

larger circle, that is, as a hypotrochoid or epitrochoid. Suppose a fixed circle has radiusk and a rolling circle has signed radius b. That is, b > 0 refers to a circle of radius brolling on the outside of the fixed circle (epitrochoid) while −k < b < 0 describes acircle of radius |b| rolling on the inside of the fixed circle (hypotrochoid). Suppose apoint P is located at signed distance d from the center of the rolling circle, starting atinitial angle φ, as in Figure 6.

P

Figure 6. Epitrochoid machine.

The equation of the resulting trochoid (epi or hypo) is found to be Rk,b,d,φ(θ) =

(k + b)γ (θ) + dγ(

k+bb θ + φ

), where γ (θ) = 〈cos θ, sin θ〉. In the special case of


φ = 0, b = − k2 , we get

Rk,− k2 ,d,0

(θ) =

(k −

k

2

)γ (θ)+ dγ

(k − k

2

−k2

θ

)=

k

2γ (θ)+ dγ (−θ)

=

⟨(k

2+ d

)cos θ,

(k

2− d

)sin θ

⟩, (4)

which is identical to the curve generated by the trammel. For later reference, the pa-rameters k and d of the trammel (or the rolling circle configuration) required to gen-erate a given ellipse ( x

a )2+ (

yb )

2= 1 are found by solving the system k

2 + d = a andk2 − d = b. Thus

k = a + b, and d =a − b

2. (5)

In this case, the initial position of the point P is on the x-axis at ( k2 + d, 0) = (a, 0).

To sum up our progress: for each point T = (t, 0) on the x-axis, there is a set ofpoints U = (u, v) that lie on a line and represent the set of destination points for whichT is the optimal solution to the refraction problem. As t varies we get a family of lineswhose envelope happens to be the evolute of a certain ellipse E . Consequently, thefamily of solution lines coincides with the family of normal lines to E . Meanwhileany ellipse, E for example, can be obtained as the path of a point P on a trammel or,equivalently, as a hypotrochoid produced by a rolling circle mechanism.

Instantaneous center of rotationOne more tool is needed for our geometric solution of the refraction problem. Thisis the instantaneous center of rotation. Consider a rigid object moving in any mannerwhatsoever in the plane. Let p and q be two points on the object and let vp and vq

be their instantaneous velocity vectors. The intersection of the normal lines to thesevelocity vectors is a point H , called the instantaneous center of rotation relative to pand q . Of interest is the instantaneous center of rotation of the trammel of Archimedes.Consider the trammel in Figure 7.

P

H

p

q

Figure 7. Instantaneous center of rotation.


The endpoints p and q travel on the x-axis and y-axis, respectively, thus their veloc-ity vectors parallel the axes. The instantaneous center of rotation is simply the intersec-tion of the vertical line at p and the horizontal line at q . In the hypotrochoid setting, theinstantaneous center of rotation is the contact point between the rolling circles. Thisis easily seen from equation (4), where p = (k cos θ, 0), q = (0, k sin θ), and conse-quently, H = (k cos θ, k sin θ), the contact point of the circles. Furthermore, a crucialfact which we use later, the ray from H to a point P on the trammel (or rolling circle)is normal to the ellipse generated by P as the trammel (or rolling circle) is moved. For,−−→H P = 〈( k

2 + d) cos θ, ( k2 − d) sin θ〉 − 〈k cos θ, k sin θ〉 = 〈(− k

2 + d) cos θ, (− k2 −

d) sin θ〉 which is orthogonal to the tangent vector 〈(− k2 − d) sin θ, ( k

2 − d) cos θ〉 tothe ellipse.

The constructive solutionWe now assemble these components into a geometric solution of the refraction prob-lem. Elvis starts swimming (as we mentioned earlier) at point S = (0, s) in the slowermedium in the upper half plane (see Figure 2). His target point, U = (u, v), is in thefaster medium in the lower half plane, with (reciprocal) index of refraction r = r2

r1> 1.

We seek the point T = (t, 0) on the x-axis which minimizes the time for Elvis to tra-verse from S to T to U . From the work above, in order for T to be the optimizingpoint, the refracted ray from T to U must coincide with a certain normal line to theellipse E with equation (

x

s√

r 2 − 1

)2

+

( y

rs

)2= 1.

We construct this normal line with a trammel/hypotrochoid machine. Once con-structed, the minimizing point T is the x-intercept of this line.

The first step is to determine the correct parameters k and d of the trammel (orrolling circle) for which a suitable point P on the trammel (or rolling circle) willgenerate E as its path under the action of the trammel (or rolling circle). From (5),

k = a + b = s(r +

√r 2 − 1

)and d =

a − b

2=

s(√

r 2 − 1− r)

2.

Thus, P is located initially at(k

2+ d, 0

)= (a, 0) =

(s√

r 2 − 1, 0).

Next consider the hypotrochoid mechanism (see Figure 8). The fixed circle C1 hasradius k = s(r +

√r 2 − 1). The interior rolling circle C2 has half this radius, and

P starts at (s√

r 2 − 1, 0) on the x-axis. Here’s what happens. The desired ellipse isgenerated by P . The mechanism rolls until the normal line to the ellipse passes throughU . How will we know when this happens? The answer is given by the instantaneouscenter of rotation, H , which is the contact point between the circles. The ray

−−→H P

is always normal to the ellipse, so the rolling circle should stop precisely when−−→H P

passes through U . Then the optimizing point T is the x-intercept of−−→H P .

The whole solution may be implemented physically. See Figure 1 for my cardboardrefraction solver. Here is how to make it. To the basic rolling circle configuration


H

S

U

C1

C2

P

Figure 8. Schematic of hypotrochoid machine.

(larger fixed circle and smaller rolling circle with half the radius of the fixed circle),pin a radial arm from the center of the fixed circle, through the center of the rollingcircle, terminating at the contact point of the two circles. Thus, the arm is a diameterof the rolling circle. Pins and/or tacks are used to mark points and make attachmentsin the model. The arm is pinned at only two points (the centers of each circle) andterminates at the instantaneous center of rotation H . Locate P appropriately, and markit with a pin. The pointer from H through P has a groove along its midline. It is pinnedat H , but straddles the pin at P via the groove. (The pointer cannot be pinned at P ,otherwise the structure won’t move.)

Figure 8 is a schematic of the model, but there is an additional factor which needsto be taken into account. The pointer needs to pass through the pin located at the centerof the rolling circle, the pin holding the arm in place. To accomplish this, I made therolling circle into a “double decker” by gluing a copy of the rolling circle above theoriginal. Cardboard struts are used to provide separation. The pointer is situated on theupper deck, and passes cleanly above the previously obstructing pin which lies below.As the interior circle (now double-decker) is rolled, the pointer swings around, alwaysnormal to the underlying, unseen ellipse traced by P . The rolling action continues untilthe pointer is over U , determining the correct refraction ray. The optimizing point Tis the intersection of the pointer with the x-axis. I took s = 2 and r = 1.25 in myconstruction. As a result, the large circle has radius 4, the rolling circle has radius 2,and the point P starts at ( 3

2 , 0).

Case IITo this point, the solution assumes the slower medium is above the faster. Thus Elvisswims first. As a result, he travels faster upon crossing the shoreline (x-axis) and theangle of refraction is greater than the angle of incidence. In the more natural situation,Elvis runs first and slows down upon entering the water. Consequently, the (recipro-cal) index of refraction is less than 1. As in case I, the family of lines containing the


refracted rays has an envelope. The calculation of the envelope is the same; the onlydifference is a sign change. The envelope is

−

xs√

1−r2

2/3

+

(ysr

)2/3

= 1.

As in case I, this is a familiar envelope, namely, the evolute of a hyperbola. The gen-eral result in this regard is the following. The hyperbola ( x

a )2− (

yb )

2= 1 has evolute

( xA )

2/3− (

yB )

2/3= 1 where Aa = Bb = a2

+ b2. This seemingly small difference be-tween the two cases of the refraction problem, in fact, translates into a large differencein terms of finding a geometric/constructive solution. There is no trammel (or rollingcircle) mechanism to generate a hyperbola.

All is not lost! If the problem is turned upside down with U as Elvis’ starting pointand S as his destination, then the slower medium is above and the faster medium isbelow. This is case I, which the trammel/hypotrochoid machine solves. So, while it’sinteresting that a hyperbola and its evolute is the result when the slower medium isbelow, it turns out that every refraction problem is solved by case I.

Summary. The refraction problem, well-known in calculus and physics, continues to revealnew insights. This paper presents a geometric solution in which the trammel of Archimedesplays the prominent role. When properly configured, the trammel generates an ellipse and itsfamily of normal lines. One normal line in particular intersects the boundary separating themedia, the intersection point being the solution. The trammel, implemented as a rolling circlemechanism, can be constructed physically and/or programmed as a computer animation.

References

1. M. Bolt and D. C. Isaksen, Dogs don’t need calculus, College Math. J. 41 (2010) 10–16; available at http://dx.doi.org/10.4169/074683410X475074

2. J. W. Bruce and P. J. Giblin, Curves and Singularities, 2nd ed., Cambridge University Press, Cambridge, 1992.3. A. Gray, Modern Differential Geometry of Curves and Surfaces, Studies in Advanced Mathematics, CRC

Press, Boca Raton FL, 1993.4. T. J. Pennings, Do dogs know calculus?, College Math. J. 34 (2003) 178–182; available at http://dx.doi.

org/10.2307/3595798

5. I. R. Porteous, Geometric Differentiation for the Intelligence of Curves and Surfaces, Cambridge UniversityPress, Cambridge, 1994.

6. J. W. Rutter, Geometry of Curves, Chapman & Hall Mathematics Series, CRC Press, Boca Raton FL, 2000.7. G. F. Simmons, Calculus with Analytic Geometry, 2nd ed., McGraw-Hill, New York,1996.8. R. C. Yates, Curves and Their Properties, Classics in Mathematics Education, vol. 4, National Council of

Teachers of Mathematics, Washington DC, 1974; reprint, J. W. Edwards, Ann Arbor MI, 1959.


http://dx.doi.org/10.4169/074683410X475074

http://dx.doi.org/10.4169/074683410X475074

http://dx.doi.org/10.2307/3595798

http://dx.doi.org/10.2307/3595798

do dogs know the trammel of archimedes?

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