farmer.dvi


Integre Technical Publishing Co., Inc. College Mathematics Journal 33:2 November 21, 2001 2:33 p.m. farmer.tex page 131

Designing a Calculus Mobile
Tom Farmer

Tom Farmer (farmerta@muohio.edu) received his Ph.D. in
mathematics at the University of Minnesota in 1976 and
ever since then has served on the faculty at Miami
University in Oxford, Ohio. In addition to gardening,
reading, and digital photography, he continues to enjoy
learning mathematics and how to teach it. The present
paper grew out of discussions of the center of mass in an
honors calculus class.

Problem: Design a mobile as in Figure 1. The parts are horizontal slices of a lamina
bounded on the left and right by smooth curves x = f (y) and x = m f (y), where
m > 1 is constant. Each part of the mobile is attached to the one above it by a single
connector located along the graph of f , and the parts are intended to hang with their
horizontal edges horizontal.

Figure 1.

This problem is connected with several topics from the latter part of first year cal-
culus including the harmonic series, the center of mass of a lamina, and separable
differential equations. The purpose of this paper is to show the connections and, in the
end, to solve the problem.

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Integre Technical Publishing Co., Inc. College Mathematics Journal 33:2 November 21, 2001 2:33 p.m. farmer.tex page 132

Many calculus instructors have used a stack of blocks to illustrate the question
of convergence or divergence of a series. Suppose n identical rectangular blocks are
stacked as in Figure 2, with the top block extending half its length out from the second
block, the second block extending one fourth its length out from the third block, then
one sixth, then one eighth, and so on. As n gets large, is there an upper bound on the
horizontal distance between the leading edge of the top block and the leading edge of
the bottom block? The answer is no, of course. The horizontal distance grows without
bound because the series

∑∞
n=1 1/2n diverges.

Figure 2.

An interesting, but somewhat hidden, aspect of the block-stacking demonstration
is the fact that the blocks are positioned with a certain center of mass property. For
example, the center of mass of the top two blocks (the two blocks being treated as
one object) is directly above the edge of the third block. In general, the collective
center of mass of the top n blocks lies directly above the edge of the next block for
any n—the stack is extending out as far horizontally as possible without toppling over.
In discussing this center of mass property, notice that, although the stack of blocks
is three dimensional, we need only look at the two dimensional side view shown in
Figure 2. As a planar lamina with constant density 1, the object represented by the
side view of the stack has the property that its center of mass (x̄, ȳ) has

x̄ = 1
2

+ 1
4

+ 1
6

+ · · · + 1
2n

,

and this is exactly where the edge of the next rectangle is positioned. This formula
for x̄ is easily proved by induction. In the induction step, we need to calculate the
moment My of n + 1 rectangles about the y-axis. But, using the induction hypothesis,
the top n rectangles, treated as one lamina, has moment (mass times distance) equal to
n( 1

2
+ 1

4
+ 1

6
+ · · · + 1

2n
). The added rectangle at the bottom, with its center ( 1

2
+ 1

4
+

1
6

+ · · · + 1
2n

) + 1
2

units from the y-axis, has moment given by this same numerical
value since its mass is 1. Thus, the moment of the n + 1 rectangles is the sum

n

(
1

2
+ 1

4
+ 1

6
+ · · · + 1

2n

)
+
(

1

2
+ 1

4
+ 1

6
+ · · · + 1

2n

)
+ 1

2

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Integre Technical Publishing Co., Inc. College Mathematics Journal 33:2 November 21, 2001 2:33 p.m. farmer.tex page 133

= (n + 1)
(

1

2
+ 1

4
+ 1

6
+ · · · + 1

2n

)
+ 1

2

= (n + 1)
(

1

2
+ 1

4
+ 1

6
+ · · · + 1

2n
+ 1

2(n + 1)
)

.

It follows that x̄ = My
M

= 1
2

+ 1
4

+ 1
6

+ · · · + 1
2(n+1) , as desired.

A good way to model physically the kinds of problems we wish to consider is to
think of a hanging mobile rather than a stack of blocks because then the center of mass
property is more apparent. Using a thin but rigid material such as cardboard or foam
board, we cut identical rectangles of unit length and hang them in a chain as in Figure 3.
The group of n rectangles at the bottom of the chain (for each n) is connected to the
lower left corner of the rectangle above by a single connector that is directly above the
center of mass of the group. With the connector at this point the group hangs with its
edges horizontal and vertical.

Figure 3.

One of the themes of first year calculus is that regions with curved boundaries can be
approximated by unions of rectangles; how about reversing the idea? Imagine that for
some large n we construct a mobile consisting of n rectangles of length 1 and width
1/n. We position the mobile in a coordinate system with the upper left hand corner
of the top rectangle at 1 on the y-axis and the lower left hand corner of the bottom
rectangle at −( 1

2
+ 1

4
+ 1

6
+ · · · + 1

2(n−1) ) on the negative x -axis as in Figure 4. Since

Figure 4.

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Integre Technical Publishing Co., Inc. College Mathematics Journal 33:2 November 21, 2001 2:33 p.m. farmer.tex page 134

n is large, this lamina is suggestive of one that is bounded between smooth curves. We
could ask what happens as n approaches infinity in this picture but, instead, we just
want to motivate the idea of using a lamina with curved boundaries and a center of
mass property consistent with what was used for a union of rectangles (see Figure 5).
Thus, we seek a lamina bounded on the left and right by x = f (y) and x = f (y) + 1,
where f is a differentiable function on (0, 1), continuous at 1, and with f (1) = 0. The
appropriate assumption on the center of mass is that if we make a horizontal cut at
y = t , for any t in (0, 1], then the part of the lamina that lies below this line should
have center of mass (x̄, ȳ) with x̄ = f (t ). What is f ? We find it as follows:

f (t ) = x̄ = My
M

=
∫ t

0
f (y)+ f (y)+1

2
(1) d y∫ t

0
(1) d y

=
∫ t

0

(
f (y) + 1

2

)
d y

t
.

Multiplying both sides by t and differentiating with respect to t yields

f (t ) + t f ′(t ) = f (t ) + 1
2
.

Thus, f ′(t ) = 1
2t

and f (t ) = 1
2

ln t + C = 1
2

ln t , since f (1) = 0. Given the connec-
tion of this problem with the series

∑∞
n=1 1/2n, our formula for f is not surprising.

However, it doesn’t seem to have been easily predictable either.
The region bounded between the graphs of x = f (y) = 1

2
ln y and x = g(y) =

1
2

ln y + 1 (y ∈ (0, 1]) causes a problem when it comes to constructing a physical
model—it is unbounded. Of course, we could cut the tail off, but the resulting lamina
may not be convincing as a representation of the center of mass property that was
promised. This drawback motivates the problem, stated at the outset, in which we
require a lamina bounded between the graphs of x = f (y) and x = g(y) = m f (y),
with m > 1 and f (0) = g(0) = 0. In order to set up and solve a differential equation
satisfied by f , we assume that f is non-negative and twice differentiable on (0, 1) with
f ′ positive. We continue to require the center of mass property just as in Figure 5: for
any horizontal cut at y = t , with t in (0, 1], the part of the lamina that lies below this
line should have center of mass (x̄, ȳ) with x̄ = f (t ).

Figure 5.

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Integre Technical Publishing Co., Inc. College Mathematics Journal 33:2 November 21, 2001 2:33 p.m. farmer.tex page 135

Since we require x̄ = My
M

= f (t ), then

f (t ) =
∫ t

0
f (y)+m f (y)

2
(m f (y) − f (y)) d y∫ t

0 (m f (y) − f (y)) d y
=

1+m
2

∫ t
0 ( f (y))

2 d y∫ t
0

f (y)d y
,

so

f (t )
∫ t

0
f (y) d y = 1 + m

2

∫ t
0

( f (y))2 d y.

Taking the derivative with respect to t on both sides,

f ′(t )
∫ t

0
f (y) d y + ( f (t ))2 = 1 + m

2
( f (t ))2.

Finally, if we collect terms, isolate the remaining integral, and differentiate once more,
then we obtain the differential equation

f (t ) = m − 1
2

(
2 f (t ) ( f ′(t ))2 − ( f (t ))2 f ′′(t )

( f ′(t ))2

)

or, equivalently,

(m − 1) f (t ) f ′′(t ) = (2m − 4) ( f ′(t ))2 . (1)
In order to solve this second-order nonlinear differential equation, we need a trick.

Note that (1) can be written as

(m − 1) f
′′(t )

f (t )
= (2m − 4)

(
f ′(t )
f (t )

)2
. (2)

So let u = f ′
f

and then u′ = f ′′ f −( f ′)2
f 2

= f ′′
f

− u2. In this way, (2) becomes

(m − 1)(u′ + u2) = (2m − 4)u2

giving us the separable first order equation

u′

u2
= m − 3

m − 1 = −Q. (3)

The general solution of (3) is u = 1
Qt+C = f

′(t)
f (t)

(for t > 0). Integrating again yields

f (t ) = D(Qt + C)1/Q and the condition f (0) = 0 determines C = 0. Finally, D
could be replaced by D/Q1/Q to provide the simpler form f (t ) = Dt 1/Q . The other
boundary of the lamina is g(t ) = m f (t ) = m Dt 1/Q , where m and Q are linked by (3)
or, equivalently, m = 3+Q

1+Q . As an example, the lamina in Figure 1 was constructed us-
ing Q = 2 and m = 5/3 so the boundary curves are x = √y and x = (5/3)√y. Other
examples of interest may include the cases

(a) Q = 1, so m = 2 and the region is bounded by x = Dy and x = 2 Dy; and
(b) Q = 1/3, so m = 5/2 and the region is bounded by x = Dy3 and x =

(5/2) Dy3.

VOL. 33, NO. 2, MARCH 2002 THE COLLEGE MATHEMATICS JOURNAL 135



Integre Technical Publishing Co., Inc. College Mathematics Journal 33:2 November 21, 2001 2:33 p.m. farmer.tex page 136

Acknowledgment. The author would like to thank Fred Gass for his contributions to this
paper.

Reference

1. Sydney C. K. Chu and Man-Keung Siu, How far can you stick out your neck?, College Math Journal 17
(1986), 122–132.

136 c© THE MATHEMATICAL ASSOCIATION OF AMERICA