Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics (Princeton Paperbacks)
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11 to 11.25 inches in length
21.5 inches circumference (round part)
28.5 inches circumference (long part)
“…it shall have the shape of a prolate spheroid”
From these specifications, it is easy to determine the following dimensions of a football: a = 5.50 in, b = 3.40 in, L = 2a = 11.0 in, and D = 2b = 6.8 in. If we substitute these numbers into equations (26.6) and (26.9), we obtain a football's volume V and surface area A, respectively. The answers are V = 266 in3 and A = 208 in2.
PROBLEM In spite of the NFL's stipulation that the football have the shape of a prolate spheroid, it simply doesn't look like one. It does not resemble a small blimp; its ends are much too pointed.
So, using one or the other of the following profile equations of a football, set up the integrals that lead to formulas for the volume and surface area of a football:
Arc of a circle
where R = 6.15 in, b = 3.40 in, R – b = 2.75 in.
Parabola
where a = 5.50 in, b = 3.40 in.
In your analysis, equation (26.10) or (26.11) would replace (26.3), which, of course, is the equation of an ellipse. Both (26.10) and (26.11) provide nice pointed ends for the football.
Doughnuts
Toward the end of the third century, a mathematician named Pappus of Alexandria discovered two rules or relationships that are very useful for the computation of the volumes and areas of solids. Unfortunately, these discoveries of Pappus were lost for over 1,200 years. Then, in the seventeenth century, the famous German mathematician Johannes Kepler (1571-1630) rediscovered them and put them to use.
These two relationships or rules of Pappus are
Volume rule: The volume of any solid generated by the revolution of a plane area about an external axis in its plane is equal to the product of the area of the generating figure and the distance its center of gravity moves
Surface area rule: The area of any surface generated by the revolution of a plane curve about an external axis in its plane is equal to the product of the generating curve and the distance its center of gravity moves
An Application of the Relationships of Pappus
In figure 26.6, a torus—a solid shaped like a doughnut or a tire—is shown with primary radius R and secondary radius A. What is its volume V and surface area A?
Employing the volume rule, we obtain
and using the surface area rule, we get
FIG. 26.6
Definition sketch for a torus.
A NUMERICAL EXAMPLE. The largest tires ever made were manufactured by Goodyear for enormous earth-moving trucks. These tires had an outer diameter of 12 feet and weighed 12,500 pounds. Assuming an inner diameter of 6 feet, it is easily established that R = 4.5 ft and a = 1.5 ft.
From equations (26.12) and (26.13), we determine that the volume of this gigantic tire was V = 200 ft3 and the area was 266 ft2.
A Goat Tied to a Fence: The Involute of a Circle
Here is an interesting little problem to think about. A goat is tied to one end of a rope of length R. The other end of the rope is tied to a post on the outside of a circular fence of radius a. In terms of R and a, what is the total area over which the goat can graze the grass outside the fence?
The problem should be clear enough. As shown in figure 26.7, in zone I the full length of the rope, R, determines the goat's grazing region. This provides the semicircular area A1 = πR2/2.
However, in zones II the effective length of the rope is reduced because a portion of it may be wrapped along the circular fence. Indeed, R, the length of the rope, is the arc length of the fence measured from post p to either of points A. The shortening of the rope results in a reduction of the grazing area. So the problem is to determine the areas of zones II. To calculate these areas, we need to know the equation of the curve AB.
FIG. 26.7
A goat grazing on the grass outside a circular fence. Curve AB is called the involute of a circle.
FIG. 26.8
Definition sketches for the involute of a circle.
A definition sketch of our problem is displayed in figure 26.8(a). It is not difficult to show that the so-called parametric equations of the curve AB are the following:
These equations define a curve called the involute of a circle. In this regard you might want to refer to books by Lawrence (1972) and by Gellert et al. (1977).
Use of the trigonometric identity sin2θ + cos2 θ = 1 and a bit of algebra reduces equations (26.14) and (26.15) to the simple form
This is the equation of curve AB in terms of polar coordinates.
With reference to figure 26.8(b), we note that the area of the small triangle is dA = (1/2)r(rdθ). Accordingly, the area swept out by the radius r as it rotates from θ = 0 to θ = θ is
Substituting (26.16) into (26.17) and integrating gives
However, the area given by this equation includes a circular area inside the fence. We need to subtract this area. Then we utilize the relationship R = aθ0, where θ0 is the value of θ corresponding to point B. This gives the area of one of the zones II. Since there are two such zones, we finally get the area of zones II, A2 = R3/3a. Adding this to the area of zone I provides the final answer:
This is the total area over which the goat can graze grass.
Suppose, for example, that the length of the rope is R = πa (i.e., the rope winds entirely around to the opposite side of the fence). Then A = (5/6)πR2. This is a rather interesting result.
More Information about the Involute of a Circle
By definition, the involute of a circle is the curve traced by the end of a string as it is tautly unwound from a circular spool of radius a. The equation of the curve, expressed in polar coordinates, is given by (26.16).
It is easy to determine the length of this curve. Over a very short distance, the length is given by . Expressing this in terms of the parameter θ, we have
Substituting the derivatives of equations (26.14) and (26.15) into this expression yields ds = aθdθ. Integrating this relationship from θ = 0 to θ = θ yields the very simple answer,
where S is the length of the curve.
EXAMPLE 1. In our goat and fence problem, suppose that a = 10 ft and that R = πa (i.e., the rope extends halfway around the fence). So θ= 180° = π radians. Then, with reference to figure 26.7 and using (26.21), we determine that the length of the curve from A to B is S = 49.3 ft.
EXAMPLE 2. Suppose a thread is tautly unwound from a spool with radius r = 1.0 in. What is the length of the curve traced by the end after two turns of thread from the spool? Since one turn of the thread corresponds to 360° = 2π radians, then θ = 2(2π) = 4π. Substituting this into (26.21), with a = 1.0 in, gives S = 79.0 in. A graphical display of this example is presented in figure 26.9.
The general definition of an involute is that it is the curve traced by the end of a thread as it is tautly unwound from a given curve.
Some examples: (a) the involute of a catenary is a tractrix, (b) the involute of a parabola is a semicubic parabola, (c) the involute of a logarithmic spiral is an identical logarithmic spiral, (d) the involute of a cycloid is an identical cycloid, and so on.
Along these lines, Hoffman (1998) presents an interesting analysis of a generalization of the goat and circular fence problem. He provides the solution corresponding to a fence defined by any smooth convex curve. As an example, he computes the grazing area for the case in which the fence has the shape of a catenary.
FIG. 26.9
The involute of a circle. a = 1.0 in. Small circles identify 30° intervals. Length of curve S = 79.0 in.
Mayon Volcano and the Hershey Chocolate Kiss
We have two more problems; they both involve integral calculus. The first is to compute the volume of the Mayon Volcano in the Philippines and the second is to calculate the volume of a Hershey chocolate kiss. Mathematically speaking, the two problems are essentially the same.
In chapter 20, we consider the geometry of the Mayon Volcano. This beautifu
l active volcano has a height of 2,460 meters and covers an area of about 300 square kilometers near sea level in the southern part of Luzon in the Philippines.
This mountain is so beautifully symmetric that, in the past, numerous mathematical equations have been proposed to describe its shape. One such equation is
FIG. 26.10
Mathematical profile of the Mayon Volcano.
where, as shown in figure 26.10, r is the horizontal distance from the vertical y-axis, y0 = 2,460 m is the elevation of the peak, y is the elevation at horizontal distance r, and c = 800 m is a constant. A plot of equation (26.22) is shown in figure 26.10.
Here is our question: assuming that equation (26.22) describes the profile of the Mayon Volcano, what is the volume of the mountain?
Once again we have a problem in integral calculus. The volume of the thin disk shown in figure 26.10 is dV = πr2dy. If we add up the volumes of all the thin disks from y = 0 to y = y0, we have the volume of our mountain. That is,
Substituting (26.22) into (26.23) gives
FIG. 26.11
Mathematical profile of a chocolate kiss.
At this point, we need a table of integrals to evaluate the right-hand side. Without much difficulty we obtain the following result:
Substituting y0 = 2,460 m = 2.460 km and c = 800 m = 0.80 km into this equation provides the desired answer: V = 91.7 km3. This is about 22.0 cubic miles.
Hershey Chocolate Kiss
A similar analysis is made to determine the volume of a chocolate kiss. The profile of the kiss is shown in figure 26.11.
Results of Laboratory Work
Some preliminary laboratory work was carried out involving a sack of Hershey's milk chocolate kisses containing 35 kisses. Utilizing a measuring cup, it was determined that the 35 kisses displaced 140 cubic centimeters of water. So the measured volume of each kiss was V = 4.0 cm3. In addition, the following geometrical features of the kiss were determined:
1. Radius of the kiss at its base (y = 0, r = 1.05 cm)
2. Height of the maximum thickness point (y = 0.3 cm)
3. Height of the inflection point (y = 1.3 cm)
4. Height of the kiss (y = 2.2 cm, r = 0)
These four points are shown as the solid dots in figure 26.11.
The Mathematical Model
Several mathematical functions were examined as possible shape equations for the chocolate kiss. They were either inadequate or too complicated. So the following polynomial expression was employed:
This equation contains four constants which are easily determined from the four measurements indicated above.
For example, the first derivative of equation (26.26) is dr/dy = k1 + 2k2y + 3k3y2. Since this quantity represents the slope of the profile, we know that dr/dy = 0 at the maximum thickness point. Likewise, the second derivative is d2r/dy2 = 2k2 + 6k3y. This quantity represents the curvature of the profile and this must be zero at the inflection point. In this way, the numerical values of the four constants were computed. The results are k0 = 1.05, k1 = 0.592, k2 = –1.115, and k3 = 0.286.
We construct our integral calculus problem in the same way as in our analysis of Mount Mayon. The volume of the thin disk shown in figure 26.11 is dV = πr2dy. Therefore, the total volume is
We now substitute equation (26.26) into (26.27) and integrate term by term. The algebra is somewhat complicated, but there is nothing difficult. For such integration it is helpful to know that
where C is a constant of integration. This is the general rule for determining the integral of a variable x when raised to a power n. This rule fails when n = –1. In this case, we have
The final algebraic answer provided by integrating (26.27) is a lengthy expression containing ascending powers of y0, the height of the kiss, and various combinations of the four numerical constants.
The numerical result of our mathematical model and computation is V = 4.15 cm3. This is remarkably close to the measured value of the volume of a Hershey's chocolate kiss, V = 4.0 cm3.
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