Chapter 4
EATING IN THE CITY
The first time I visited a really big city on more than a day trip, I wondered where people found groceries. I had just left home to start my time as an undergraduate in London, but at home I could either ride my bicycle or use my father’s car to drive to the village store or visit my friends (not always in that order). In London I used the “tube”—the underground transport system (the subway) and sometimes the bus, but I walked pretty much everywhere else. I soon found out where to get groceries, and in retrospect realized that there are some quite interesting and amusing mathematics problems associated with food, whether it’s in a city or in the middle of the country. What follows is a short collection of such items connected by a common theme—eating!
X = W: WATERMELON WEIGHT
A farmer harvested ten tons of watermelons and had them delivered by truck to a town 30 miles away. The trip was a hot and dusty one, and by the time the destination was reached, the watermelons had dried out somewhat; in fact their water content had decreased by one percentage point from 99% water by weight to 98%.
Question: What was the weight of the watermelons by the time they arrived in the town?
W1 = 0.99W1 + 0.01W1 (water weight plus pith weight) and
W2 = 0.98W2 + 0.02W2, but the pith weight is unchanged; therefore 0.01W1 = 0.02W2, and so
W2 = W1/2;
the weight of the watermelons upon arrival is now only five tons. Very surprising, but it shows that a small percentage of a large number can make quite a difference . . .
X = V: HOW MUCH OF THAT FRUIT IS FRUIT?
Suppose that Kate feels like having a healthy snack, and decides to eat a banana. Mathematically, imagine it to be a cylinder in which the length L is large compared with the radius r. Suppose also that the peel is about 10% of the radius of the original banana. Since the volume of her (now) idealized right circular banalinder (or cylinana) is πr2L, she loses 19% of the original volume when she peels it (1 − (0.9)2 = 0.81). Okay, now let’s do the same thing for a spherical orange of volume 4πr3/3. The same arguments with the peel being about 10% of the radius give a 27% reduction in the volume (1 − (0.9)3 = 0.73). We might draw the conclusion in view of this that it is not very cost-effective to buy bananas and oranges, so she turns her stomach’s attention to a peach. Now we’re going to ignore the thickness of the peach skin (which I eat anyway) in favor of the pit. We’ll assume that it is a sphere, with radius 10% of the peach radius. Then the volume of the pit is 10−3 of the volume of the original peach; a loss of only 0.1%. What if it were 20% of the peach radius? The corresponding volume loss would be only 0.8 %. These figures are perhaps initially surprising until we carry out these simple calculations [11]. But is a banana a fruit or an herb? Inquiring minds want to know.
Meanwhile, the neighbor’s hotdogs are cooking. How much of the overall volume of a hotdog is the meat? Consider a cylindrical wiener of length L and radius r surrounded by a bun of the same length and radius R = ar, where a > 1. If the bun fits tightly, then its volume is
where Vm is the volume of the wiener. If a = 3, for example, then Vb = 8Vm. But a cheap hotdog bun is mostly air; about 90% air in fact!
X = tc: TURKEY TIME
Question: How long does it take to cook a turkey (without solving an equation)?
Let’s consider a one-dimensional turkey; these are difficult to find in the grocery store. Furthermore, you may object that a spherical turkey is much more realistic than a “slab” of turkey, and you’d be correct! A spherical turkey might be a considerable improvement. However, the equation describing the diffusion of heat from the exterior of a sphere (the oven) to the interior can be easily converted by a suitable change of variables to the equation of a slab heated at both ends, so we’ll stick with the simpler version.
The governing equation is the so-called heat or diffusion equation (discussed in more detail in Appendix 10)
where T is the temperature at any distance x within the slab at any time t; κ is the coefficient of thermal diffusivity (assumed constant), which depends on the thermal properties and density of the bird; and L is the size (length) of the turkey. This equation, supplemented by information on the temperature of the turkey when it is put in the oven and the oven temperature, can be solved using standard mathematical tools, but the interesting thing for our purposes is that we can get all the information we need without doing that. In this case, the information is obtained by making the equation above dimensionless. This means that we define new variables for which (i) the dimensions of time and length are “canceled,” so to speak, and (ii) the temperature is defined relative to the interior temperature of the bird when it is fully cooked. We’ll call this temperature Tc. We’ll also define tc as the time required to attain this temperature Tc—the cooking time. It is this quantity we wish to determine as a function of the size of the bird. The advantage of this formulation is that we don’t have to repeat this calculation for each and every turkey we cook: indeed, as we will see, with a little more sophistication we can express the result in terms that are independent of the size of the turkey.
To proceed with the “nondimensionalization” let T′ = T/Tc, t′ = t/tc, and x′ = x/L. Using the chain rule for the partial derivatives, equation (4.2) in the new variables takes the form
Has anything at all been accomplished? Indeed it has. Since both derivatives are expressed in dimensionless form, then so must be the constant κtc/L2. Let’s call this constant a. It follows that
But the weight W of anything is (for a given mean density) proportional to its volume, and its volume is proportional to the cube of its size, that is, W ∝ L3, so L2 ∝ W2/3, from which we infer that tc ∝ W2/3. And that is our basic result: the time necessary to adequately cook our turkey is proportional to the two-thirds power of its weight. We have in fact made use of a very powerful technique in applied mathematics in general (and mathematical modeling in particular): dimensional analysis. As seen above, this involves scaling quantities by characteristic units of a system, and in so doing to reveal some fundamental properties of that system.
Where can we go from here? One possible option is to determine the unknown constant of proportionality a/κ; in principle κ can be found (but probably not in any cookbook you possess), but of course a is defined in terms of tc, which doesn’t help us! However, if we have a “standard turkey” of weight W0 and known cooking time t0, then for any other turkey of weight W, a simple proportion gives us
From this the cooking time can be calculated directly. Note that tc is not a linear function of weight; in fact the cooking time per pound of turkey decreases as the inverse cube root of weight, since
Hence doubling the cooking time t0 for a turkey of weight 2W0 may result in an overcooked bird; the result (4.4) implies that tc = 22/3t0 ≈ 1.6t0 should suffice. However, always check the bird to be on the safe side. Note that the units used here, pounds, are really pounds-force, a unit of weight. Generally, pounds proper are units of mass, not weight, but in common usage the word is used to mean weight.
X = N: HOW MANY TOMATOES ARE CONSUMED
BY CITY-DWELLERS EACH YEAR?
Of course, this is a rather ill-posed question. Are we referring to large tomatoes, which we slice nd put in salads, or those little ones that we find in salad bars? Of course, we can find both sizes, and those in between, in any supermarket or produce store. And I suspect that more tomatoes are consumed during the summer than the rest of the year, for obvious reasons. (Note that we are not including canned tomatoes used in pasta sauce.) Now big tomatoes and little tomatoes share a very import characteristic: they are both tomatoes! I’m going to work with a typical timescale of one week; that is, I might use one large tomato per week in my sandwiches, or consume more of the smaller tomatoes in the same period of time. Therefore I will take the average in the following sense. A fairly big tomato 3 inches in diameter is about 30 times as large by volume as one that is one inch in diameter, so we can use the Goldilocks principle referred to earlier—is it too large, too small, or just right? To implement this, we merely take the geometric mean of the volumes. (A reminder: the geometric mean of two positive numbers is the most appropriate measure of “average” to take when the numbers differ by orders of magnitude.) The geometric mean of 1 and 30 is about 5, so accounting for the range of sizes, we’ll work with 5 “generic” tomatoes eaten per week in the calculations below. Let’s suppose that about a third of the U.S. population of 300 million eat tomatoes regularly, at least during the summer. (I am therefore ignoring those adults and children who do not eat them by dividing the population into three roughly equal groups, again using the Goldilocks principle: fewer than 100% but more than 10% of the population eat tomatoes; the geometric mean is 
, or about 1/3.)
Therefore, if on average one third of the population eat 5 tomatoes per week, then halving this to reflect a smaller consumption (probably) during the winter months (except in Florida 
), the approximate number of tomatoes eaten every year is
that is, three billion tomatoes. We multiply this by about 2/3 (to account for the proportion of city-dwelling population in the U.S.) to get roughly two billion (with no offense to non-city dwellers, I trust).
X = Pr: PROBABILITY OF BITING INTO . . .
I love apples, don’t you? Sometimes though, they contain “visitors.” Suppose that there is a “bug” of some kind in a large spherical apple of radius, say, two inches. We will assume that it is equally likely to go anywhere within the apple (we shall ignore the core). What is the probability that it will be found within a typical “bite-depth” of the surface? Based on my lunchtime observations, I shall take this as ¾ inch, but as always, feel free to make your own assumptions.
The probability P of finding the bug within one bite-depth of the surface is therefore the following ratio of volumes (recall that the volume of a sphere is 4π/3 times the cube of its radius):
or about 76%. Ewwwhhh!