18.01 Problem Set 4 solutions
Part II: 15 points
1a) Do problem 2A-5 from the supplementary notes.
I will write out an argument, since the point of the rest of the problem is to criticize the argument. The
central idea is that if you expand a solid without changing its shape, the volume is proportional to the cube
of the height (or the width, or the waistline, or any other linear measure). If the density of humans is more
a less a universal constant (a bit less than the density of water), it follows that the mass of a human (of a
fixed shape) should be proportional to the cube of the height:
m(h) = Ch3 .
Now we vary h a little near some basic height h0 :
h = h0 + ∆h.
Since the derivative of m is 3Ch2 , the linear approximation to m near h0 is
m(h0 + ∆h) ≈ m(h0 ) + 3Ch20 ∆h = m(h0 ) + 3Ch30
or
∆h
,
h0
∆h
∆m
.
=3
m(h0 )
h0
That is, the relative change in mass (the dimensionless quantity change in mass divided by mass) should be
three times the relative change in height.
Changing from 60 inches to 61 inches is a relative change in height of 1/60, so it should correspond to a
relative change in mass of 3/60 = 1/20. If the mass at height 60 inches is 120 pounds, then a relative change
of 1/20 is 6 pounds; so the conclusion is that the mass for a height of 61 inches should be 126 pounds.
b) Look on the web to find some kind of table describing average weights as a function of
height. (It’s easy to find some old “ideal weight” tables like this, and one of those would be
OK; it would be more interesting to find tables recording averages of measurements of real
people.) Write the URL where you found the table, and the weights in the table associated
to heights of 60 and 61 inches. (If the table you found doesn’t include those heights, or is
calibrated in centimeters and kilograms, then you should also repeat the calculation in 1(a) to
guess the number for the larger height from the number for the smaller height.)
There is a table recording maximum allowed weights as a function of height for members of the US
Army at
http://usmilitary.about.com/od/army/l/blmaleweight.htm
or in the file AR-600-9 at
http://www.army.mil/usapa/epubs/600 Series Collection 1.html
Some entries near h = 60 are
m(60) = 132,
m(61) = 136,
m(62) = 141.
c) Is the data you found compatible with the result of 2A-5 (also available in the supplementary notes)? Can you offer an explanation for any difference?
The approximation from 1(a) would have suggested a value close to 132(1 + 3/60) ≈ 139 for m(61) and
(132)(1 + 6/60) ≈ 145 for m(62).
Since the weight figures are given as a whole number of pounds, it is better to think of the changes in
mass as ranges like (going from 60 inches to 61 inches)
(3 to 5)/(131.5 to 132.5) = (1.36 to 2.28)(1/60).
No matter how generously these numbers are interpreted, they are not consistent with the idea that mass
should be proportional to the cube of height; the factor that ought to be three is more like two in the tables,
suggesting that mass is proportional to the square of height. (For example, the chart has four weights listed
for a height of 60 inches, corresponding to various minimum and maximum allowed weights. Each entry for
a height of 80 inches can be obtained by multiplying the entry for 60 by (80/60)2 = 1.777 . . ., to within a
pound or two.
I don’t know any good explanation of this; it says that taller people are relatively skinnier, but I don’t
know enough about physiology to guess why. There is a lot of interesting biology done about the relative
sizes of various organs in similar animals of different dimensions. (Brain mass in similar animals, for example,
is closer to being proportional to the square of height than to the cube. This discovery led to a reevaluation
of the intelligence of dinosaurs, whose brains suddenly looked comparable to those of modern reptiles. Read
about it in Harry Jerison, “Brain evolution and dinosaur brains,” The American Naturalist 103 (1969).)
2a) Three chess players A, B, and C live on the x axis, at coordinates 0, 1, and 20. They plan
to hold a chess club meeting at some coordinate x along the axis (to be chosen). How should
they choose x so as to minimize the total distance they need to travel to get to the meeting?
Players A and C will have to travel a total of at least 20 to get to the meeting, wherever it is held. (If
it’s held to the left of 0 or to the right of 20, then they will need to travel even further.) So the total travel
needed is at least 20. If the meeting is held at 1, then B need not travel at all; so the total travel required
is 20, and this is the unique minimum.
b) Same as (a), but now the coordinates of the chess players are just given as three numbers
a, b, and c; and your answer should be expressed in terms of a, b, and c.
One of the three players must live between the other two. The meeting should be held at the middle
player’s coordinate:
(
a if b ≤ a ≤ c or c ≤ a ≤ b
x = b if a ≤ b ≤ c or c ≤ b ≤ a
c if a ≤ c ≤ b or b ≤ c ≤ a.
The reasoning is the same as for (a).
c) Same as (b), but now minimize the sum of the squares of the travel distances for the three
chess players.
The function to be minimized is
f (x) = (x − a)2 + (x − b)2 + (x − c)2 = 3x2 − 2(a + b + c)x + (a2 + b2 + c2 ).
This is an upward pointing parabola, so it has a unique minimum at its unique critical point
0 = f ′ (x) = 6x − 2(a + b + c),
x = (a + b + c)/3.
The meeting should be held at the coordinate given by the average of the three.
3. The N students in a class take an exam, receiving scores of a1 , a2 ,. . . , aN . The professor
wishes to report to the class some number x that reflects a typical score.
a) Suppose that the professor wants to report a number that minimizes the sum of the
absolute values of the differences from actual scores. This means that if he reports x, he
wants to minimize
M (x) = |x − a1 | + |x − a2 | + · · · + |x − aN |.
What value of x minimizes M (x)? (Hints: this problem is more or less the same as Problem
2, except that 3 has become N . Solving it by calculus is hard, because the absolute value
function doesn’t have a derivative at 0. The answer to this question is not difficult or deep,
but finding that answer is not something you “should” be able to do.)
The function M is continuous, and it tends to +∞ as x tends to ±∞. So it has a minimum value. The
function |x| is differentiable except at 0, with derivative +1 for positive x and −1 for negative x. So M is
differentiable except at x equal to one of the ai . Its derivative is
M ′ (x) = number of ai smaller than x − number of aj bigger than x.
So M (x) is decreasing as long as more than half of the ai are smaller than x, and increasing when more than
half of the aj are bigger than x. (If N is even, then there can be an interval between the two middle ai ; on
this interval M ′ is zero.)
The conclusion is that if we arrange the ai in increasing order, then the minimum value of M is attained
exactly at the middle one x = a(N +1)/2 if N is odd, and all along the middle interval aN/2 ≤ x ≤ a(N/2)+1
if N is even. In particular, the minimum is attained at the “median” of the ai , which is defined to be the
midpoint (aN/2 + a(N/2)+1 )/2 of the middle interval if N is even.
b) Suppose that the professor wants to report a number that minimizes the sum of the squares
of the differences from actual scores. This means that if he reports x, he wants to minimize
A(x) = (x − a1 )2 + (x − a2 )2 + · · · + (x − aN )2 .
What value of x minimizes A(x)?
This question appears a couple of times in the text (4.4/30 and 4.additional/66). If you expand all the
squares in the definition of A(x), you find
A(x) = (x2 − 2a1 x + a21 ) + (x2 − 2a2 x + a22 ) + · · · (x2 − 2aN x + a2N )
= N x2 − 2(a1 + a2 + · · · + aN )x + (a21 + a22 + · · · + a2N ).
This is clearly a parabola, opening up since N > 0; so there is exactly one critical point, which is a minimum.
The derivative is
A′ (x) = 2N x − 2(a1 + a2 + · · · + aN ),
so the critical point occurs when
N x = (a1 + a2 + · · · + aN ).
This is the same as
x = (a1 + a2 + · · · + aN )/N.
That is, the number minimizing the sum of the squares of the differences is the average of the scores ai .
4. This problem is about the function cos(x) and its quadratic approximation at zero Q(x) =
1 − x2 /2.
a) Sketch on the same axes the graphs of cos and of Q in the range −π ≤ x ≤ π.
b) Is it true that cos(x) ≥ Q(x) for all x?
Look at the function f (x) = cos(x) − Q(x). The question asks whether f is always non-negative.
Certainly f (0) = 0; so if we can show that
f is increasing for positive x and decreasing for negative x,
(∗)
then it will follow that f is always non-negative. (That’s more or less what the graph suggests: that f
decreases toward 0 as x approaches zero from below, then increases away from 0 as x moves away from 0 in
the positive direction.)
Now (∗) is a question about the derivative f ′ (x) = − sin(x) + x: we want to know that f ′ is negative
for negative x, and positive for positive x. Since f ′ (0) = 0, it is enough to show that
f ′ is increasing for all x.
(∗∗)
(This sounds like we’re going in circles, but it’s progress because f ′ is a simpler function than f .)
Now (∗∗) is a question about the derivative of f ′ , which is f ′′ (x) = − cos x + 1. Since cos is always
between −1 and 1, we deduce that
0 ≤ f ′′ (x) ≤ 2;
so f ′ is always increasing, as we wished to show. The conclusion is (first) that f ′ (x) has the same sign as x,
and (second) that f is always non-negative. So it’s true.