Chapter 16
Solutions to Exercises
16.1 Introduction
No exercises
16.2 Symmetry and Physics
Exercise 1. Calculate the length D of roadway in Fig. 2.3 as a function
of l and a, and verify that it is minimized by the value of l given in Eq.(2.1)
above.
Solution:
Each diagonal segment on the right hand side of Fig. 2.3 has length:
s
a 2 a − l 2
d(l) = + (16.1)
2 2
The total length of pavement D(l) as a function of l is:
s
a 2 a − l 2
D(l) = l + 4d = l + 4 + (16.2)
2 2
496
,CHAPTER 16. SOLUTIONS TO EXERCISES 497
To find the value of lmin that minimizes the function D(l) we have to solve:
dD(l) 2(a − l)(−1)
= 1+ p
dl lmin a2 + (l − a)2
= 0 (16.3)
A bit of algebra yields:
a2
(a − lmin )2 = (16.4)
3
Since l must be less than a the relevant solution is:
1
lmin = a 1 − √ (16.5)
3
Plugging this into the expression for D(l) we find:
1
D(lmin ) = a 1 + √ (16.6)
3
which is less than the length of pavement required by building two diagonal
roads directly through the center of the square, namely:
√ √
Ddiagonals = 2 a2 + a2 = 2 2a (16.7)
A straightforward calculation verifies that
d2 D(l)
<0 (16.8)
dl2 lmin
so that lmin is indeed a minimum as required.
Exercise 2. Consider three towns, called N (for North), SW and SE,
respectively, located a distance a apart at the vertices of an equilateral tri-
angle, as shown in Fig.[2.4]. We wish to build a network of roads connecting
all three towns such that the roads (shown in blue) meet at an arbitrary
point P along the line joining the center of the triangle to the northern town
at some distance l from N . Show that the minimum total length for such a
,CHAPTER 16. SOLUTIONS TO EXERCISES 498
√
configuration of roads occurs when l = a/ 3, so that the least expensive way
to join the towns is to have the three segments of road meet at the center,
C. Is there symmetry breaking in this case? Explain.
Hint: Use the law of cosines to figure out the distance from P to the other
two towns SW and SE. This should give you an expression for the total
length of the three roads that join P to N , SW and SE. Finally, minimize
the total length of pavement with
√ respect to the parameter l and show that
the minimum occurs at l = a/ 3.
Figure 16.1: Three towns to be joined by shortest road.
Solution:
Consider the triangle with vertices N-P-SE. The angle at each vertex in the
equilateral triangle is 60◦ , so that the angle between the lines N-P and N-SE
is 30◦ . Applying the cosine law:
d2 (l) = l2 + a2 − 2al cos(30)
√
q
→ d(l) = l2 + a2 − 3al (16.9)
The total length of road D(l) is:
D(l) = 2d(l) + l
√
q
= 2 l2 + a2 − 3al + l (16.10)
, CHAPTER 16. SOLUTIONS TO EXERCISES 499
We want to find the value lmin of l that minimizes D(l):
√
dD(l) 2l − 3a
= 2 p √ +1
dl lmin 2 l2 + a2 − 3al
= 0q
√ √
2
→ 2lmin − 3a = lmin + a2 − 3al (16.11)
Squaring both sides of the above gives a quadratic equation for lmin in terms
of a whose solution is:
√
(3 ± 1) 3
lmin = a (16.12)
2
Since we need lmin to be less than a the physical solution is the minus sign,
so that: √
lmin = 3a (16.13)
A straightforward calculation verifies that
d2 D(l)
>0 (16.14)
dl2 lmin
so that lmin is indeed a minimum as required.
Exercise 3. Derive Equation (2.2) using the diagram in Fig. 2.5. The
distances a, b and c are given. You can use the fact that the total time Ttot
taken along Path 2 is:
p p
c21 + b2 c22 + a2
Ttot = + (16.15)
VG VS
and find the equation for c1 that minimizes Ttot . Hint: in order to solve the
resulting equation, you may need:
c1
= tan(θG )
b
c2
= tan(θS ) (16.16)
a
Solutions to Exercises
16.1 Introduction
No exercises
16.2 Symmetry and Physics
Exercise 1. Calculate the length D of roadway in Fig. 2.3 as a function
of l and a, and verify that it is minimized by the value of l given in Eq.(2.1)
above.
Solution:
Each diagonal segment on the right hand side of Fig. 2.3 has length:
s
a 2 a − l 2
d(l) = + (16.1)
2 2
The total length of pavement D(l) as a function of l is:
s
a 2 a − l 2
D(l) = l + 4d = l + 4 + (16.2)
2 2
496
,CHAPTER 16. SOLUTIONS TO EXERCISES 497
To find the value of lmin that minimizes the function D(l) we have to solve:
dD(l) 2(a − l)(−1)
= 1+ p
dl lmin a2 + (l − a)2
= 0 (16.3)
A bit of algebra yields:
a2
(a − lmin )2 = (16.4)
3
Since l must be less than a the relevant solution is:
1
lmin = a 1 − √ (16.5)
3
Plugging this into the expression for D(l) we find:
1
D(lmin ) = a 1 + √ (16.6)
3
which is less than the length of pavement required by building two diagonal
roads directly through the center of the square, namely:
√ √
Ddiagonals = 2 a2 + a2 = 2 2a (16.7)
A straightforward calculation verifies that
d2 D(l)
<0 (16.8)
dl2 lmin
so that lmin is indeed a minimum as required.
Exercise 2. Consider three towns, called N (for North), SW and SE,
respectively, located a distance a apart at the vertices of an equilateral tri-
angle, as shown in Fig.[2.4]. We wish to build a network of roads connecting
all three towns such that the roads (shown in blue) meet at an arbitrary
point P along the line joining the center of the triangle to the northern town
at some distance l from N . Show that the minimum total length for such a
,CHAPTER 16. SOLUTIONS TO EXERCISES 498
√
configuration of roads occurs when l = a/ 3, so that the least expensive way
to join the towns is to have the three segments of road meet at the center,
C. Is there symmetry breaking in this case? Explain.
Hint: Use the law of cosines to figure out the distance from P to the other
two towns SW and SE. This should give you an expression for the total
length of the three roads that join P to N , SW and SE. Finally, minimize
the total length of pavement with
√ respect to the parameter l and show that
the minimum occurs at l = a/ 3.
Figure 16.1: Three towns to be joined by shortest road.
Solution:
Consider the triangle with vertices N-P-SE. The angle at each vertex in the
equilateral triangle is 60◦ , so that the angle between the lines N-P and N-SE
is 30◦ . Applying the cosine law:
d2 (l) = l2 + a2 − 2al cos(30)
√
q
→ d(l) = l2 + a2 − 3al (16.9)
The total length of road D(l) is:
D(l) = 2d(l) + l
√
q
= 2 l2 + a2 − 3al + l (16.10)
, CHAPTER 16. SOLUTIONS TO EXERCISES 499
We want to find the value lmin of l that minimizes D(l):
√
dD(l) 2l − 3a
= 2 p √ +1
dl lmin 2 l2 + a2 − 3al
= 0q
√ √
2
→ 2lmin − 3a = lmin + a2 − 3al (16.11)
Squaring both sides of the above gives a quadratic equation for lmin in terms
of a whose solution is:
√
(3 ± 1) 3
lmin = a (16.12)
2
Since we need lmin to be less than a the physical solution is the minus sign,
so that: √
lmin = 3a (16.13)
A straightforward calculation verifies that
d2 D(l)
>0 (16.14)
dl2 lmin
so that lmin is indeed a minimum as required.
Exercise 3. Derive Equation (2.2) using the diagram in Fig. 2.5. The
distances a, b and c are given. You can use the fact that the total time Ttot
taken along Path 2 is:
p p
c21 + b2 c22 + a2
Ttot = + (16.15)
VG VS
and find the equation for c1 that minimizes Ttot . Hint: in order to solve the
resulting equation, you may need:
c1
= tan(θG )
b
c2
= tan(θS ) (16.16)
a
