2
Figure 2. To nd
p
1
0
1
p
2 2
2 on the real line you draw a square of sides 1 and drop the diagonal onto the real line.
Almost every equation involving variables x, y, etc. we write down in this course will be true for some
values of x but not for other
By the chain rule we therefore get
dax
d2xlog2 a
=
dx
dx
dx log2 a
dx
xlog2 a
= (C log2 a) 2
= C 2xlog2 a
= (C log2 a) ax .
So the derivative of ax is just some constant times ax , the constant being C log2 a. This is essentially our
formula for the deriv
Solutions
a) path: b-a-c-e-d; no circuit: rule 1 at b, c, d yields 3 edges at a and e,
b) path: a-d-g-h-e-b-c-f-i, no circuit: rule 1 at d, e, f, by symmetry at b choose b-a
and delete b-c (rule 3), at c by rule 1 must use c-a, and now a has degree 3,
c)
Solutions
13. A directed graph has an Euler trail if and only if at all but two vertices indegree =
outdegree and at those two, indegree and outdegree differ by one; proof: add on extra edge
so that indegree = outdegree at two unbalanced vertices and resu
Solutions
13. A directed graph has an Euler trail if and only if at all but two vertices indegree =
outdegree and at those two, indegree and outdegree differ by one; proof: add on extra edge
so that indegree = outdegree at two unbalanced vertices and resu
Solutions
b) each edge of Kn is incident to n-2 other edges at each endvertex, 2(n-2) incidences in all,
c) can be no vertices of degree 0 or 1 and so all degrees 2; since e =
1
2
deg, then to have
e = v (so that G and L(G) to have same number of vertices
Solutions
13. A directed graph has an Euler trail if and only if at all but two vertices indegree =
outdegree and at those two, indegree and outdegree differ by one; proof: add on extra edge
so that indegree = outdegree at two unbalanced vertices and resu
Solutions
Chapter Two Solutions
Section 2.1: 1a) many possibilities, b) many possibilities with b and f as end vertices;
2a) n odd,
b) yes, K2; c) r and s even
3. yes, make degree even at all other vertices
and degree also even at new vertex (or else grap
Solutions
16. e = v - t: for each tree, ei = vi - 1 and summing e = ei = (vi -1) = n - t;
17. Unbalancing a tree:
makes sum of level numbers larger and so
smallest sum occurs when binary tree is balanced in which case each level number is
log2l or log2l +
Solutions
Chapter Three Solutions
Section 3.1: 1a)
b)
c)
2. 21
3. no odd circuits bipartite (Theorem 2 of Sect. 1.3) 2-colorable;
4. no circuits
no K3,3 or K5 configuration (or just draw tree as in Figure 3.1 in planar fashion);
5a) no circuits means uni
Solutions
18. The new graph Gk+1 has a vertex adjacent to every possible combination of one
vertex in each of k copies of Gk; if k colors are needed to color Gk, then in some vertex
in Gk+1 is adjacent to vertices with k different colors and so Gk+1 requi
Solutions
to b-c) and delete b-c and use c-d, delete a-e and use d-e-j, delete f-g (rule 2) and use f-j,
now have subcircuit a-b-g-h-c-d-e-j-f-a;
7a) rule 1 at f, h, j forces subcircuit e-f-g-h-i-j-e,
b) by symmetry at p, use m-p-n and delete p-o forcing
Solutions
12. Follows immediately from interpretation of vertices of hypercube as binary n-tuples
and x adjacent to y if they differ in one just bit.
13. See graph in Odd Solutions of text;
14. Exer 7 in Section 1.2 has all graphs (see Exer.12 in Section
Solutions
12. Follows immediately from interpretation of vertices of hypercube as binary n-tuples
and x adjacent to y if they differ in one just bit.
13. See graph in Odd Solutions of text;
14. Exer 7 in Section 1.2 has all graphs (see Exer.12 in Section
caps, then (using the same 100in2 os sheet metal), then
which choice of radius and height of the cylinder give you
the container with the largest volume?
277. What are the smallest and largest values that
(sin x)(sin y) can have if x + y = and if x and y
7.2. Velocity from acceleration. The acceleration of the object is by denition the rate of change of
its velocity,
a(t) = v 0 (t),
so you have
v(t) = v(0) +
Z
t
a(t)dt.
0
Conclusion: If you know the acceleration a(t) at all times t, and also the velocit
r = f (x)
3.4. Solids of revolution. In principle, formula (63) allows you to compute the volume of any solid,
provided you can compute the areas A(x) of all cross sections. One class of solids for which the areas of the
cross sections are easy are the so
f 0 (x) > 0
f 0 (x) < 0
x=
1
3
p
f 0 (x) > 0
x=
3
1
3
p
Figure 4. The graph of f (x) = x
3
3
x.
6.4. A function whose tangent turns up and down innitely often near the origin. We end
with a weird example. Somewhere in the mathematicians zoo of curious fun
1
x+
x
x
0
1
1
Figure 2. The slice at height x is a square with side 1
Therefore there is some ck in the interval [xk
1 , xk ]
volume of k
th
x.
such that
slice = (1
c k ) 2 xk .
Adding the volumes of the slices we nd that the volume V of the pyramid is g
y(t + t)
7.5. The velocity of an object moving in the plane. We have seen
that the velocity of an object which is moving along a line is the derivative of
its position. If the object is allowed to move in the plane, so that its motion y(t)
is described by
332 The Riemann-sum is the total area of the rectangles,
so to get the smallest Riemann-sum you must make the
rectangles as small as possible. You cant change their
widths, but you can change their heights by changing
the ci . To get the smallest area we
CHAPTER 5
Graph Sketching and Max-Min Problems
The signs of the rst and second derivatives of a function tell us something about the shape of its graph.
In this chapter we learn how to nd that information.
1. Tangent and Normal lines to a graph
The slope
254. y = 4 sin x + sin2 x
264. In the following two problems it is not possible to solve
the equation f 0 (x) = 0, but you can tell something from
the second derivative.
255. y = 2 cos x + cos2 x
256. y =
4
2 + sin x
257. y = 2 + sin x
(a) Show that the f
p
p
p
6.3. Example Find limx!2 x . Of course, you would think that limx!2 x = 2 and you can
indeed prove this using & " (See problem 44.) But is there an easier way? There is nothing in the limit
properties which tells us how to deal with a square root, a
CHAPTER 8
Applications of the integral
The integral appears as the answer to many dierent questions. In this chapter we will describe a number
of things which are an integral. In each example there is a quantity we want to compute, and which we can
approx
Solutions
i) 4, a,c,g,i form a K4,
j) 2,
k) 2,
l) 4, an attempt to 3-color vertices clockwise
around the circle starting at a forces j to have same color as a m) 4, graph has 5-wheel,
n) 4, graph has 5-wheel,
o) 3, outer pentagon (odd-length circuit) cann
Solutions
11a) for n = 1, (G) + (G ) = 2; assume for n-1 and consider an n-vertex graph G; by
induction we can color G-x and G -x, for any given vertex x, with a total of n colors for
both graphs (possibly less); x has a total of n-1 edges in G and G and