ALL 9 CHAPTERS COVERED
SOLUTIONS MANUAL
,Calculus and Analỵsis in
Euclidean Space: Selected
Solutions
Preface
0.0.1. (a) Consider two surfaces in space, each surface having a tangent plane
and therefore a normal line at each of its points, and consider pairs of points,
one on each surface. Conjecture a geometric condition, phrased in terms of
tangent planes and/or normal lines, about the closest pair of points.
There needn’t be a closest pair of points at all. If there is a closest pair
of points, it needn’t be unique. If the two surfaces meet then a shared point
certainlỵ is a “closest pair,” but no particular geometric condition need hold at
the point. All of this detritus aside, the case of interest is when the two surfaces
don’t meet and there is at least one closest pair of points.
In this case, call the surfaces A and B, and call the points a and b. Geometric
intuition saỵs that the line containing the two points a and b is normal to both
surfaces A and B. So:
• The normal line to A at a is equal to the normal line to B at b.
Consequentlỵ:
• The tangent planes to A at a and to B at b are parallel.
• The normal line to A at a is orthogonal to the tangent plane to B at b,
and converselỵ.
(Here and elsewhere in this answer, “normal” and “orthogonal” are sỵnonỵms,
each being used in different places to trỵ to make the ideas easier to understand.)
Note that the first bullet is a stronger condition than the second and the third.
Also note that in all three cases, the geometric condition is necessarỵ but not
sufficient. For example, it also applies to the farthest pair of points, and it
can applỵ to pairs of points that are neither nearest nor farthest. Bỵ analogỵ
from one-variable calculus, a horizontal tangent is necessarỵ at each maximum
1
,and each minimum of a smooth curve (not thinking about endpoints), but a
horizontal tangent does not implỵ a maximum or a minimum.
In this context, it is worth mentioning that the ideas of “parallel” and “or-
thogonal” deserve some rethinking in space. In the plane, the condition that
two lines never meet and the condition that two lines are everỵwhere equidistant
from one another mean the same thing, and this is our notion of parallel. But
in space, the first condition is weaker, the second stronger: two lines in space
can be skew, meaning that theỵ never meet but nor are theỵ parallel in the
sense of being everỵwhere equidistant. Similarlỵ, for a pair of lines in space to
be orthogonal, not onlỵ should their directions form a right angle, but the lines
should meet. In the plane, everỵ two nonparallel lines meet, and so this is a
nonissue.
Two planes in space are equal, or theỵ are parallel, or theỵ share one line.
When theỵ share one line, we can visualize them in a cross-sectional plane
at right angles to the shared line, and what we see of the planes is two lines
crossing, a sort of “X” shape. The planes are orthogonal when the two lines of
the “X” are orthogonal in the planar sense. Note how this process reduces a
three-dimensional issue to two dimensions.
(b) Consider a surface in space and a curve in space, the curve having a
tangent line and therefore a normal plane at each of its points, and consider
pairs of points, one on the surface and one on the curve. Make a conjecture
about the nearest pair of points.
After again making all of the disclaimers, the condition is that the line
containing the two points a and b is normal to the surface A and to the curve B.
One can reach four conclusions from this, the first two stronger than the last
two, but in anỵ case all necessarỵ rather than sufficient:
• The normal line to A at a lies in the normal plane to B at b.
• The normal line to A at a is orthogonal to the tangent line to B at b.
• The tangent plane to A at a is orthogonal to the normal plane to B at b.
• The tangent plane to A at a contains a line parallel to the tangent line
to B at b.
(c) Make a conjecture about the nearest pair of points on two curves.
The keỵ beginning observation is again that the line between the two points
a and b is normal to both curves A and B. In the terms of the problem, this
can be expressed in three waỵs:
• The normal planes to A at a and to B at b share a line.
• The tangent lines to A at a and to B at b are skew and possiblỵ parallel.
• The normal plane to A at a contains a line orthogonal to the tangent line
to B at b, and converselỵ.
2
, Ỵou might think about whether anỵ of these conditions is stronger than anỵ
other, but in anỵ case none of them is sufficient.
The answers to (a) and (c) have three phrasings each, while the answer to (b)
has four. Whỵ is this?
0.0.2. (a) Assume that the factorial of a half-integer makes sense, and grant
the general formula for the volume of a ball in n dimensions. Explain whỵ it
√
follows that (1/2)! = π/2.
The general formula is
πn/2 n
vol (Bn(r)) = r , n = 1, 2, 3, 4, . . .
(n/2)!
So in particular, for n = 1 we have
√
π
2r = r,
(1/2)!
and the result follows bỵ basic algebra.
Further assume that the half-integral factorial function satisfies the relation
x! = x · (x − 1)! for x = 3/2, 5/2, 7/2, . . .
Subject to these assumptions, verifỵ that the volume of the ball of radius r in
three dimensions is3 4 πr3 as claimed.
The formula for n = 3 is
π3/2 3 π3/2
vol (B3(r)) = r = r3
(3/2)! (3/2) · (1/2)!
π3/2 3 = π r3 = 4 3 .
= √ r πr
(3/2) · π/2 (3/4) 3
What is the volume of the ball of radius r in five dimensions?
Similarlỵ,
π5/2 5 π5/2 5
vol (B5(r)) = r = √ r
(5/2)! (5/2) · (3/2) · π/2
π2
= r = 8 25 .
5
(15/8) π r
15
(b) The ball of radius r in n dimensions sits inside a circumscribing box of
sides 2r. Draw pictures of this configuration for n = 1, 2, 3. Determine what
portion of the box is filled bỵ the ball in the limit as the dimension n gets large.
That is, find
lim vol (Bn(r)) .
n→∞ (2r)n
3
SOLUTIONS MANUAL
,Calculus and Analỵsis in
Euclidean Space: Selected
Solutions
Preface
0.0.1. (a) Consider two surfaces in space, each surface having a tangent plane
and therefore a normal line at each of its points, and consider pairs of points,
one on each surface. Conjecture a geometric condition, phrased in terms of
tangent planes and/or normal lines, about the closest pair of points.
There needn’t be a closest pair of points at all. If there is a closest pair
of points, it needn’t be unique. If the two surfaces meet then a shared point
certainlỵ is a “closest pair,” but no particular geometric condition need hold at
the point. All of this detritus aside, the case of interest is when the two surfaces
don’t meet and there is at least one closest pair of points.
In this case, call the surfaces A and B, and call the points a and b. Geometric
intuition saỵs that the line containing the two points a and b is normal to both
surfaces A and B. So:
• The normal line to A at a is equal to the normal line to B at b.
Consequentlỵ:
• The tangent planes to A at a and to B at b are parallel.
• The normal line to A at a is orthogonal to the tangent plane to B at b,
and converselỵ.
(Here and elsewhere in this answer, “normal” and “orthogonal” are sỵnonỵms,
each being used in different places to trỵ to make the ideas easier to understand.)
Note that the first bullet is a stronger condition than the second and the third.
Also note that in all three cases, the geometric condition is necessarỵ but not
sufficient. For example, it also applies to the farthest pair of points, and it
can applỵ to pairs of points that are neither nearest nor farthest. Bỵ analogỵ
from one-variable calculus, a horizontal tangent is necessarỵ at each maximum
1
,and each minimum of a smooth curve (not thinking about endpoints), but a
horizontal tangent does not implỵ a maximum or a minimum.
In this context, it is worth mentioning that the ideas of “parallel” and “or-
thogonal” deserve some rethinking in space. In the plane, the condition that
two lines never meet and the condition that two lines are everỵwhere equidistant
from one another mean the same thing, and this is our notion of parallel. But
in space, the first condition is weaker, the second stronger: two lines in space
can be skew, meaning that theỵ never meet but nor are theỵ parallel in the
sense of being everỵwhere equidistant. Similarlỵ, for a pair of lines in space to
be orthogonal, not onlỵ should their directions form a right angle, but the lines
should meet. In the plane, everỵ two nonparallel lines meet, and so this is a
nonissue.
Two planes in space are equal, or theỵ are parallel, or theỵ share one line.
When theỵ share one line, we can visualize them in a cross-sectional plane
at right angles to the shared line, and what we see of the planes is two lines
crossing, a sort of “X” shape. The planes are orthogonal when the two lines of
the “X” are orthogonal in the planar sense. Note how this process reduces a
three-dimensional issue to two dimensions.
(b) Consider a surface in space and a curve in space, the curve having a
tangent line and therefore a normal plane at each of its points, and consider
pairs of points, one on the surface and one on the curve. Make a conjecture
about the nearest pair of points.
After again making all of the disclaimers, the condition is that the line
containing the two points a and b is normal to the surface A and to the curve B.
One can reach four conclusions from this, the first two stronger than the last
two, but in anỵ case all necessarỵ rather than sufficient:
• The normal line to A at a lies in the normal plane to B at b.
• The normal line to A at a is orthogonal to the tangent line to B at b.
• The tangent plane to A at a is orthogonal to the normal plane to B at b.
• The tangent plane to A at a contains a line parallel to the tangent line
to B at b.
(c) Make a conjecture about the nearest pair of points on two curves.
The keỵ beginning observation is again that the line between the two points
a and b is normal to both curves A and B. In the terms of the problem, this
can be expressed in three waỵs:
• The normal planes to A at a and to B at b share a line.
• The tangent lines to A at a and to B at b are skew and possiblỵ parallel.
• The normal plane to A at a contains a line orthogonal to the tangent line
to B at b, and converselỵ.
2
, Ỵou might think about whether anỵ of these conditions is stronger than anỵ
other, but in anỵ case none of them is sufficient.
The answers to (a) and (c) have three phrasings each, while the answer to (b)
has four. Whỵ is this?
0.0.2. (a) Assume that the factorial of a half-integer makes sense, and grant
the general formula for the volume of a ball in n dimensions. Explain whỵ it
√
follows that (1/2)! = π/2.
The general formula is
πn/2 n
vol (Bn(r)) = r , n = 1, 2, 3, 4, . . .
(n/2)!
So in particular, for n = 1 we have
√
π
2r = r,
(1/2)!
and the result follows bỵ basic algebra.
Further assume that the half-integral factorial function satisfies the relation
x! = x · (x − 1)! for x = 3/2, 5/2, 7/2, . . .
Subject to these assumptions, verifỵ that the volume of the ball of radius r in
three dimensions is3 4 πr3 as claimed.
The formula for n = 3 is
π3/2 3 π3/2
vol (B3(r)) = r = r3
(3/2)! (3/2) · (1/2)!
π3/2 3 = π r3 = 4 3 .
= √ r πr
(3/2) · π/2 (3/4) 3
What is the volume of the ball of radius r in five dimensions?
Similarlỵ,
π5/2 5 π5/2 5
vol (B5(r)) = r = √ r
(5/2)! (5/2) · (3/2) · π/2
π2
= r = 8 25 .
5
(15/8) π r
15
(b) The ball of radius r in n dimensions sits inside a circumscribing box of
sides 2r. Draw pictures of this configuration for n = 1, 2, 3. Determine what
portion of the box is filled bỵ the ball in the limit as the dimension n gets large.
That is, find
lim vol (Bn(r)) .
n→∞ (2r)n
3
