1.
[4 marks]
Find the shortest distance from the point
)
4
,
1
,
1
(
−
−
P
to the line tangent
to the curve
at the point
.
)
6
,
3
,
2
(
)
(
t
t
e
e
t
t
−
+
−
=
r
)
5
,
2
,
0
(
0
−
P
The point
corresponds to the parameter value
)
5
,
2
,
0
(
0
−
P
0
=
t
. So the tangent vector to
the curve at this point is
If we construct the vector
, then the shortest distance will be the height
h
of the parallelogram spanned by the
)
1
,
1
,
2
(
)
,
,
2
(
)
0
(
0
0
−
=
−
=
′
=
e
e
r
v.
→
P
P
0
vectors
v
and
which is: area of the parallelogram
A
divided by the length of the base.
→
P
P
0
Now
2
3
)
3
,
3
,
0
(
)
1
,
1
,
,
1
(
)
1
,
1
,
2
(
0
=
−
−
=
−
×
−
=
×
=
→
P
P
A
v
and
6
=
v
, so the
shortest distance is
3
6
2
3
=
=
h
.
2.
(a)
[5 marks]
Let
S
be the subset of
2
R
defined by
.
}
1
:
)
,
0
{(
\
}
4
:
)
,
{(
2
2
>
<
+
=
y
y
y
x
y
x
S
(i)
Is
S
open, closed or neither ? Justify your choice, no marks for guessing.
This is just a version of the homework question 1(f),sec.1.2.
The set
S
is neither closed nor open. It contains the boundary point (0,1) but no other
boundary points.
(ii)
Describe in the set notation:
=
int
S
)}
1
,
0
{(
\
S
=
S
}
4
:
)
,
{(
2
2
≤
+
y
x
y
x
=
∂
S
}
2
1
:
)
,
0
{(
}
4
:
)
,
{(
2
2
≤
≤
∪
=
+
y
y
y
x
y
x
(b)
[3 marks]
Prove that if
x
is in the closure of
then there is a sequence of points
n
R
S
⊂
in
S
that converges to
x.
This is the second part of the proof of Th. 1.14 in the text (assignment).
2