1.
Let the surface
S
be parametrized by
)
,
,
2
(
)
,
(
2
v
u
e
u
v
u
v
u
v
+
+
+
=
−
f
and let
be a point in
)
1
,
2
,
2
(
0
P
3
R
.
(a)
[2 marks]
Show that the point
lies on the surface
S
.
0
P
We should be able to find (
u
,
v
) such that
)
1
,
2
,
2
(
)
,
(
=
v
u
f
. We have the system of
equations
1
,
2
,
2
2
2
=
+
=
+
=
+
−
v
u
e
u
v
u
v
. From the second equation we get
since on the right hand side we do not have
e
and also
u
=1, to satisfy that equation.
0
=
v
Its easy to see that the other two equations are satisfied also. So
lies on the surface
S
0
P
and it corresponds to
)
0
,
1
(
)
,
(
=
v
u
(b)
[5 marks]
Write the equation for the tangent plane to
S
at the point
in the
0
P
form
d
cz
by
ax
=
+
+
, where
are constants.
d
c
b
a
,
,
,
Vector normal to the surface
S
at the point
is
0
P
)
0
,
1
)(
(
v
u
∂
∂
×
∂
∂
=
f
f
N
,
)
2
,
1
,
1
(
)
2
,
1
,
1
(
)
0
,
1
)(
(
0
1
=
+
=
∂
∂
=
=
v
u
u
v
u
u
f
,
)
1
,
1
,
1
(
)
1
,
,
1
(
)
0
,
1
)(
(
0
1
−
=
−
+
=
∂
∂
=
=
v
u
v
e
v
u
v
f
.
So
N =
=
)
1
,
1
,
1
(
)
2
,
1
,
1
(
−
×
)
2
,
1
,
3
(
−
and the equation of the tangent plane to
S
at the
point
is
0
P
0
)
1
)(
2
(
)
2
(
1
)
2
(
3
=
−
−
+
−
+
−
z
y
x
or
6
2
3
=
−
+
z
y
x
.
2

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