Tuesday, October 19, 2010
1
st
Midterm Exam MTH 164 (Multivariable Calculus)
1. (20 points)
Consider the surface given by the equation
x
2
−
2
y
2
+
z
2
+
yz
=2
(a) Find an equation of the tangent plane to the surface at the point
P
(2
,
1
,
−
1).
Answer:
The equation is a level set of
g
(
x, y
)=
x
2
−
2
y
2
+
z
2
hence the gradient of this function
is normal to the surface.
∇
g
=
°
2
x,
−
4
y
+
z,
2
z
+
y
±
and
∇
g
(2
,
1
,
−
1) =
°
4
,
−
5
,
−
1
±
.
A plane is de±ned by a point and its normal vector, so with
X
=
°
x, y, z
±
the tangent
plane is de±ned by
∇
g
·
(
X
−
(2
,
1
,
−
1)) = 0 or 4
x
−
5
y
−
z
=4
If you use coordinates
°
dx, dy, dz
±
=
°
x
−
2
,y
−
1
,z
+1
±
relative to the coordinate
system whose origin is at (2
,
1
,
−
1) then you get the slightly simpler equation of a
plane through the this new origin 4
d
−
5
dy
−
dz
=0
.
(b) If the point
Q
(2
.
1
,
0
.
9
,k
) lies near
P
(2
,
1
,
−
1) and on the surface, use linear approxi-
mation to approximate the value of
k
.
Answer:
On the tangent plane we have 4(
.
1)
−
5(
−
.
1)
−
dz
= 0 hence
.
9=
dz
=
z
+1or
z
=
−
.
1. Hence the
z
coordinate of the corresponding point on the surface will be
z
=
−
.
1+
error
and we are assuming that the changes are small enough so that the
error term is negligible and our linear approximation is fairly accurate.
Another approach is to assume that
z
=
z
(
x, y
) is a function of
x
and
y
de±ned
implicitly by the equation above. Then using implicit di²erentiation on the equation
above we can calculate that
∂z
∂x
(2
,
1) = 4 and
∂y
(2
,
1) =
−
5so
dz
=
dx
+
dy
=
4(
.
1)
−