MATH 253 WORKSHEET 10
TANGENT PLANES
Find the equation of the plane tangent to the following surfaces at the following points:
2
2
(1) z = e−x −y at (0, 0, 1).
2
2
2
2
∂z
∂z
∂z
∂z
Solution: ∂x
= −2xe−x −y , ∂y
= −2ye−x −y , ∂x
(0, 0) = 0, ∂y
(0, 0) = 0, and the equation is
z = 1.
(2) z = e
at 1, 1, e .
∂z
(1, 1) = −2e−2 ,
Solution: Now ∂x
−x2 −y 2
−2
∂z
∂y (1, 1)
= −2e−2 , and the equation is
z − e−2 = −2e−2 (x − 1) − 2e−2 (y − 1)
which we can rearrange to
2x + 2y + e2 z = 5 .
(3) Two planes tangent to the surface z = 1 − x2 − y 2 meet the x-axis at
What
are they? Where are the points of tangency?
Solution: Suppose the point of tangency is (x0 , y0 , z0 ). The equation of the tangent plane is
21
16
and the y -axis at
21
8 .
z = z0 − 2x0 (x − x0 ) − 2y0 (y − y0 )
that is
2x0 x + 2y0 y + z
=
1 − x20 − y02 + 2x20 + 2y02
1 + x20 + y02 .
21
, 0, 0 and 0, 21
We are given that this plane contains 16
8 , 0 , that is that:
(
2x0 21
= 1 + x20 + y02
16
.
21
2y0 8 = 1 + x20 + y02
=
In particular we have
that is
21
8 x0
=
21
4 y0
so x0 = 2y0 . Plugging into the rst equation we nd
5y02 −
By the quadratic formula,
21
4 y0
= 1+5y02 ,
21
y0 + 1 = 0 .
4
√
441 − 16 · 20
21 ± 121
21 ± 11
y0 =
=
=
=
10
40
40
40
4
10
1
=
and
y
=
=
.
The
points
of
tangency are
so the two possibilities are y0 = 32
0
40
5
40
4
therefore
8 4
11
16
8
21
1 1 11
,
,
−
where
the
plane
is
x
+
y
+
z
=
or
16x
+
8y
+
5z
=
21
and
,
,
5 5
5
5
5
5
2 4 16 where the
1
21
plane is x + 2 y + z = 16 .
21
4
Date
±
q
212
42
− 20
21 ±
√
: 27/9/2013.
1