SECTION
15.4
Differentiability, Linear Approximation, and Tangent Planes
(ET Section 14.4)
665
We substitute these values in (1) to obtain the following equation of the tangent plane:
z
=
5
+
4
(
r
−
2
)
−
5
(
s
−
1
)
=
4
r
−
5
s
+
2
.
16.
H
(
s
,
t
)
=
se
st
,
(
0
,
0
)
SOLUTION
The equation of the tangent plane at
(
0
,
0
)
is
z
=
f
(
0
,
0
)
+
f
s
(
0
,
0
)
s
+
f
t
(
0
,
0
)
t
(1)
We compute the following values:
f
(
s
,
t
)
=
f
(
0
,
0
)
=
0
f
s
(
s
,
t
)
=
1
·
e
+
s
·
te
=
e
(
1
+
)
⇒
f
s
(
0
,
0
)
=
1
f
t
(
s
,
t
)
=
s
2
e
f
t
(
0
,
0
)
=
0
Substituting in (1) gives the following equation of the tangent plane:
z
=
0
+
1
·
s
+
0
·
t
=
s
That is,
z
=
s
.
17.
f
(
x
,
y
)
=
e
x
ln
y
,
(
0
,
1
)
The equation of the tangent plane at
(
0
,
1
)
is
z
=
f
(
0
,
1
)
+
f
x
(
0
,
1
)
x
+
f
y
(
0
,
1
)(
y
−
1
)
(1)
We compute the values of
f
and its partial derivatives at
(
0
,
1
)
:
f
(
x
,
y
)
=
e
x
ln
yf
(
0
,
1
)
=
0
f
x
(
x
,
y
)
=
e
x
ln
y
⇒
f
x
(
0
,
1
)
=
0
f
y
(
x
,
y
)
=
e
x
y
f
y
(
0
,
1
)
=
1
Substituting in (1) gives the following equation of the tangent plane:
z
=
0
+
0
x
+
1
(
y
−
1
)
=
y
−
1
That is,
z
=
y
−
1.
18.
f
(
x
,
y
)
=
ln
(
x
2
+
y
2
)
,
(
1
,
1
)
The equation of the tangent plane at
(
1
,
1
)
is
z
=
f
(
1
,
1
)
+
f
x
(
1
,
1
)(
x
−
1
)
+
f
y
(
1
,
1
)(
y
−
1
)
(1)
We compute the following values, using the Chain Rule:
f
(
x
,
y
)
=
ln
(
x
2
+
y
2
)
f
(
1
,
1
)
=
ln 2
f
x
(
x
,
y
)
=
2
x
x
2
+
y
2
⇒
f
x
(
1
,
1
)
=
1
f
y
(
x
,
y
)
=
2
y
x
2
+
y
2
f
y
(
1
,
1
)
=
1
Substituting in (1) gives the following equation of the tangent plane:

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