Problem Set 4
MH2100
Solutions
E±²³´µ¶² · Let
±
:
R
3
²³´³µ
→
R
3
¶³·³¸
be a map given by
¶
=
¹
(
²³´³µ
)
³
·
=
º
(
)
³
¸
=
»
(
)
³
where
¹
,
º
, and
»
are polynomials in the variables
.
(a) Prove that
±
is continuous on
R
3
.
(b) Is it true that
±
is also diﬀerentiable on
R
3
? Justify your answer.
S¸¹º»µ¸¼
(a) Since all component functions are polynomials, they are elementary func
tions deﬁned on the whole
R
3
and are therefor continuous everywhere.
(b) All partial derivatives of all component functions are also polynomials. They are
elementary functions deﬁned on the whole
R
3
and continuous everywhere and
hence the map
±
is diﬀerentiable at any point.
E±²³´µ¶² ½ What is the Jacobian matrix of a linear map
±
(
x
) =
A
x
,
±
:
R
¼
→
R
½
. Here,
A
is an
½
by
¼
matrix and
x
∈
R
¼
is a vectorcolumn.
S¸¹º»µ¸¼
It’s
A
itself at any point. Indeed, the deﬁnition of diﬀerentiability applied to
±
(
x
) =
A
x
with (supposedly)
±
¾
(
x
) =
A
says
lim
Δ
x
→
0
¿A
(
x
+ Δ
x
)
 A
x
 A
Δ
x
¿
¿
Δ
x
¿
= 0
³
which is true.
:)
1
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View Full Document1 Questions on calculation
E±²³´µ¶² · Write down the equation of the tangent plane to each of the following sur
faces at given points
(a)
±
=
²
2
+
³
2
at the point (1
´
2
´
5).
(b)
²
2
+
³
2
+
±
2
= 169 at the point (3
´
4
´
12).
(c)
±
=
³
+ ln
²
±
at the point (1
´
1
´
1).
S¸¹º»µ¸¼
(a) Here,
±
²
= 2
²
,
±
³
= 2
³
, so
±
²
(1
´
2) = 2 and
±
³
(1
´
2) = 4. Thus the tangent
plane is
±
= 5 + 2(
² 
1) + 4(
³ 
2), which is 2
²
+ 4
³  ± 
5 = 0.
(b) Here,
±
=
µ
½
169
 ²
2
 ³
2
. Since at the given point we have
±
= 12
>
0, we use
±
=
½
169
 ²
2
 ³
2
. Further,
±
²
=

²
½
169
 ²
2
 ³
2
=

²
±
´
±
³
=

³
½
169
 ²
2
 ³
2
=

³
±
¶
Hence
±
²
(3
´
4
´
12) =

1
4
and
±
³
(3
´
4
´
12) =

1
3
. Thus, the answer is
±
= 12

² 
3
4

³ 
4
3
⇔
3
²
+ 4
³
+ 12
±
= 169
¶
(c) Here, it’s diﬃcult to express
±
as a function of
²´³
, so we express
³
in
²´±
instead.
We have
³
=
±
ln
²
±
=
ln
²
+ln
±
,
³
²
=

1
²
and
³
²
(1
´
1) =

1. Further,
³
±
= 1+
1
±
and
³
±
(1
´
1) = 2. Thus the answer is
³
= 1

(
² 
1) + 2(
± 
1), that is,
²
+
³ 
2
±
= 0.
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 Spring '13
 Calculus, Derivative, Limit, lim

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