 
2.
Partial Differentiation
2A. Functions and Partial Derivatives
2A1
In the pictures below, not d l of the level curves are labeled. In (c) and (d), the
picture is the same, but the labelings are different. In more detail:
b) the origin is the level curve 0; the other two unlabeled level curves are .5 and 1.5;
c) on the left, two level curves are labeled; the unlabeled ones are 2 and 3; the origin is
the level curve 0;
d) on the right, two level curves are labeled; the unlabeled ones are
1
and 2; the origin
is the level curve
1;
The crude sketches of the graph in the first octant are at the right.
2A3
a) both sides are mnxm' yn'
xY
.
x
b)
f,
=
(x
+
yI2
'
fx,
=
(fx),
=

f,
=

f y x
=
(Y

x)
(x
+
yI3
'
(x
+
yI2
'
(x
+
Y ) ~
c)
f,
=
2xsin(x2+y),
f,,
=
(f,),
=
2xcos(x2 +y);
f,
=
sin(x2+y),
f,,
=
cos(x2+y).2x.
d) both sides are fl(x)g'(y).
2A4
(f,),
=
ax
+
6y,
(f,),
=
22
+
6y; therefore f,,
=
f,,
H
a
=
2.
By
inspection,
one sees that if a
=
2,
f
(x, y)
=
x2y
+
3xy2 is a function with the given
f,
and f,.
2A5
a) w,
=
aeax sin ay,
w,,
=
a2eax
sin ay;
W,
=
eaxa cos ay,
w,,
=
eaxa2
(
sin ay);
therefore wyy
=
w,,.
22
2(y2

x2)
b) We have w,
=

WXX
=
(x2
+
y2)2
'
If we interchange x and y, the function
x2
+
y2
'
w
=
ln(x2
+
y2) remains the same, while w,,
gets turned into w,,;
since the interchange
just changes the sign of the right hand side, it follows that w,,
=
w,,.
2B. Tangent Plane; Linear Approximation
2B1
a) z,
=
y2, z,
=
2xy; therefore at (1,1,1), we get
z,
=
1,
z,
=
2, so that the
tangent plane is z
=
1
+
(x

1)
+
2(y

I), or z
=
x
+
2y

2.
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1
2.
PARTIAL DIFFERENTIATION
b) w,
=
y2/x2,
W,
=
2ylx; therefore at (1,2,4), we get w,
=
4,
w,
=
4, so
that the tangent plane is w
=
4

4(x

1)
+
4(y

2), or w
=
4x
+
4y.
x
x
2B2 a) z,
=

by symmetry (interchanging x and Y), z,
=
;
Y
then the
I
/

z
z
xo
Yo
xo
Yo
tangentplaneisz=zo+(xxo)+(yyo),
or z =
x+Y
,sincexi+~:=z:.
zo
20
zo
zo
b) The line is x
=
xot, y
=
yot, z
=
zot; substituting into the equations of the cone
and the tangent plane, both are satisfied for all values of t; this shows the line lies on both
the cone and tangent plane (this can also be seen geometrically).
2B3
Letting x, y, z be respectively the lengths of the two legs and the hypotenuse, we
have z
=
I
/

thus the calculation of partial derivatives is the same as in 2B2, and
3
4
7
we get Az
M
Ax
+
Ay.
Taking Ax
=
Ay
=
.01, we get Az
M
(.01)
=
.014.
5
5
5
2B4 From the formula, we get R
=
R1R2
.
From this we calculate
R1+ R2
dR
2
,
and by symmetry,
4
4
1
Substituting R1
=
1,
R2
=
2 the approximation formula then gives AR
=
ARl+ AR2.
9
9
4
1
5
By hypothesis, lARil
5
.l, for
i
=
1,2, so that
1
ARl
5
(.I)
+
(.I)
=
(.I)
M
.06; thus
9
9
9
n
2B5 a) Wehave f ( x , y ) = ( ~ + y + 2 ) ~ ,
fX=2(x+y+2), fy=2(x+y+2). Therefore
at (0,0), f,(O, 0)
=
f,
(0,O)
=
4,
f
(0,O)
=
4;
linearization is 4
+
42
+
4y;
at (1,2), fx(l,2)
=
fy(l,2)
=
10,
f
(l,2)
=
25;
linearization is 10(x

1)
+
10(y

2)
+
25, or lox
+
10y

5.
linearization at (0,O):
1
+
x;
linearization at ( 0 , ~ / 2 ) :y
dV
dV
dV
2B6 We have V
=
rr2h,

=
2 ~ r h ,

"
AT+
dr
a h = r r 2 ;
A V M ( ~ ) ,
( = ) , ~ h .
Evaluating the partials at
T
=
2, h
=
3,
we get
Assuming the same accuracy lArl
5
6,
[Ah1
<
6
for both measurements, we get
1
+
47r
1 r ~
=
1
6
~
which is
<
.1 if
6
<
1
<
.002
.
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 Spring '08
 Auroux
 Critical Point, Derivative, Optimization, Partial differential equation

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