

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
Yo
tangentplaneisz=zo+(xxo)+(yyo),
or z= x+Y
,sincexi+~:=z:.
zo
20
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
=
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)
+
=
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

+
10(y

2)
+
25, or lox
+
10y

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

=
2~rh,

"
AT+
dr
ah =rr2; AVM(~),
(=),~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
1r~
=
16~ which is
<
.1 if
6
<
1
<
.002
.
lAVl
5
12~6
160~
dT
X
y
2B7 We have
T
=
Jw,
X
ax
TI
dy
T'
8 =tan
y
;






Therefore at (3,4),
T
=
5, and AT
M
;Ax
+
$Ay. If lAxl and lAyl are both
5
.01, then
1A.l
5
~
I
A
x
~
+
=
;(.0l)
=
.014 (or .02).