Thomas Calculus 13th [Solutions]

1078 Chapter 14 Partial Derivatives
6. (a)
(b)
(c)
lim sin 6r
sin
0
6
lim t
1,
r
r
t 0
t
where t 6r
sin 6h
1
6h
r h h 2
h
(0, 0) lim f (0 h, 0) f (0, 0) lim lim sin 6h 6h lim cos 6h 6 36 sin 6
0 0 0 6
0
12 lim h
f
0
12
0
h h h h h h
(applying 1Hôpitals rule twice)
sin 6r
sin 6r
6r
6r
f ( r, h) f ( r, )
f ( r, ) lim lim lim 0 0
h 0
h
h 0
h
h 0
h
7. (a)
(b)
2 2 2
r xi yj zk r r x y z and
n 2 2 2 n
r x y z
r
x
y
i j z k
2 2 2 2 2 2 2 2 2
x y z x y z x y z
r
r
n 2 2 2
( n/2) 1
2 2 2
( n/2) 1
2 2 2
( n/2) 1
n 2
( r ) nx x y z i ny x y z j nz x y z k nr r
(c) Let n 2 in part (b). Then
(d) dr dxi dyj dzk r dr x dx y dy z dz,
and
r dr x dx y dy z dz r dr
2
1 2 1 2 1 2 2 2
( r ) r r r r x y z is the function.
2 2 2 2
(e) A ai bj ck A r ax by cz ( A r)
ai bj ck A
x y
dr r z
x dx ry dy rz dz dx dy dz
r r r
df f f dy f f dy
8. f ( g( t), h( t)) c 0 dx
i j dx i j , where
dt x dt y dt x y dt dt
f is orthogonal to the tangent vector
dx
dt
i
dy
j is the tangent vector
dt
9.
2 2
f ( x, y, z) xy yz cos xy 1 f z y sin xy i ( z x sin xy) j (2 xz y) k f (0, 0, 1) i j
the tangent plane is x y 0; r (ln t) i ( t ln t) j tk r 1 i (ln t 1) j k;
x y 0, z 1
t
t 1 r (1) i j k . Since ( i j k) ( i j) r (1) f 0, r is parallel to the plane, and
r(1) 0i 0j k r is contained in the plane.
10. Let
3 3 3 2 2 2
f ( x, y, z) x y z xyz f 3x yz i 3y xz j 3z xy k
f (0, 1, 1) i 3j 3k the tangent plane is x 3y 3z
0;
2
3
r t 2 i 4 3 j cos( t 2) k
4 t
r 3t
i 4 j (sin( t 2)) k ; x 0, y 1, z 1 t 2 r (2) 3 i j . Since r (2) f 0
4 2
t
parallel to the plane, and r(2)
i k r is contained in the plane.
r is
11.
z
x
2
3x 9y 0 and
z
x
2 1 2
2
2
4
3y 9x 0 y x and 3 1 x 9x 0 1 x 9x
0
3
3 3
3
x x 27 0 x 0 or x 3. Now x 0 y 0 or (0, 0) and x 3 y 3 or (3, 3). Next
2 2
z
x
6 x, z 6 y , and
y
2 2
2
z
x y
2 2 2 2
for (3, 3), z z z 243 0 and
2 2
x y x y
2 2 2 2
9. For (0, 0), z z z 81
2 2
x y x y
2
z 18 0
2
x
a local minimum.
no extremum (a saddle point), and
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