17. r(t)
=
,t
5t1
K
,
2
—
Iv(t)
and
161
—
at t
=
64
=
(8t
0 or t
v(t)
161)
—
42
(2t, 5,21— 16), v(t)
641 ± 281)_1/2(16/
Iv(t)
4. Since
=
=
<0
—
for 1
+ 25+412
—
641 + 256
2
8t
—
61! + 281
erator is zero, that is,
64). This is zero if and only if the num
d of 3 is attained
> 0 fort > 4, the minimum spee
Iv(t)
<4 and
4 units of time.
18. Since r(t)
=
3
3 k, a(t)
I + 12 j + t
F(t)l
=
Law,
6t i + 2j -b 61k. By Newton’s Second
r”(t)
=
.
6rn1 i-I- 2mj + Grntk is the required force
rn a(t)
F(t)
19.
VECTOR FUNCTIONS
CHAPTER 10
212
is, so F(t)
20 N in the direction of the positive z-ax
Since 20k
=
P(t)
speed is fv(t)
=
4a(t), a(t)
=
5k. Then v(t)
=
=
20k. Also n
5tk+ c where Ci
2t ‘\/25t2 + 2.AIso r(t)
\/1 ± 1 +25
Ii
tj
—
i—j.
4kg, r(O)
=
0 and v(O)
1—j so vQ)
=
i—i -4- 5tkand the
=
2 andO
k -4-c
=
r(O), 50 c
=
=
0
a=tJTh.
r
by a(t).
r(t) replaced by v(t) and r’(t) replaced
ithIiIie as that in Example 10.2.4 with
20. The argumerIe
of the velocity is
tion is 600, a unit vector in the direction
200 rn/s and, since the angle of eleva
and ifwe set up the axes so that
= 1001 + 100
j. Thus v(O) = 200(i +
(cos60°)i + (sin60°)j = i +
, the only force is that due to gravity, so
= 0. Ignoring air resistance
r(0)
then
n,
origi
the
s
at
start
ctile
proje
the
rating, we have v(t) = —9.Stj + C. But
. Thus a(1) = —9.8j and, integ
2
F(t) = n2a(t) = —mgj where g 9.8 rn/s
\/
9.81) j and then (integrating again)
= 1001 + (100
1001 + 100 /i = v(D) = C, so v(t)
ctile is
D. Thus the position function of the proje
4.912)
j + D where 0 = r(0) =
r(t) = loot i-j-- (100 /t
21. v(0)
=
.
—
—
r(t)
lOOt I + (100 \/t
—
t
j.
2
4.9
)
/
are r(t)
(a) Parametric equations for the projectile
=
/t—4.9t
2
100
=
y(t) =0(andt
=
>0)
ioo(i)
1001, y(i)
=
100
—
4.912.
t(1O0/—4.9t) =0
I
The projectile reaches the ground when
=
35.3s. Sotherangeis
3.535 m.
when the vertical component
has a critical number (or equivalently,
(b) The maximum height is reached when y(1)
17.7 s. Thus the maximum height is
=
9Sf = 0
of velocity is 0): y’(t) = 0 = 100
.
—
(ioor) =
\i—(r)
—
(c) From part (a), impact occurs at I
(iochJS)2
=
1001 + [100\/—
lv
(io_
l90. s. Thus, the velocity at impact is
_)] j
9
(
8
.
x/i00+30,0O0
1531 m.
=
1001
-
100\/j and the speed is
200 rn/s.
Io ooboho, i,, hoIr ,
. .-opiod. orduphdtd.orpoooodo opubhdy ooosth
2UIO(ooo, Lro,ng AU Roh,, Roo,r,,d ‘.ho. ,o ho ,.rno,,j
SECTION 10.4 MOTION INSPACE: VELOCITY ANO ACCELERMION
22. As inExercise 21, v(t)
Butr(0)
lOOj, soD
(a) p = 0
=
lOOi ± (i00— 9.8t)j andr(t)
=
lOOj and r(t)= lOOti+ (100+ )
2
100t—
j.
4.9t
100 + 100
100± (_100)2
t
2
4.9t
—
—
=
or
0
2
4.9t
4(4.9)(—100)
2(4.9)
=
‘°°‘°
—
lOOti + (100t— 4.9t
)j +D.
2
100 xt
100 = 0. From the quadratic formula we have
—
100± 31,960
Taking the positive t-vahie gives
9.8
.
=
35.9 s. Thus the range isx = 100.
(b) The maximum height is attained when
maximum height is 100 + 100
0
/(1OOv)
3.592
100
rn.
t=
9.8t = 0
—
49(1OO)
—
1]
17.7 s and the
1631 m.
Alternate solution: Because the projectile is fired in the same direction and with the same velocity
as in Exercise 21,
but from a point 100 rn higher, the maximum height reached is 100 m higher than that found in Exercise
21, that is,
1531 m+ 10Dm
1631 m.
(c) From part (a), impact occurs at i = 1Oo’T
v (ioo oo) = 100 i +
vj = /10,D00 + 31,960
23. As in Example 5, r(t) =
96
1
4
v/
O
,
(Vo cos
45°)t I +
y=0(andt >0)
t =
velocity is
30 rn/s.
=
00
24. As in Example 5, r(t) =
v(t) = r’(t) =
—
—
—
s. Thus the velocity at impact is
(b00
9.8
t6o)]
j
= lDOi
cos 30°)ti +
—
{(VO
sin45°)t
—
]
2
gt
j
=
[voa
/
5
t i + (vo/t
0 Sin
[(V
30°)t
—
]
2
gt
j
—
) i]. The ball lands when
2
gt
ox ort,2 =90g andthe initial
[uo/ti + (u t
—
gt
)
2
j] and then
2gt)j]. The shell reaches its maximum height when the vertical component of velocity
t =
2gt) = 0
/31,960j and the speed is
—
205 rn/s.
s.Nowsinceitlands90maway, 90 =
[vox/i + (vo
is zero, so (v
0
so
(V
[ioo x/
/3L96O
The vertical height of the shell at that time is 500 m,
g()2]
[uo()
500
= 500
vo
=
= ‘4DDO(9.8)
198 rn/s.
Let a be the angle of elevation. Then Vo = 150 rn/s and from Example
5, the horizontal distance traveled by the projectile is
d=
vsin2a
g
Thus
1502 sin2a
p
800
Two angles of elevation then are a
Here v
0
sin2a =
10.2° and a
100
0.3484
2a
20.1° or 180—20.4 = 159.6°.
79.8°.
115 ft/s, the angle of elevation is a = .50°, and if we place the origin at home plate. then r(0) =
Asiri Example 5,
so r(t)
cos
r
wehaver(t)
a)t i +
= —gt
j +tvo + D whereD
2
[(VO sin
a)t
—
2 + 3]
gt
j.
r(D) = 3j and
v
00
cosOl +vn
Sin (IJ,
Thus, parametric equations for the trajectory of the ball are
n.e
‘
3.1.
0_DID Cengage Learning All Rights Resened.
May not be scanned, copied, or duplicated, or posted to a publicly accessible icabsite, in whole or in past.
CHAPTER 10 VECTOR FUNCTIONS
D
214
x
=
(VU
y
(vo cos cs)t, y
400
coscs)t
= (VO
= (VU
sina)t
=.
sina)t
=
VQCOSCS
(115 sin 50°)(5.41)
400
11.2 ft. Since the fence is lOft high, the ball
(32)(5.41)2 —I- 3
—
=
is
5.41 s. At this time, the height of the ball
400
ll5cos5O°
400
t
2+3
gt
—
x
2 + 3. The ball reaches the fence when
gt
—
clears the fence.
the city lies between (100,0)
e the catapult is 100 meters from the city, so
27. Place the catapult at the origin and assum
As in Example 5, the trajectory’ of
and let 9 be the angle the catapult is set at.
and (600,0). The initial speed is V = 80 rn/s
(100, 15),
4.9t j. The top of the near city wall is at
]
cos 8)t i + [(80 sin 8)t 2
the catapulted rock is given by r (t) = (80
—
which the rock will hit when (80 cos 0) t
8OsinO
4.9(
4cosO
20
7.65625 tan
—
=
100 tan 0 + 22.62625
=
4cos8
lOOtanO
15
4cosOj
t
100
=
—
and (80 sin O)t
625sec 0
2
7.65
=
—
2
4.9t
15
=
2 8 + 1 gives
2 8 with tan
15. Replacing sec
0.230324, 12.8309
tan 0
0. Using the quadratic fonnula, we have
The base of the far wall is
the rock will land beyond the near city wall.
13.0°, 85.5°. So for 13.0° < 6 < 85.5°,
15
and (80 sin &)t 4.912 = 0
cos 0)1 = 600
=
8
located at (600,0) which the rock hits if (80
&
—
8OsinO
15
\2cos&J
10
2cos0
28
275.625 tan
—
600 tan 6
-
=
0.658678. 1.51819
0. Solutions are tan 0
275.625
0
625sec 0
2
275.
öOOtan&
0
8
=
33.4°, 56.6°. Thus tine
allow the rock to land on city ground
for 33.4° < 8 < 56.6°, and the angles that
rock lands beyond the enclosed city ground
into the cky, we
rock can hit the far wall and bounce back
3340, 56.60
< 8 < 85.5°. If you consider that the
are 13.0 < 0 <
and
= 600
0
2
the top of the wall at (600, 15): (80 cos &)t
calculate the angles that cause the rock to hit
(80 sin 6)1
—
4.912 =
600 tan8
15
—
0.727506, 1.44936
Solutions are tan 0
28
275.625 sec
15
28
275.625 tan
z
—
600 tanO + 290.625
0.
with angle 8 where
36.0°, 55.4°, so the catapult should be set
8
=.
=
13.0° < 0 < 36.0°, 55.4° <8 < 85.5°.
g east and k pointing
pointing in the northward direction with i pointin
28. Place the ball at the origin and considerj to be
a constant acceleration of
acceleration = force/mass, so the wind applies
upward. Force = mass x acceleration
ation acting
the acceleration due to gravity, the acceler
2 in the easterly direction. Combined with
4 N/0.8 kg = 5 rn/s
nt vector.
a(t) cit = 5t1 9.81 k + C where C is a consta
5 j 9.8k. Then v(t)
on the ball is a/f)
—15vj+ l5kand
C
15k
—3Ocos3O°j +3Osin3O°k= —15\/j +
Weknowv(0)=C
0
4.9t2)
k + D butr(0) =D
2 i 15tj + (151
15 j + (15— 9.8t) k. r(t) = v(t)dt = 2.5t
511
3.0612s,
t = 0, t = 15/4.9
2 = 0
2.512
) k. The ball lands when 15t 4.9t
2
4.9t
is tj + (151
SC) r(t)
speed is
the direction S 16.4°E. Its
23.43 i 79.53j which is 82.9 in away in
so the ball lands at approximately r(3.0612)
f
—
—
f
—
—
—
—
—
—
—
approximately v(3.0612)
15.306j
—
15
\/j
—
15k1
33.68 rn/s.
on poOnd
a Lnaan.ng All Rl0a R on.nd Ma no Sn soaaannal. aopnd. on doplootnd.
o
0
02010 C nn
a publiol) aooaas,hIo noobabo,
0
oholn on in port
216
t
CHAPTER 10 VECTOR FUNCTIO
5t cos a’
0) we need
In order to land at point B(40.
-—.
—pcosa
t
40
=
24
24
cos(tcosa)+ 7rcosa =0
tsincs—
z
5
2
ircosa
48
40 sin a’ +
r/8’\ cos a1 + 24
I
I
cos
8\cosaJ
J 7rcosa
24
ircosa
18’\.
I sin a
51
\coscs)
=
0
sina
=
=
1/. a
I 40 sin
coscs\
0
—
6
Thus a
5ir
.
—
=
=
——.
./I
sin
=
6t
I
5rj
—22.5
———
—
24’
24
cos ir -tirJ
ir
—
—
=
0
south of east.
so the boat should head 22.5°
,
the direction of motion, so if
Remember that r’(t) points in
.
r(t)
c
and
h
bot
to
lar
dicu
c x r(t) then r’(t) is perpen
also perpendicular
perpendicular to c. But r’(t) is
31. If r’(t)
st lie in a plane
c, the path of the particle mu
r’(t) is always perpendicular to
ions, the path
origin. Considering both restrict
path to a sphere centered at the
the
s
fine
con
ch
whi
r(t)
tor
vec
in in the
to the position
centered on a line through the orig
pendicular to c, and the circle is
per
e
plan
a
in
lies
that
le
circ
must be contained in a
direction of c.
Section 10.3], so the
g a straight line, then is = 0 [see
N. If a particle moves alon
2
= v’T + v
a
have
we
7
n
32. (a) From Equatio
le of the unit tangent vector, it is
eleration vector is a scalar multip
acc
the
e
aus
Bec
.
v’T
=
a
es
acceleration vector becom
parallel to the tangent vector.
. Thus the acceleration vector is
iv
N
2
the curve is
and points in the direction that
pendicular to the tangent vector
per
is
ich
(wh
tor
vec
mal
parallel to the unit nor
e is constant, then v’
(b) If the speed of the particl
j
2
tI+t
3
=
r”(t)
=
r’(t)
:t.
r’(t)I
2
4 + 4t
9t
[or by Equation 8, aT
I’(t)
a—
x
=
r”(t)
2j,
aN
—
—
=to
2t)j
r’(t) x r”(t)
=
=
2k.
r’(t)
Then
—
costj
=
. Then Equation 9 gives
2
6t
3t + 4t)/V9t4 +
(18
—
sintj,
=
I + (2t
and Equation 10 gives
4t2]
—
2)j,
r’(t)
=
,.,/12 ± (2t
—
2)2
= v/4t2
2(2t—2)
r’(t).r”(t)
—8t-I-5,
and Equation 10
aT
2
8t + 5
r’(t)
=
r’(t) x r”(t)
—sinti+costj +k,
sinti
cost
r’Q) r”(t) = sint cost— sint
--
r’(t)j
=
\/sin2t+cos2t+ 1
=
cuslj + lc.
-
UT
=
,
2
4t + 4t
v’9
3 + 4t
18t
2
4 + 4L
9t
Then Equation 9 gives
=
costi+sintj +tkz.
r”(t)
x r”(t)
=
2
6t
—
35. r(t)
Ir’(t)
(2t)
(3f + 2
2
)
=
/9t4i_4t2
r’(t) x r”(L)
s
=
—
2
(1 +t)i+ (t
r”(t)
—
2
+ 4t
=
—
Ir’(t)
=
v’
r’(t)
2t k,
—6
=
t + (2t)(2)
(6t)
2
(3
)
—
give
2 I + 2tj,
3•L
r’(t) r”(t)
aT
34. r(t)
=
r’(t) x r”(t)
6ti + 2j,
.
—
0 and Equation 7 gives a
p.
turning).
33. r(t)
=
—
-
0 and a N
—
-
Jr’(t)
x r”(t)
r’(t)
—
1
2
1
j
5
I
+s
2
co
+t
-
pOot
rbl, ,rohror. o soIrol, or or
dopLorrd. or porord ro o prrbhcly oooos,
r,d Moy ‘orb, rorrorod. ropori. or
C OIO C,og Lrrm,no All Rrghrr Rr.o
1.
SEC11ON 10.4 MOTION iN SPACE: VELOCITY AND ACCELERATION
36. j
2
r(t)
+3
=ti
tk
+t
r”(t)
2j,
=
r’(i) =i-4-2tj±3k,
r’(t) x r”(i)
41
.
,Jit_l(2t)2+32 2
=...
±
10
/4t
,
2
4
./
r’Q) x r”(i)
r’(t)
and ar
=
Ir’(t)i
=
—61 + 2k.
r’(t) r”(t)
Then aT
r’(t)I
± 10
/i
2
+ 10
./4t2
37. The tangential component of a is the length of the projection of a onto
T, so we sketch
the scalar projection of a in the tangential direction to the curve and
estimate its length to
be 4.5 (using the fact that a has length 10 as a guide). Similarly, the norma
l component of
a is the length of the projection of a onto N, so we sketch the scalar project
ion of a in the
normal direction to the curve and estimate its length to be 9.0. Thus
aT
a
D
Y
aN
4.5 cm/s
2 and
0
x
9.0 cm/s
.
2
38. L(t)=inr(t) xv(t)
L’(t)
m[r’(t) x v(t) + r (I) x v’(t)j
m[v(t) x v(t) + r(t) x v’(t)}
=
So if the torque is always 0, then L’(t)
[by Formula 5 of Theorem 10.2.3]
=
rn[0 + r(t) x a (t)]
=
0 for all t, and so L(t) is constant.
39. If the engines are turned off at time I, then the spacecraft will continue
to travel in the direction of v(t), so we need at such
that for some scalars > 0, r(t) + sv(t)
=
(6, 4,9).
v(t)
=
r’(t)
=
I +
+
2
r(f)
1
+
O
1
)
+sv(i)=
2
(3±t+s2
+1n
(
1
)
t+,7_
so 7
±
=
—
It is easily seen that t
40. (a) Tn dv
di
rim
di
—
t
]
—
dv
rim
du
—
v(t)
—
(b) v(t)I
2 IVel
=
v(0)
=
=
v
=
dv
=
—
di
r 1 din
J
m do
1dm
mdi
—
dii
t + 8t
2
—
Ve.
Thus Tn(0)e7fl(O)
in(0)
.
dv
=
0
v
v(O)
=
rt
v(i)
=
J
—
4, so t
0.
=
1 is the desired solution.
.
rim
—
[Substitution Rule]
=t
m(O)
v(0)
—
0. Therefore, by part (a), 2
EN0te:
=
121 + 3
s3—t,
Integrating both sides of this equation with respect to t gives
J
—
2 0
v
)
, and v(0)
.
ye.
1
m
(t)
—
in()v.
isi()
tt
3+t+s=6
1 is a root of this polynomial. Also 2 + ln 1 +
42
Ve
2
k
9 +1)2
(t
_1n()
m(0) > in(t) so that ln() >
o]
rn(t)
=
e
m
2
(0).
2 is the fraction of the initial mass that is burned as fuel.
e
02010 Cnttgogn Lannntg
All Rights Rtanrs,d May not ho sontnnd. nopid, ordoplinotod, or perIod to a pnbhnly
aorosatblo woiroto, rn whole or in port
27
p
SECTION 103 PAMETRIC SURFACES
10.5
219
Parametric Surfaces
I. P(7. 10.4) lies on the parametric surface r(u. v)
and v where 2u + 3v
u
2, v
=
=
=
7, 1 + 5u
—
v
(2u + 3v. 1 + 5u
10, and 2 +
=
5, 1 + 5n
first two equations simultaneously gives u
4, v
=
2. P(3, —1,5) lies on the parametric surface r(u, v)
=
±v
11
2
3, u
—
v
=
—1, and
=
2
±v
it
=
2
second
equationgivesu —3±u=
v
=
5
or
Q(—1,
+v
=
v, 2 + v
v) if and only if there are values for
-—
11
4. But solving the first two equations simultaneously gives
1 and these values do not satisfy the third equation, so P does not lie
on the surface.
Q(5, 22, 5) lies on the surface if 2u + 3u
where
ii
—
it =
1
it =
=.
—2, v
=
1
v
22, and 2 + u + v
=
—1 and these values satisfy the third equation, so
Ku + v,
—
1,
it
it =
=
—1, u
2
—
v
=
Q
lies on the surface.
vu + v
) ifand only if there are values for u and v
2
H-v--2=0
3
=
u and substituting into the
(v+2)(u— 1)=0,sou
2. The third equation is satisified by
or
5 for some values ofu and v. Solving the
=
5. From the first equation we have v
2
3,4) lies on r(u, v) if and only if it + v
second gives u
—
it =
1, v
=
—2
2 so P does lie on the surface.
3, and u -b v
2
=
4, but substituting the first equation into the
—2, and neither of these pairs satisfies the third equation. Thus,
Q does not lie
on the surface,
3. r(u,
it)
(it
+ v) 1 ± (3— v)j + (1
4u + 5u)k
-
=
(0,3,1) + u (1,0,4) +
it
(1, —1,5). From Example 3, we recognize
this as a vector equation ofa plane through the point (0,3, 1) and containing vectors
a
=
(1,0,4) and b
wish to find a more conventional equation for the plane, a normal vector to
the plane is a x b
=
I
jk
1
0 4
1—1
and an equation of the plane is 4(x —0)
4. r(u,
the
z
)
= it.
=
2 Sin
—
(p —3)
—
(z —1)
=
0 or4a’
—
y
—
z
=
the yz-plane are all ellipses. Since z
= it
=
2
sin
it
1, —1,5). lfwe
=
41
j
—
—
k
5
—4.
I + 3 cos nj + v k, so the corresponding parametric equations for the surface
are x
For any point (x, p z) on the surface, we have (.r/2)
2 + (y/3)
2
=
cos u
+2
=
2 sin it, p
=
3 cos it,
1, so cross-sections parallel to
with 0 < v < 2, the surface is the portion of the elliptical cylinder x
/4 + g
2
/9
2
=
1
forO < z <2.
5. r(a, t)
(s, t, t
2
—
s
)
2
, so the corresponding parametric equations for the surface are x
Point (x, p. z) on the surface, we have z
We
=
2
p
—
=
a, p
=
t, z
=
. With no restrictions on the parameters, the surface is z
2
x
—
2
p
2,
—
For any
, which
2
.r
recognize as a hyperbolic paraboloid.
6. r(.s. i)
s sin 2t I +
2
i ± a cos 2t k,
so the corresponding parametric equations for the surface are x
sin 2t,
=
= a cos 2t. For any point (x,
y, z) on the surface, we have x
2 +z
2 = 2
2t + 2 co
2 2t
= p. Since no
5 are placed
TStriction
on the parameters, the surface is p = x
2+z
, which we recognize as a circular paraboloid whose axis
2
-
2010 Urogogo loro,,o
All R,ohto Rosrr,d. Iloy not
noormod. copird. or dopliotod,
0, po,lnd too poblely oro,r,bl, nobron, to
obolo or ot port
.
U
SECTlONlO.5PARfJJRlCSlJRFACES
II. x
sinv, p
=
cosusin4v,
Note that if v
=
Vo
sin2zisin4v, 0<
z
is constant, then x
=
sin
u
< 27r,
—*
<V
<
is constant, so the
La
corresponding grid curves must be parallel to the yz-plane. These
are the vertically oriented grid curves we see, each shaped like a
“figure-eight.” When
u =
equations become x
no is held constant, the parametric
sin
sin 2uo sin 4v. Since
u,
z
cos no sin 4v,
y
is a constant multiple ofy, the
—i
corresponding grid curves are the curves contained in planes
z =
12. x
=
ky that pass through the i-axis.
usinu cosv,y
ncosu cosu.
z =
‘usinv.
We graph the portion of the surface with parametric domain
0
u
< 4ir, 0 <
v
equations become x
< 2. Note that ifv
n sin u cos u, y
=
= vO
=
is constant, the parametric
n cos n cos to,
z = u
sin tO.
The equations for x and y show that the projections onto the iy-plane give
a spiral shape, so the corresponding grid curves are the almost-horizontal
spiral curves we see. The vertical grid curves, which look approximately
circular. correspond to
0 (‘050,
U05iflU
taft
IIi III;
13. r(u, v)
=
u cos v i +
zi =
no being held constant, giving
p
=
at
sin vi +
UO C05L1o COSV, Z
v
‘LL
Sin
k. The parametric equations for the surface are x
the grid curves first; if we fix v, then x and y parametrize a straight line in the plane
held constant, the projection onto the xy-plane is circular; with
z = at,
= at
cos v, p
z
v
=
u sin v, z
=
v. We look at
which intersects the z-axis. If at is
each grid curve is a helix. The surface is a spiraling
ramp, graph IV.
14. r(u, v)
z
=
ucvi+
sinu, —ir <
‘it
<
u
sin vj + sin u k. The corresponding parametric equations for the surface are
‘ir. Ifu
0 in the horizontal plane
u
=
z =
no is held constant, then x
sinuo. Ifv
grid curves lie in vertical planes y
=
vO
no cosv, y
is constant, then x
=
at
sinv
,p
0
zicosv
i =
u cos
i’,
p
= at
sin v,
so each grid curve is a circle of radius
=
usinvo
p
=to
=
(tan vo).r, so the
ki through the z-axis. In fact, since x and p are constant multiples of
it
and
z =
sin at,
each of these traces is a sine wave. The surface is graph 1.
15. r(u, u)
Z =
sin
=
it
sin v i + cos at sin 2vj + sin at sin 2v k. Parametric equations for the surface are
sin2v. Ifv
=
VO
is fixed, then x
=
0 is constant, andy
sinv
(sin2vo)cosu
.t’ =
and z
sin at, p
=
=
(os at sin 2v,
(sin2vu)sinu describe a
f circle of radius sin 2voI, so each corresponding grid curve is a circle contained in the vertical plane .r
sin v parallel to the
liz-plane. The only possible surface is graph II. The grid curves we see running lengthwise along the surface correspond to
holding
fl@
at
—
constant, in which case y
iij
that includes the
(cos no) sin 2v, z
=
(sin uO) sin 2v
oz.
z
=
(tan uo)y, so each grid curve lies in a
,7-a\Is.
C 20101 ngogn Lnsoning All Rh,n
nd. tlo not bo toaonnd, oopnd, or duplicotod, or ponond to a pubhcb 000nbuibin svnba,tr. in t.hol, or to pant.
16. x
VECTOR FUNCTIONS
CHAPTER 10
U
222
(1
(1 — u)(3 + cos v) cos 4iru, y
=
graph V: when
a =
0, then
x =
centered at (3, 0,0). When a
radius
=
—
u)(3 + cos
3 + cos v, y
then x
,
in the ia-plane centered at
0, and
=
sin 4wa,
0, and
1, then x
(, 0, ). When a
— a) sin v. These equations correspond to
3u + (1
z =
e with radius 1 in the xz-plane
sin v, which are equations of a circl
a =
cos v, y
+
=
v)
—f-
z =
y
=
=
e with
sin a, which are equations of a circl
-
0 and z
=
n in the
3, giving the topmost point show
le on the surface. The spiralling
the vertically oriented circles visib
are
tant
cons
u
with
es
curv
grid
the
graph. This suggests that
constant.
grid curves correspond to keeping a
17.
3
3 a cos
cos
v,
p
=
3 v. If V
sin
3 v, z
3 u cos
sin
is held constant then a
= Vo
=
3 tio is constant, so the
sin
ontal grid curves, but the curves for this
Several of the graphs exhibit horiz
e.
plan
l
onta
horiz
a
in
lies
e
curv
corresponding grid
curves here are
ibility. (In fact, the horizontal grid
ght lines, so graph III is the only poss
surface are neither circles nor strai
cal grid curves we see on the surface
and are called astroids.) The verti
u
3
= a sin
p
u,
s
aco
3
=
x
ly
fami
members of the
3 v so the corresponding grid curve
3 no cos
3 v, p = sin
cos
3
as then we have x = cos
tant,
cons
held
U
=
u
to
d
spon
corre
3 aO)x through the z-axis.
lies in the vertical plane p = (tan
x =
nx
)
)
he
T
l
u.l
=D
,z
u
u
u
nv
a
s
l
si
l
l
f
lnD
n
-—
y
i
(1
—
=
—
sv,y
o
i
1
co
1
—
o
(
uD
(
l
-l
-.
(
(1
=
s
+
=
.x
18
±
=
c
v
s
v
,
2
The graph then must be graph VI.
al) in the horizontal plane z a.
circle of radius (1
is held constant, each grid curve is a
al) sinvo. Then p = (tan vo)x, so the grid
(1 — al) COSVO andy = (1
z
Ifv is held constant, so v = Vo, we have
spond to keeping v constant.
the surface in the planes y = kx corre
g
alon
cally
verti
ing
runn
see
we
curves
i> and
that contains the vectors a = (1, 1,—
plane through the point (1,2, —3)
the
for
tions
equa
ic
metr
para
3,
19. From Example
v(—1)
v(1) +u+vy = 2 + u(1) +
b = (1, —1, 1) are x = 1 + u(1) +
—
—
z =
—3 +
u(—1) + v(1)
20. Solving the equation for
z
gives
2 =
z
1
—
2
2x
—
2
4y
=t-
—
z =
2x2
—
are x
parameters, parametric equations
ellipsoid). If we letx andy be the
+
rewrien as
Alternate solution: The equation can be
p
=
asinv, then
z
—/i
2
— 2x
2
zip
=
_\/iZ
2
u
s
2
co
v
=
x, p
2
+ a
(2r
—
2 (since we want the lower half of
— zip
in
2
a2s
=
=
v =
y,
z =
—/i
—
2
2x
1, and if we let w
/1
—
—
.
2
4y
a
=
, where 0 <
2
n
the
a
cosv and
< land
—
0 < a < 2w.
since we want the
. (We choose the positive root
2
2 —I.- a
x
11
p
2
a
2
x
+
1
=
s
2
p
21. Solving the equation for p give
X, Z = Z,
parametric equations are x =
we let x and z be the parameters,
If
0.)
to
ds
spon
corre
id
that
rbolo
part of the hype
—
—
y.1_x2z2.
22.
4
—
2
y
—
,p
2
2a
=
y,
z
z
ciated vector equation is
2 < 4 since x > 0. Then the asso
2 + 2z
where p
2a yj + zk.
i+
2 )
2
—
r(y,z)= (4— p
2
2 +p
23. Since the cone Intersects the sphere in the circle a
can parametrize the surface as .r
x, y
=
y, a
=
Alternate solution: Using spherical coordinates, x
=
v/Z2
=
2,
—
z =
and we
2 where
y
2 sin th cos 8, y
2
i
+
want
e this, we
the portion of the sphere abov
2.
2 sin 4’ sin0, z
=
2 cos 4’ where 0
0<0<2w.
ro pobl,oly aocroo,blr wrb,,tr.
bfl mam,,d. flopfld. o, daplmfld. or portrd
02010 Cgfl L,,,ng All R,gh Rr,d M. ‘ml
fl
oholo or n port
4’
and
11
PARTIAL DERIVATIVES
11.1
Functions of Several Variables
1. (a) From Table I, f(—15, 40)
—27, which means that if the temperature is —15°C
and the wind speed is 40 km/h, then the
air would feel equivalent to approximate’ —27°C withou
t wind.
(b) The question is asking: when the temperature is —20°C
, what wind speed gives a wind-chill index of —30°C
? From
S4
Table 1, the speed is 20 km/h.
(c) The question is asking: when the wind speed is 20 km/h,
what temperature gives a wind-chill index of _9 C?
From
Table 1, the temperature is —35° C.
(d) The function
))7
= f(—5, v) means that we fix Tat —5 and allow v to
vary, resulting in a function of one variable. In
other words, the function gives wind-chill index values
for different wind speeds when the temperature is
—5° C. From
Table I (look at the
corresponding to T = —5), the function decreases and
appears to approach a constant value as v
increases.
row
(e) The function IV = f(T, 50) means that we fix v at 50
and allow T to vary, again giving a function of one variabl
e. In
other words, the function gives wind-chill index values for
different temperatures when the wind speed is 50 km/h.
From
Table I (look at the column corresponding to v = 50),
the function increases almost linearly as T increases.
2. (a) From Table 3, f(95, 70) = 124, which means that
when the actual temperature is 95°F and the relative humid
ity is 70%,
the perceived air temperature is approximately 124° F.
(h) Looking at the row corresponding to T = 90, we
see that f(90, Ii) = 100 when h = 60.
(c) Looking at the column corresponding to ii
= 50, we see that f(T, 50) = 88 when T = 85.
(d) I = f(80. h) means that T is fixed at 80
and ii is allowed to vary, resulting in a function of h that
gives the humidex values
for different relative humidities when the actual
temperature is 80°F. Similarly, I = f(100, h) is a functio
n of one
variable that gives the hum jdex values for
different relative humidities when the actual temperature
is 100° F. Looking at
the rows of the table corresponding to
T = 80 and T = 100, we see that f(80, h) increases at
a relatively constant rate of
approximately 1°F per 10% relative humid
ity, while f(100, h) increases more quickly (at first with
an average rate of
change of 5°F per 10% relative humid
ity) and at an increasing rate (approximately 12°F per
10% relative humidity for
larger values of h).
..
—
.
If the amounts of
labor and capital are both doubled, we replace
L, K in the function with 2L, 2K, giving
P(2L.2K) = (2
2
1.O
7
K
1(2
5
)° = K
L)°
1.(
)
2
)
7
L
01
2
5
(2
°
°
°
° = 2
(2’)1.O
7
K
5
°
1L°
Thus the production is
doubled. It is also true for the general case
P(L, K) = bL°K’
2P(L,K)
—
b(2L)”(2K)’
1
b(2)(
2
)LK’
(
)
1
bL
2
K = 2P(L,K).
r
239
C 2010
Cnnnagn Lnoonrn
0
All Righto Rn,ortnd 01
oy not On noom,d. coptnd, or duplioatnti o, postnd tn
pohitoly nonnsibl wrbntt. in otholn
or tn p’ot.
SECTION 11.1
FUNCtiONS OFSEVERA]1I
Ct
241
9. The point (—3, 3) lies between the level curves
with z-values 50 and 60. Since the point is a
little closer to the level curve with
z = 60, we estimate that f(—3, 3)
56. The point (3, —2) appears to be just about halfwa
y between the level curves with
z-values 30 and 40, so we estimate f(3, —2)
35. The graph rises as we approach the origin, gradua
lly from above, steeply
from below.
10. (a) C (Chicago) lies between level curves with pressu
res 1012 and 1016 mb, and since C appears
to be located about
one-fourth the distance from the 1012 mb isobar
to the 1016 nib isobar, we estimate the pressu
re at Chicago to be about
1013 mb. N lies very close to a level curve with
pressure 1012mb so we estimate the pressure
at Nashville to be
approximately 1012 mb. S appears to be just about
half\vay between level curves with pressures
1008 and 1012 mb, so we
estimate the pressure at San Francisco to be about
1010 mb. V lies close to a level curve with
pressure 1016mb but we
can’t see a level curve to its left so it is more difficu
lt to make an accurate estimate. There are lower
pressures to the right
of V and V is a short distance to the left of the level
curve with pressure 1016 mb, so we might estima
te that the pressure at
Vancouver is about 1017 nib.
d
flu
f
(b) Winds are stronger where the isobars are closer togeth
er (see Figure 6), and the level curves are closer
near S than at the
other locations, so the winds were strongest at San
Francisco.
11. The point (160, 10), corresponding to day 160 and
a depth of 10 m, lies between the isothermals with
temperature values
of 8 and 12° C. Since the point appears to be located
about three-fourths the distance from the 8°C
isothermal to the 12°C
isothermal, we estimate the temperature at that point
to be approximately 11° C. The point (180
,5) lies between the 16 and
20CC isothermals, very close to the 20°C level curve,
so we estimate the temperature there to be about
19.5°C.
12. If we start at the origin and move along the
x-axis, for example, the z-values of a cone centere
d at the origin increase at a
constant rate, so we would expect its level curves
to be eqially spaced. A paraboloid with vertex
the origin, on the other hand,
has z-values which change slowly near the
origin and more quickly as we move farther away.
ThuS, we would expect its level
curves near the origin to be spaced more widely
apart than those farther from the origin. Therefore
contour map I must
correspond to the paraboloid, and contou
r map lithe cone.
13. Near A, the level curves
are very close together, indicating that the terrain
is quite steep. At B, the level curves are much
farther apart, so we would expect
the terrain to be much less steep than near A, perhap
s almost flat.
‘4
2010 Crno, Lrtrrnng.
ku lOghIt R,tMvd Moy not bo sooon,d. copIrd.
or dophootrd, or pottnd too pobholy ooc,,nbl, n,bntt, n whom
.
o, n paot
C]
244
2
9x
2
4y
= 36
The contour map consists of the level curves k
, Ic 0, a family of ellipses with major axis the
2
2 = 36 k
2 + 4y
Ox
—
y
28.
I
$flJ_
PARI1AL DERIVATIVES
CHAPTER 11
—
—
p-axis. (Or, if Ic
=
6, the origin.)
‘4
z4
3
=,
The graph of f(x, y) is the surface z
=
./36
—
2
9x
—
id
, or equivalently the upper half of the ellipso
2
4y
36. If we visualize lifting each ellipse Ic
2
2 +z
2 + 4y
9x
=
\/36
—
2
Ox
—
2 of the contour map to the plane z
4y
=
k,
the graph of f.
we have horizontal traces that indicate the shape of
29, The isothermals are given by k
2
2 + 2y
x
=
(100
—
=
) or
2
2 + 2y
100/(1 ± x
k)/k [0 < k < 1001, a family of ellipses.
30. The equipotential curves are k
C
=
./r2
or
.2
—
—
/
2
2 +p
x
=
Note: As k
2
r
—
2
(C)
—* 00,
a family of circles (k
c/r).
the radius of the circle approaches r.
.3
y
x
f(x,y)=x
—
2
31. 3
2
—2
—2
—2
l to the xzplane
nt trace in the graph above) are parabolas; those paralle
The traces parallel to the yz-plane (such as the left-fro
sitting on the stirfa
surface is called a monkey saddle because a monkey
(such as the right-front trace) are cubic curves. The
near the origin has places for both legs and tail to rest.
nonnslbIn o.eboitn, tn whom orn pOt.
not bn oannnd, topind, or dpIioatnd, o, potnd ton pblloIy
C 2010 Cnonagn Lnrning. Alt Riht Rnsor.nd Moy
SECTION UI
32. f(r, y)
FUNCTIONS OF SEVERAI VARIABLES
0
3
iy
—
The traces parallel to either the
yz-plane or the xzplane are cubic
curves.
33. f(x, y)
=
’
2
e
2 )
2
(sin
(x + )
2
cos
)
(y
I
4
V?
4
—4
34. f(x,y)
cosx cosy
=
2
0 0
The traces parallel to either the
yz- or xz-plane are cosine curves
zO
2ir
with amplitudes that vary
0 0
from Oto 1.
-2ir
35. z
=
sin(xy)
(a) C
(b) II
Reasons: This function is periodic in
both x and y, and the function is the same
when x is interchanged with y, so its grap
h is
symmetric about the plane p = x. In
addition, the function is 0 along the x- and
p-axes. These conditions are satisfied only
by
(‘and II.
36.
.
Z =
cot, y
(a) A
(b) IV
Reasons: This function is perio
dic in p but not x, a condition satisfied only
by A and IV. Also, note that traces in .r
Cosine curves with amp
litude that increases as ‘x increases.
37. z
=
sin(x
y)
z
=
slur
sing
(a) E
(b) III
Reasons This func
tion is periodic in both
11
k are
(a) F
(b) I
Reasons: This function
is periodic in both a and p but is cons
tant along the lines
by F and I.
—
=
=
x + k, a condition satisfied only
—
Ia
.
and
v’ but unlike the ILInCL on in Exeicise 37, it iS not constant along lines such
Z + r
5 so the contour
map is III Also notice that traces in
p
2o the graph
must be E.
=
k are vertically shifted copies of the sine wave z
C O
2
tO(fl5
Lraartno AH Rights Rrsrs,,
,d. M not ho arnn,d. oopid, ordtplirsgod,
orpostsd too pobIicy occnsib5 srrbsit, in ohol, or
in pars.
as
sinr,
245
DjCHAPThRI1]ARTIA
246
39. z
(b)Vl
(a) B
(1 —y
—x
)
(1 2
)
2
=
.—
Reasons: This function is 0 along the lines x
=
±1 and p
note that the trace in the az-plane is the parabola z
=
1
—
is VI. Also
±1. The only contour map in which this could occur
.r and the trace in the yz-plane is the parabola a
=
1
—
so the
graph is B.
.z=
1
4O
9
y
(b)V
(a)D
ch 0 as we use
the graphs in A, C, E, and F. Also, the values of z approa
Reasons: This function is not periodic, ruling out
shows this behavior is D, which corresponds to V.
points farther from the origin. The only graph that
41. k
=
normal vector (1,3,5>.
a + 3y + 5z is a family of parallel planes with
fork
2 is a family of ellipsoids fork > 0 and the origin
2 + 5z
2 + 3y
x
42. k
43. Equations for the level surfaces are k
radius ‘/>. When k
=
0.
ers with axis the .r-axis and
. For k > 0, we have a family of circular cylind
2
2 +z
p
level surfaces for k < 0.)
0 the level surface is the x-axis. (There are no
44. Equations for the level surfaces are
l2
—
—
2
a
=
k. For k
=
2 +z
0, the equation becomes p
=
2 and the surface is a
x
oloids of two sheets with
the a-axis. For k > 0, we have a family of hyperb
right circular cone with vertex the origin and axis
of hyperboloids of one sheet with axis the x-axis.
axis the x-axis, and fork < 0, we have a family
upward 2 units.
45. (a) The graph of p is the graph of f shifted
vertically by a factor of 2.
(b) The graph of p is the graph of f stretched
the xp-plane.
(c) The graph of g is the graph of f reflected about
(d) The graph of g(a’, y)
=
—f(x, y)
xy-plane and then shifted upward 2 units.
+ 2 is the graph of f reflected about the
2 units in the positive a-direction.
46. (a) The graph of p is the graph of f shifted
in the negative p-direction.
(b) The graph of g is the graph of f shifted 2 units
positive p-direction.
in the negative x-direction and 4 units in the
(c) The graph ofg is the graph of f shifted 3 units
47. f(x,y)
a
=
=
e
. First, if c
e’
°
2
0, the graph is the cylindrical surface
(whose level curves are parallel lines). V/hen
c>
0, the vertical trace
in the x-direction
above the p-axis remains fixed while the sides of the surface
paraboloid. The
“curl” upward, giving the graph a shape resembling an elliptic
Ji
level curves of the surface are ellipses centered at the origin.
ForD <
c<
r1cty increases as
1, the ellipses have major axis the x-axis and the eccent
C 2010 (ng Lrnng. All Rght
‘.1 ,e1 b aond.
o,
dphd. o, pod
too
c —i
0.
pubholy aot,IbI, wbtt,
0 oholfi