Unformatted text preview: 176 Section 4.6 4. Use implicit differentiation.
d
(x sin y)
dx (x)(cos y) dy
dx (x (sin y)(1)
x cos y) 3. (a) Since
d
(1
dx
dy
x
dx dy
dx
dy
dx
dy
dx y 2x
dy
dx
dy
dx dV dh
dV
, we have
dh dt
dt (b) Since dV
dt dV dr
dV
, we have
dr dt
dt xy)
y(1)
sin y (c) y sin y
x x cos y
y sin y
x x cos y 5. Use implicit differentiation.
d2
x
dx dV
dt 4. (a) d
tan y
dx
dy
sec2 y
dx
2x
sec2 y 2x cos2 y dV
dt
dV
dt
dV
dt d2
rh
dt
dh
r2
dt
dh
r2
dt dP
dt
dP
dt
dP
dt
dP
dt d
ln (x y)
dx
dy
1
1
dx
xy
dy
1
dx
dy
dx dh
.
dt
dr
dt 2 rh . d
(RI 2)
dt
dR
d
R I2 I2
dt
dt
dI
dR
R 2I
I2
dt
dt
dI
dR
2RI
I2
dt
dt d2
(r h)
dt
dr
h(2r)
dt
dr
2 rh
dt (b) If P is constant, we have 6. Use implicit differentiation. r2 2RI d
(2x)
dx dI
dt dR
dt 0, or y2 I2 dP
dt 0, which means dR
dt 2R dI
I dt 2P dI
.
I 3 dt z2 2
2(x
2x 5. y)
2y 8. Using A(0, 4), we create the parametric equations
x 0 at and y
4 bt, which determine a line
passing through A at t 0. We now determine a and b so
that the line passes through B(5, 0) at t 1. Since
5 0 a, we have a 5, and since 0
4 b, we have
b 4. Thus, one parametrization for the line segment is
x 5t, y
4 4t, 0 t 1. (Other answers are
possible.) 10. One possible answer: 2
3
2 d
dt x2 1
ds
dt 7. Using A( 2, 1) we create the parametric equations
x
2 at and y 1 bt, which determine a line
passing through A at t 0. We determine a and b so that
the line passes through B(4, 3) at t 1. Since
4
2 a, we have a 6, and since 3 1 b, we
have b
4. Thus, one parametrization for the line
segment is x
2 6t, y 1 4t, 0 t 1. (Other
answers are possible.) 9. One possible answer: ds
dt t dA
dt dA dr
dA
, we have
dr dt
dt 2. Since dS
dt dS dr
dS
, we have
dr dt
dt 6. 2x 2 y y 2 2 dx
x
dt dy
y
dt x2 dA
dt
dA
dt
dA
dt y2 (x 2 y2 dx
dt 2y z 2) z dt 1
z
z 2 2x dy
dt 2z dz
dt dz
dt z2 d1
ab sin
dt 2
1 da
db
b sin
a
2 dt
dt
1
da
db
b sin
a sin
2
dt
dt dV
dt 2 sin ab ab cos d
sin
dt d
dt 1 volt/sec. (b) Since I is decreasing at the rate of
dI
dt Section 4.6 Exercises
1. Since ds
dt 2x d
2 7. (a) Since V is increasing at the rate of 1 volt/sec, 3
2 t ds
dt 1
2 dr
dt 2r .
dr
dt 8r . 1
amp/sec.
3 (c) Differentiating both sides of V
dV
dt 1
amp/sec,
3 I dR
dt dI
dt R. IR, we have Section 4.6
(d) Note that V IR gives 12 2R, so R 6 ohms. Now
substitute the known values into the equation in (c).
1 2 dR
dt (b) P 1
3 6 dR
2
dt
3
ohms/sec
2 3
dR
dt 3
ohms/sec. Since this
2 At the instant in question, dr
dt 0.01 cm/sec, r dA
.
dt Step 5:
2r dr
dt 52 At the instant in question,
cm2/sec 2 (50)(0.01) At the instant in question, the area is increasing at the rate
cm2/sec. At the instant in question,
dw
dt 2 cm/sec, 2 cm/sec, l 12 cm, 5 cm. dy
dt y = 3 m, and z We want to find 2 m/sec, dz
dt 1 m/sec, x 4 m, 2 m. dA dP
dD
, , and
.
dt dt
dt dV dS
ds
, , and .
dt dt
dt Steps 4, 5, and 6:
(a) V xyz
xy dz
dt xz (4)(3)(1) dy
dt yz dx
dt (4)(2)( 2) dS
dt
dS
dt 2(xy xz dy
2x
dt yz)
dx
y
dt x dz
dt z dx
dt y lw 2[(4)( 2) (3)(1)
(3)(1) (2)( 2)] dz
dt z dy
dt (4)(1) (2)(1) Steps 4, 5, and 6: 2 m3/sec (3)(2)(1) The rate of change of the volume is 2 m3/sec.
(b) S Step 3: dw
l
dt 1 m/sec, dV
dt
dV
dt Step 2: We want to find dx
dt Step 3: 9. Step 1:
l length of rectangle
w width of rectangle
A area of rectangle
P perimeter of rectangle
D length of a diagonal of the rectangle dA
dt
dA
dt dl
2l
2
2
dt
2l
w
(12)( 2) (5)(2)
122 dw l
w
dt
dt
dw
2w
2
dt
l
w2
14
cm/sec
13 1 Step 2: Step 6: (a) A w2 10. Step 1:
x, y, z edge lengths of the box
V volume of the box
S surface area of the box
s diagonal length of the box r2 and w 0 cm/sec (d) The area is increasing, because its derivative is positive.
The perimeter is not changing, because its derivative is
zero. The diagonal length is decreasing, because its
derivative is negative. Step 4: dl
dt 2(2) 14
cm/sec.
13 50 cm. Step 3: of 2( 2) The rate of change of the length of the diameter is Step 2: dA
dt dw
dt dl dD
dt
dD
dt 8. Step 1:
r radius of plate
A area of plate dA
dt 2 l2 (c) D value is positive, R is increasing. A 2w dl
2
dt The rate of change of the perimeter is 0 cm/sec. R is changing at the rate of We want to find 2l dP
dt
dP
dt 177 0 m2/sec The rate of change of the surface area is 0 m2/sec.
dl
w
dt (12)(2) (5)( 2) 14 cm2/sec The rate of change of the area is 14 cm2/sec. 178 Section 4.6
13. Step 1:
x distance from wall to base of ladder
y height of top of ladder
A area of triangle formed by the ladder, wall, and ground
angle between the ladder and the ground 10. continued
x2 (c) s y2 z2 1 ds
dt 2 x 2 y2 2x z2 dx
dt 2y dy
dt 2z dz
dy Step 2:
dx
x
dt dy
y
dt x 2 y dz
z
dt 2 z (4)(1) ds
dt At the instant in question, x 2 (2)(1) 32 0 22 29 Step 4, 5, and 6: 11. Step 1:
s (diagonal) distance from antenna to airplane
x horizontal distance from antenna to airplane ds
dt Step 4:
x 2 49 y 300 mph. dx
.
dt s2 x2 169 169 122 5. rate of 12 ft/sec. (Note that the downward rate of 49 motion is positive.)
2s s2 49 ds
dt s
s2 ds
49 dt 10
102 (300) 3000 49 mph 1
xy
2
1 dy
x
2 dt (b) A Step 6:
dx
dt 0 The top of the ladder is sliding down the wall at the s 2 or x 1
2 169
dy
2y
dt dy
Then 2(12)(5) 2(5)
0
dt
dy
dy
12 ft/sec or
12 ft/sec
dt
dt Step 5:
dx
dt y2 To evaluate, note that, at the instant in question, Step 3:
We want to find (a) x 2
dx
2x
dt Step 2:
At the instant in question,
10 mi and 5 ft/sec. dy dA
d
, , and .
dt dt
dt We want to find 0 m/sec The rate of change of the diagonal length is 0 m/sec. s dx
dt Step 3: (3)( 2)
42 12 ft and dA
dt 420.08 mph y dx
dt Using the results from step 2 and from part (a), we 51 The speed of the airplane is about 420.08 mph. have 12. Step 1:
h height (or depth) of the water in the trough
V volume of water in the trough dA
dt 1
[(12)( 12)
2 (5)(5)] 119
ft/sec.
2 The area of the triangle is changing at the rate of
59.5 ft2/sec. Step 2:
At the instant in question, dV
dt 2.5 ft3/min and h 2 ft. y
x (c) tan
2 x d
dt Step 3: sec dh
We want to find .
dt Since tan Step 4:
4
3 cos The width of the top surface of the water is h, so we have
V 1
4
(h) h (15), or V
2
3 10h 2 20h y dx
dt x2
5
, we have for 0
12
1 12
and so sec2
13 dh
dt d
dt Step 6: of
dh
dt 2.5 20(2) dh
dt 0.0625 1
ft/min
16 The water level is increasing at the rate of 1
ft/min.
16 2 12 2
13 169
.
144 Combining this result with the results from step 2 and
from part (a), we have Step 5:
dV
dt dy
dt 169 d
144 dt (12)( 12) (5)(5)
, so
122 1 radian/sec. The angle is changing at the rate
1 radian/sec. Section 4.6
14. Step 1: 179 Step 5:
Kite
s Inge dV
dt 12 r dr
dt Step 6: 300 ft dV
dt x 1
3000 12 (1.900) 19
2500 0.0076 3 0.0239 in /min
s
x length of kite string
horizontal distance from Inge to kite The volume is increasing at the rate of approximately
0.0239 in3/min. Step 2:
At the instant in question, dx
dt 25 ft/sec and s 500 ft 16. Step 1: Step 3:
We want to find h ds
.
dt Step 4:
x 2 r 300 2 s 2 r
h
V Step 5:
2x dx
dt 2s ds
dx
or x
dt
dt s ds
dt base radius of cone
height of cone
volume of cone Step 2: Step 6:
At the instant in question, since x 2
x s2 3002 5002
ds
dt Thus (400)(25) (500) , so 3002 3002
ds
dt s 2, we have At the instant in question, h 4 m and dV
dt 10 m3/min. Step 3: 400.
20 ft/sec. Inge must let the string out at the rate of 20 ft/sec. We want to find dh
dr
and .
dt
dt Step 4: 15. Step 1: Since the height is 3
of the base diameter, we have
8 4
h.
3
12
142
16 h 3
We also have V
rh
hh
. We will
3
33
27
16 h 3
4
use the equations V
and r
h.
27
3 h
6 in. 3
(2r) or r
8 Steps 5 and 6:
r (a) The cylinder shown represents the shape of the hole.
r radius of cylinder
V volume of cylinder
Step 2:
At the instant in question, dr
dt 0.001 in.
3 min and (since the diameter is 3.800 in.), r
Step 3: Step 4:
r 2(6) 1.900 in. 10
dh
dt 16 h 2 dh
9
dt
16 (4)2 dh
9
dt
45
m/min
128 1125
cm/min
32 The height is changing at the rate of
1125
32 11.19 cm/min. (b) Using the results from Step 4 and part (a), we have dV
We want to find .
dt V 1
in./min
3000 dV
dt dr
dt 4 dh
3 dt 4 1125
3 32 375
cm/min.
8 The radius is changing at the rate of
6 r2 375
8 14.92 cm/min. 180 Section 4.6 17. Step 1: Step 6:
45 m
r 6
6 6m 144 dy
dt h or
r
h
V radius of top surface of water
depth of water in reservoir
volume of water in reservoir r dV
dt At the instant in question, 50 m3/min and h 5 m. Step 3:
We want to find 0.01326 m/min 25
6 1.326 cm/min
y)2 (13 169 132, y)2 (13 26y dh
dr
and .
dt
dt Step 2:
At the instant in question, Note that
Then V h
6
by similar cones, so r 7.5h.
r
45
12
1
rh
(7.5h)2h 18.75 h3
3
3 We want to find the value of
18.75 h 3, dV
dt 56.25 h 2 dh
.
dt From part (b), r 7.5h, dr
dt dr
dt dh
dt 80
cm/min. The rate
3 or of change of the radius of the water’s surface is
80
3 18. (a) Step 1:
y depth of water in bowl
V volume of water in bowl 1
y2 26y 2 (26 2y) dy
dt 13
26y y dy
y 2 dt 13 8 26(8) 5
288
125
72 1
24 82 5
12 1
24 0.00553 m/min
0.553 cm/min Step 2:
No numerical information is given. Step 2:
dV
dt y 2. 26y 19. Step 1:
r radius of spherical droplet
S surface area of spherical droplet
V volume of spherical droplet 8.49 cm/min. At the instant in question, dr
.
dt Step 6:
dr
dt 7.5 1
m/min.
24 Step 5: falling is positive.)
(b) Since r dy
dt 8 m, Step 4: 56.25 (52) 50 6 m3/min, y Step 3: dh
, and so
dt
dh
8
32
m/min
cm/min.
dt
225
9
32
The water level is falling by
1.13 cm/min.
9
dh
(Since
0, the rate at which the water level is
dt Thus dV
dt and therefore (from part (a)) Step 5 and 6:
(a) Since V y 2. (c) Step 1:
y depth of water
r radius of water surface
V volume of water in bowl Step 4: y dy
dt dy
dt 1
24 (b) Since r 2 Step 2: (82)] [26 (8) 6 m3/min and Step 3:
We want to show that 8 m. Step 3: Step 4: dy
We want to find the value of .
dt S Step 4: dr
is constant.
dt Steps 5 and 6: V 3 y 2(39 y) or V Step 5:
dV
dt (26 y y 2) dy
dt 13 y 2 3 y3 4 r 2, V 4 3 dV
r,
3
dt kS for some constant k 43
dV
dr
r , we have
4 r2 .
3
dt
dt
dV
Substituting kS for
and S for 4 r 2, we have kS
dt
dr
or
k.
dt Differentiating V dr
dt S, ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.
 Spring '08
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