45. x = 10, 000 " p2 = (10,000 - p2)1/2
d
d
x =
(10,000 - p2)1/2
dx
dx
1
d
!
1 =
(10,000 - p2)-1/2
[10,000 - p2]
2
dx
1
1 =
· (!2 pp')
2(10, 000 " p2)1 2
" pp #
1 =
10, 000 " p2
!
" 10, 000 " p2
p' =
p
!
47. (L + m)(V + n) = k
dL
(L + m)(1) + (V + n)
= 0
!
dV
dL
(L + m)
= dV
V + n
EXERCISE 4-6
!
!
Things
to remember:
!
1. SUGGESTIONS FOR SOLVING RELATED RATE PROBLEMS
Step 1. Sketch a figure.
Step 2. Identify all relevant variables, including those
whose rates are given and those whose rates are to
be found.
Step 3. Express all given rates and rates to be found as
derivatives.
Step 4. Find an equation connecting the variables in step 2.
Step 5. Implicitly differentiate the equation found in
step 4, using the chain rule where appropriate,
and substitute in all given values.
Step 6. Solve for the derivative that will give the unknown
rate.
1. y = x2 + 2
Differentiating with respect to t:
dy
dx
dy
dx
= 2x
;
= 2(5)(3) = 30 when x = 5,
= 3
dt
dt
dt
dt
EXERCISE 4-6
191
3. x2 + y2 = 1
Differentiating with respect to t:
dx
dy
2x
+ 2y
= 0
dt
dt
dx
dy
2x
= -2y
dt
dt
0.8
dx
y dy
dx
16
= ;
= (-4) = - ,
(!0.6)
dt
x dt
dt
3
dy
when x = -0.6, y = 0.8,
= -4
dt
5. x2 + 3xy + y2 = 11
Differentiating with respect to t:
dx
dy
dx
dy
2x
+ 2x
+ 3y
+ 2y
= 0
dt
dt
dt
dt
dy
dx
(3x + 2y)
= -(2x + 3y)
dt
dt
(2x + 3y) dx
(2 ! 1 + 3 ! 2)
dy
dy
16
= ;
= 2 = 3x + 2y dt
(3 ! 1 + 2 ! 2)
dt
dt
7
dx
when x = 1, y = 2,
= 2
dt
7. xy = 36
Differentiate with respect to t:
d(xy ) d(36)
=
dt
dt
dy
dx
x
+ y
= 0
dt
dt
dx
Given:
= 4 when x = 4 and y = 9. Therefore,
dt
dy
4
+ 9(4) = 0
dt
dy
4
= -36
dt
dy
and
= -9.
dt
The y coordinate is decreasing at 9 units per second.
9.
z = rope
y = 4
x
From the triangle,
x2 + y2 = z2
or x2 + 16 = z2, since y = 4.
Differentiate with respect to t:
dx
dz
2x
= 2z
dt
dt
dx
dz
or
x
= z
dt
dt
192
CHAPTER 4
ADDITIONAL DERIVATIVE TOPICS
dz
= -3.
dt
Therefore,
Given:
30
dx
=
dt
916 (-3)
Also, when x = 30, 900 + 16 = z2 or z =
and
916 .
dx
"30.27
"3 916
" 916
!
=
≈
=
dt
10
30
10
≈ -3.03 feet/second.
[Note:
The negative sign indicates that the distance between the boat
!
and the dock is decreasing.]
!
!
11. Area: A = πR2
d!R 2
dA
dR
=
= π·2R
dt
dt
dt
dR
Given:
= 2 ft/sec
dt
dA
= 2πR·2 = 4πR
dt
4 3
πR
3
dV
4
dR
dR
=
π3R2
= 4πR2
dt
3
dt
dt
dR
Given:
= 3 cm/min
dt
dV
= 4πR23 = 12πR2
dt
13. V =
dA
= 4π(10) = 40π ft2/sec
dt R =10 ft
≈ 126 ft2/sec
15.
dV
= 12π(10)2 = 1200π
dt R =10 cm ≈ 3768 cm3/min
P
= k
(1)
T
P = kT
Differentiate with respect to t:
dP
dT
= k
dt
dt
dT
Given:
= 3 degrees per hour, T = 250˚, P = 500 pounds per square
dt
inch.
From (1), for T = 250 and P = 500,
500
k = 250 = 2.
Thus, we have
dP
dT
= 2
dt
dt
dP
= 2(3) = 6
dt
Pressure increases at 6 pounds per square inch per hour.
17. By the Pythagorean theorem,
x2 + y2 = 102
or x2 + y2 = 100
(1)
Differentiate with respect to t:
dx
dy
2x
+ 2y
= 0
dt
dt
10
y
x
EXERCISE 4-6
193
x dx
dy
dx
= . Given:
= 3.
y dt
dt
dt
From (1), y2 = 100 - x2 and, when x = 6,
y2 = 100 - 62
= 100 -36 = 64.
Thus, y = 8 when x = 6, and
dy
!3(6)
!18
!9
=
ft/sec.
=
=
dt (6,8)
8
8
4
Therefore,
Thus,
!3x
dy
=
.
y
dt
19. y = length of shadow
x = distance of man from light
z = distance of tip of shadow from light
dz
We want to compute
. Triangles ABE and CDE are similar
dt
triangles; thus, the ratios of corresponding sides are
equal.
z
y
z ! x
Therefore,
=
=
[Note: y = z - x.]
20
5
5
z
z ! x
or
=
20
5
z = 4(z - x)
z = 4z - 4x
4x = 3z
Differentiate with respect to t:
dx
dz
4
= 3
dt
dt
dz
4 dx
=
dt
3 dt
dx
dz
4
20
Given:
= 5. Thus,
=
(5) =
ft/sec.
dt
dt
3
3
4 3
πr
(1)
3
Differentiate with respect to t:
21. V =
1
dV
dr
dr
dV
= 4πr2
and
=
2 ·
4!r
dt
dt
dt
dt
dV
Since
= 4 cu ft/sec,
dt
1
1
dr
=
ft/sec
(2)
2 (4) =
4!r
!r 2
dt
At t = 1 minute = 60 seconds,
V = 4(60) = 240 cu ft and, from (1),
3V
3(240)
180
" 180 $ 1/3
r3 =
; r = #
≈ 3.855.
=
=
! %
4!
4!
!
194
CHAPTER 4
ADDITIONAL DERIVATIVE TOPICS
A
20
ft
C
B
5 ft
D
E
y
x
z
From (2)
1
dr
=
≈ 0.0214 ft/sec
!(3.855)2
dt
At t = 2 minutes = 120 seconds,
V = 4(120) = 480 cu ft and
3V
3(480)
360
" 360 $ 1/3
r3 =
; r = #
≈ 4.857
=
=
! %
4!
4!
!
From (2),
1
dr
=
≈ 0.0135 ft/sec
!(4.857)2
dt
To find the time at which
1
= 100
!r 2
1
r2 =
100!
1
1
r=
=
100"
10 "
Now, when r =
!
V=
!
1
,
10 "
4 # 1 &3
π%
(
3 $10 " '
=
4
1
·
3 1000 "
=
1
750 "
!
dr
= 100 ft/sec, solve
dt
!
Since the volume at time t is 4t, we have
! 4t =
t=
1
and
750 "
1
≈ 0.00019 secs.
3000 "
!
!
EXERCISE 4-6
195
dx
= 3.
dt
Differentiate with respect to t:
23. y = e x + x + 1;
dy
dx
dx
= ex
+
= e x (3) + 3 = 3(e x + 1)
dt
dt
dt
To find where the point crosses the x axis, use a graphing utility to
solve
ex + x + 1 = 0
The result is x ≈ -1.278.
Now, at x = -1.278,
dy
= 3(e-1.278 + 1) ≈ 3.835 units/sec.
dt
25. C = 90,000 + 30x
x2
R = 300x 30
P = R - C
(1)
(2)
(3)
(A) Differentiating (1) with respect to t:
dC
d(90, 000) d(30x)
=
+
dt
dt
dt
dC
dx
= 30
dt
dt
" dx
%
dC
= 500'
= 30(500) $
!Thus,
# dt
&
dt
= $15,000 per week.
Costs are increasing at the rate of $15,000 per week at this
production level.
!
(B) Differentiating (2) with respect to t:
x2
d
dR
d(300x)
30
=
dt
dt
dt
dx
2x dx
= 300
dt
30 dt
#
&
x dx
= %300 "
(
$
15' dt
#
"
%
dR
6000&
dx
= 500'
Thus,
= %300 "
( (500)
$x = 6000,
$
#
&
dt
15 '
dt
=
(-100)500
!
= -$50,000 per week.
!
! the rate of $50,000 per week at this
Revenue
is decreasing at
production level.
196
CHAPTER 4
ADDITIONAL DERIVATIVE TOPICS
(C) Differentiating (3) with respect to t:
dP
dR
dC
=
dt
dt
dt
Thus, from parts (A) and (B), we have:
dP
= -50,000 - 15,000 = -$65,000
dt
Profits are decreasing at the rate of $65,000 per week at this
production level.
27. S = 60,000 - 40,000e-0.0005x
Differentiating implicitly with respect to t, we have
ds
dx
ds
dx
= -40,000(-0.0005)e-0.0005x
and
= 20e-0.0005x
dt
dt
dt
dt
dx
Now, for x = 2000 and
= 300, we have
dt
ds
= 20(300)e-0.0005(2000)
dt
= 6000e-1 = 2,207
Thus, sales are increasing at the rate of $2,207 per week.
29. Price p and demand x are related by the equation
2x2 + 5xp + 50p2 = 80,000
(1)
Differentiating implicitly with respect to t, we have
dx
dp
dx
dp
+ 5x
+ 5p
+ 100p
= 0
dt
dt
dt
dt
dp
!(5x + 100p)
dx
dt
(A) From (2),
=
4x + 5 p
dt
4x
(2)
Setting p = 30 in (1), we get
2x2 + 150x + 45,000 = 80,000
or
x2 + 75x - 17,500 = 0
Thus, x =
"75 ±
!75 ± 275
(75)2 + 70, 000
=
= 100, -175
2
2
Since x ≥ 0, x = 100
dp
Now, for x = 100, p = 30 and
= 2, we have
!
dt
![5(100) + 100(30)] " 2
dx
dx
7000
=
= and
= -12.73
4(100) + 5(30)
dt
dt
550
The demand is decreasing at the rate of -12.73 units/month.
EXERCISE 4-6
197