BC Calculus
Related Rates
1.
A ladder 15 feet long is leaning against a building so that end X is on
level ground and end Y is on the wall as shown in the figure. X is
moved away from the building at the constant rate of 1/2 foot per
second.
(a)
Find the rate at which the length OY is changing when X is 9 feet from
the building.
2
2
x + y = 225
2⋅y⋅dy
2⋅x⋅dx
+
=0
dt
dt
⎛ 1 ⎞ 2⋅⎛12⎞ ⋅dy
=0
2⋅⎛⎝ 9⎞⎠ ⋅ ⎜ ⎟ + ⎝ ⎠
⎜⎝ 2 ⎟⎠
dt
-3
dy
=
dt
8
(b)
A=
Find the rate of change of the area of triangle XOY when X is 9 feet from the building.
1
⋅x⋅y
2
1
1
⋅x⋅dy
⋅y⋅dx
dA
2
2
=
+
dt
dt
dt
21
dA
1
⎛ -3 ⎞ 1
⎛1⎞
= ⋅⎛⎝ 9⎞⎠ ⋅ ⎜ ⎟ + ⋅⎛⎝12⎞⎠ ⋅ ⎜ ⎟ =
⎜⎝ 8 ⎟⎠ 2
⎜⎝ 2 ⎟⎠
dt
2
16
2.
A container has the shape of an open right circular cone, as shown in the
acco mpanying figure. The height of the container is 10 cm and the
diameter of the opening is also 10 cm. Water in the container is
evaporating so that its depth h is changing at the constant rate of -3/10
cm/hr.
-3
dh
=
dt
10
(a)
Find the rate of change of the volume of water in the container, with respect to time, when
the depth of the water is 5 cm.
π 2
⋅h ⋅dh
dV
4
π 3
⋅h -->
h = 2 r --> V =
=
dt
dt
12
-15⋅π
dV
π 2 ⎛ -3 ⎞ dv
=
= ⋅5 ⋅ ⎜
=
⎟
⎜⎝ 10 ⎟⎠
dt
4
dt
8
(b)
Show that the rate of change in the volume of the water in the container due to evaporation
is directly proportional to the exposed surface of the water. What is the constant of
proportionality?
π 2
⋅h ⋅dh
dV
4
π 3
⋅h -->
V=
=
12
dt
dt
Since 2r = h,
dV
2 ⎛ -3 ⎞
= π⋅r ⋅ ⎜
⎜⎝ 10 ⎟⎟⎠
dt
Therefore, the rate of change of the volume is -3/10 (the constant) times the area of
the circle (the surface).
3.
Ship A is traveling due west toward Lightning Rock (R)
at a speed of 15 km/hr. Ship B is traveling due north
away from Lightning Rock at a speed of 10 km/hr. Let
x be the distance between Ship A and Lightning Rock
and let y be the distance between Ship B and Lightning
Rock.
dy
dx
= -15
= 10
dt
dt
(a)
Find the rate of change, in km/hr, of the distance between the two ships when x = 4 km
and y = 3 km.
2
2
x +y =z
2
2⋅y⋅dy
2⋅z⋅dz
2⋅x⋅dx
+
=
dt
dt
dt
2⋅⎛⎝4⎞⎠ ⋅⎛⎝ -15⎞⎠ + 2⋅⎛⎝ 3⎞⎠ ⋅⎛⎝10⎞⎠ =
2⋅⎛⎝5⎞⎠ ⋅dz
dt
dz
= -6
dt
(b)
Let θ be the angle as shown in the figure. Find the rate of change θ of, in radians per
hour, when x = 4 km and y = 3 km.
x⋅dy y⋅dx
2
⋅ ⎝⎛⎜ θ⎠⎞⎟ ⋅d⋅θ
y
dt
dt
sec
tan⎝⎛⎜ θ⎠⎞⎟ =
=
2
x
dt
x
5
When x = 4 and y = 3, z = 5, sec ⎛⎝⎜θ⎞⎠⎟ =
4
d⋅θ
4⋅⎛10⎞ - ⎛3⎞ ⋅⎛-15⎞
= ⎝ ⎠ ⎝ ⎠⎝ ⎠
2
dt
4
⎛4⎞
⋅⎜ ⎟
⎜⎝ 5 ⎟⎠
2
=
17
5
Therefore,
4.
A tanker accident has spilled oil in Pristine Bay. Oil-eating bacteria are gobbling at 5
ft3/hr. The oil slick has the form of a cylinder. When the radius of the cylinder is 500 ft,
the thickness of the slick is 0.01 ft and decreasing at the rate of 0.001 ft/hr. At what rate is
the area of the slick changing at this time? Is the area of the slick increasing or
decreasing?
dV
= -5
dt
V = π⋅r ⋅h When r = 500, h = 0.01 and
2
dh
= -0.001
dt
π⋅r ⋅dh
2⋅π⋅r⋅dr
dV
⋅h +
=
dt
dt
dt
2
-5 =
2⋅π⋅⎛⎝500⎞⎠ ⋅⎛⎝.01⎞⎠ ⋅dr
2
+ π⋅500 ⋅⎛⎝ -0.001⎞⎠
dt
dr
250⋅π - 5
=
dt
10⋅π
2
A = 2⋅π⋅r + 2⋅π⋅r⋅h
⎛ h⋅dr
dA
4⋅π⋅r⋅dr
r⋅dh ⎞
=
+ 2⋅π⋅ ⎜
+
⎟
⎜⎝ dt
dt
dt
dt ⎟⎠
⎛ 250⋅π - 5 ⎞
⎛
⎛ 250⋅π - 5 ⎞
⎞
dA
+ 2⋅π⋅ ⎜ .01⋅ ⎜
+ 500⋅⎛⎝ -.001⎞⎠ ⎟
= 4⋅π⋅⎛⎝ 500⎞⎠ ⋅ ⎜
⎟
⎟
⎜⎜⎝ 10⋅π ⎟⎟⎠
⎜⎜⎝
⎜⎜⎝ 10⋅π ⎟⎟⎠
⎟⎟⎠
dt
dA
= 156078.
dt
5.
A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness.
4⋅π
3
⋅ ⎛⎝4 + r⎞⎠
V=
3
(a)
If the ice melts at the rate of 10 in3/min, how fast is the thickness of the ice decreasing
when it is 2 in. thick?
2
4⋅π⋅ ⎝⎛ 4 + r⎞⎠ ⋅dr
dV
=
dt
dt
4⋅π⋅ ⎛⎝ 4 + 2⎞⎠ ⋅dr
10 =
dt
2
-->
dr
10
=
dt
144⋅π
(b)
How fast is the outer surface area of the ice decreasing at this time?
dA
8⋅π⋅⎛⎝ 4 + r⎞⎠ ⋅dr
2
=
A = 4⋅π⋅ ⎛⎝ 4 + r⎞⎠ -->
dt
dt
10
dA
8⋅π⋅⎛⎝ 4 + 2⎞⎠ ⋅10
=
=
dt
3
144⋅π
6.
An inverted cone has a height 10 cm and radius 2 cm. It is partially filled with liquid,
which is oozing through the sides at a rate proportional to the area of the cone in contact
with the liquid. Liquid is being poured in the top of the cone at a rate of 1 cm3/min.
When the depth of the liquid in the cone is 4 cm, the depth is decreasing at a rate of
0.1cm/min. At what rate must liquid be poured into the top of the cone to keep the liquid
at a constant depth of 4 cm?
r
1
=
h
5
2
π 3
π ⎛h⎞
⋅h
V = ⋅ ⎜ ⎟ ⋅h =
75
3 ⎜⎝ 5 ⎟⎠
dV
= 1 - rate⋅of⋅escape
dt
π 2
⋅h ⋅dh
25
1 - esc =
dt
1 - esc =
-16⋅π
250
When h = 4, dh/dt = -0.1
--> 1 - esc =
--> esc = 1 +
π 2
⋅4 ⋅ ⎛⎝ -0.1⎞⎠
25
8⋅π
125
Increase the flow into the cone at the rate 1 +
8⋅π
cm^3/min
125
7.
A Ferris wheel, 50 feet in diameter, revolves at the rate of 10 radians per minute. How
fast is a passenger rising when the passenger is 15 feet higher than the center of the wheel
and is rising?
50 foot diameter yields a 25 foot radius. The position of the rider is a sinusoid of the form
h = 25⋅sin⎝⎛⎜ 10⋅t⎠⎞⎟ + h0 where h0 is the height of the hub of the wheel. When h = h0 + 15 , 15
= 25 sin(10t), 3/5 = sin(10t) so t = 0.1 invsin(3/5).
dh
dh
⎛
⎛ 3 ⎞⎞
= 250⋅cos ⎛⎝⎜10⋅t⎞⎠⎟ which yields
= 250⋅cos ⎜10⋅0.1⋅invsin⎜ ⎟ ⎟
⎜⎝
⎜⎝ 5 ⎟⎠ ⎟⎠
dt
dt
dh
= 200. feet/minute
dt
8.
The ends of a horizontal trough 10 ft long are isosceles trapezoids with lower base 3 ft,
upper base 5 ft, and altitude 2 ft. If the water level is rising at the rate of 0.25 in/min when
the depth is 1 ft, how fast is water entering the trough?
V=
h
⋅⎛b1 + b2⎞⎠ ⋅10
⎟
2 ⎝⎜
dh
1
=
dt
48
Lower base is a constant 3 ft. At any point in time, the upper
base can be represented as 3 + 2x, where x is the length of
the horizontal dotted segment in the diagram above. The
length of the horizontal segment is 1/2 h (Similarity
relationship). Therefore, V = 5⋅h⋅⎛⎝ 3 + h + 3⎞⎠ -->
2
V = 5⋅h⋅⎛⎝ 6 + h⎞⎠ = V = 30⋅h + 5⋅h
dV
30⋅dh
10⋅h⋅dh
=
+
dt
dt
dt
dV
⎛ 1 ⎞
⎛ 1 ⎞
⋅
+
10⋅⎛
1⎞
= 30⋅ ⎜
⎝
⎠
⎜⎜ 48 ⎟⎟
⎜⎝ 48 ⎟⎟⎠
dt
⎝
⎠
9.
5
6
An airplane flying east at 400 mph goes over a certain town at 11:30 AM, and a second
plane, flying northeast at 500 mph goes over the same town at noon. How fast are they
separating at 1:00 PM?
z = n + e - 2⋅cos⎛⎝ 45⎞⎠ ⋅ne
de
= 400
dt
2
2
2
z = n + e - 2 ⋅ne
dn
= 500
dt
2
=
2
2
2
2
+
- 2 ⋅⎛⎝ 500⎞⎠ ⋅⎛⎝ 600⎞⎠ = 430.971
500
600
At 1PM, n = 500 and e = 600. Therefore z =
⎛ n⋅de
2⋅z⋅dz
2⋅n⋅dn
2⋅e⋅de
e⋅dn ⎞
=
+
- 2 ⋅⎜
+
⎟
⎜⎝ dt
dt
dt
dt
dt ⎟⎠
2⋅⎛ 500⎞⎠ ⋅⎛⎝ 500⎞⎠ + 2⋅⎛⎝ 600⎞⎠ ⋅⎛⎝400⎞⎠ - 2 ⋅ ⎛⎝⎜ ⎛⎝ 500⎞⎠ ⋅⎛⎝ 400⎞⎠ + ⎛⎝ 600⎞⎠ ⋅⎛⎝ 500⎞⎠ ⎞⎠⎟
dz
= ⎝
dt
2⋅⎛⎝430.971⎞⎠
dz
= 316.603 miles/hour
dt
B
10.
The accompanying figure represents an observer at point A watching
balloon B as it rises from point C. The balloon is rising at a constant
rate of 3 meters per second and the observer is 100 meters from point
C.
dy
=3
dt
(a)
x
y
theta
A
AC = 100
Find the rate of change of x at the instant when y = 50.
2⋅x⋅dx
2⋅y⋅dy
2
2
2
2
2
-->
When y = 50, x = 50 + 100 = x = 50⋅ 5
=
100 + y = x
dt
dt
C
50⋅⎛⎝3⎞⎠
3⋅ 5
dx
=
=
dt
5
50⋅ 5
(b)
Find the rate of change in the area of the right triangle BCA at the instant when y = 50.
2
50⋅dy
dA
150⋅m
dA
=
A = 50⋅y
=
dt
dt
dt
sec
(c)
Find the rate of change in ø at the instant when y = 50.
x = 100⋅sec ⎛⎝⎜ θ⎞⎠⎟
100⋅sec ⎛⎝⎜ θ⎞⎠⎟ ⋅tan⎛⎝⎜θ⎞⎠⎟ ⋅d⋅θ
dx
=
dt
dt
θ := arctan⎛⎝ .5⎞⎠
when y = 50
3
5
d⋅θ
= .024
=
dt
100⋅sec ⎝⎛⎜θ⎠⎞⎟ ⋅tan⎝⎛⎜θ⎠⎞⎟
11.
A rectangle with fixed area of 60 square units has a width which increases at the rate of 2
units per minute.
dw
LW = 60 and
=2
dt
(a)
Determine the rate of change of the length at the instant the width is 10 units.
W⋅dL
L⋅dw
When W = 10, L = 6
+
=0
dt
dt
6⋅⎛⎝ 2⎞⎠ +
(b)
10⋅dL
=0
dt
dL
-6
=
dt
5
Determine the rate of change of the length of the diagonal at the instant when the width is
10 units.
2
2
2
L + W = z When W = 10 and L = 6, z = 136
2⋅W⋅dW
2⋅z⋅dz
2⋅L⋅dL
+
=
dt
dt
dt
⎛ -6 ⎞
⎛⎝ 6⎞⎠ ⋅ ⎜ ⎟ + ⎛⎝ 10⎞⎠ ⋅⎛⎝ 2⎞⎠
⎜⎝ 5 ⎟⎠
16⋅ 34
dz
=
=
dt
85
136