Math 131.
Related Rates
Larson Section 2.6
There are many natural situations when there are related variables that are changing with
respect to time. For example, a spherical balloon is being inflated with air. The volume and
4
radius are related by the volume equation V = πr3 , and both the volume and radius are
3
changing with respect to time. The volume and radius are typically implicit functions of t
since we may not know their explicit expressions. Then we can implicitly differentiate this
volume equation with respect to time (the right side will require the chain rule); this will allow
dV
dr
us to determine how the rates of change
and
are related:
dt
dt
d
dV
d 4 3
d 4 3 dr
dV
dr
⇒
[V ] =
πr
=
πr ·
⇒
= 4πr2
dt
dt 3
dt
dr 3
dt
dt
dt
Example 1. (a) A spherical balloon is being inflated with air so that its radius is increasing
at a constant rate of 2 centimiters per second. Find the rate of change of the volume when the
radius is 10 centimeters.
(b) Another spherical balloon is being inflated at a constant rate of 100 cubic centimeters per
second. Find the rate at which the radius is increasing when the radius is 5 centimeters.
dr
dV
= 4πr2
and the problem gives us that
Solution: (a) From the above, we know that
dt
dt
dr
= 2cm/s so when r = 10 cm, we find
dt
cm
cm3
dV
dr
= 4πr2 = 4π(10cm)2 · 2
= 800π
dt
dt
s
s
dV
dr
(b) For this part, we also use
= 4πr2
and we are given that
dt
dt
when r = 5 cm, we have
100 = 4π(52 )
dr
dt
and so
dV
dt
= 100 cm3 /s. Then
dr
100
1 cm
=
=
dt
100π
π s
In a typical related rates problem you will:
• identify the given quantities and relevant quantities that are changing and label them
with variables;
• write an equation that relates the variables;
• use the chain rule to implicitly differentiate both sides of the equation with respect to t;
• use the information given to solve for the required rate of change.
The remainder of these notes present several classic examples of related rates problems.
Example 2. A stone dropped into a still pond sends out a circular ripple whose radius
increases at a constant rate of 3.2 feet per second.
(a) How rapidly is the area enclosed by the ripple increasing when the radius is 3 feet?
(b) How rapidly is the area enclosed by the ripple increasing at the end of 6.2 seconds?
Solution: The relevant quantiies are the area of the circle which we call A and the radius
of the circle which we call r.
These quantities are related by the geometric formula A = πr2 .
dr
dA
dA
= 2πr
where
is the rate of
Then differentiating this with respect to t yields
dt
dt
dt
dr
change of the area of the circle, and
is the rate of change of the radius of the circle.
dt
With this, we can answer both (a) and (b).
(a) When r = 3 we have
dA
= 2 · π · 3 · 3.2 = 19.2π ft2 /s
dt
(b) When t = 6.2 seconds we have r = 19.84 feet (since it started at 0 and increased at a
rate of 3.2 feet per second for 6.2 seconds so r = (3.2)(6.2) = 19.84) and thus
dA
= 2 · π · 19.84 · 3.2 = 126.976π ft2 /s
dt
Example 3. A man 6 feet tall walks at a rate of 5 feet per second away from a light that is
15 feet above the ground (see the figure below). At what rate is the tip of his shadow moving
when he is 10 feet from the base of the light?
Solution: Let the man be x feet from the base of the lamp post and let his shadow be s
x+s
s
feet long. Then =
Then
6
15
1 ds
1 dx
1 ds
=
+
6 dt
15 dt
15 dt
1 dx
1 ds
= 15
and so ds
= 23 dx
and we know dx
= 5. Thus the length of his shadow
and so 10
dt
dt
dt
dt
dt
10
is changing at 3 feet per second, and the tip of his shadow is moving away from the lamp
post at a rate of 25
feet per second.
3
Example 4. An airplane flies at an altitude of 5 miles toward a point directly over an observer
(see the figure below). The speed of the plane is 600 miles per hour. Find the rates at which
the angle of elevation θ is changing when the angle is (i) θ = 30◦ , and (ii) θ = 60◦ . Express
answer in radians per minute.
Solution: Let x represent the length of the base of the triangle in the figure. Then
x
cot θ = ; differentiating with respect to t yields
5
− csc2 θ ·
we know
dθ
1 dx
= ·
dt
5 dt
dx
= −10 miles per minute because the jet is traveling 600 miles per hour. Thus
dt
dθ
1
dx
= − sin2 θ ·
dt
5
dt
(i) When θ = 30◦ , sin θ = 1/2 and so
1 1
1
dθ
= − · 2 · (−10) =
dt
5 2
2
radians per minute.
(ii) When θ = 60◦ , sin θ =
√
3/2 and so
√
dθ
1 ( 3)2
3
=− ·
· (−10) =
2
dt
5
2
2
radians per minute.
Example 5. A ladder 15 feet long is leaning against the wall of a house (see the figure below).
The base of the ladder is pulled away from the wall at a rate of 2 feet per second. How fast
is the top of the ladder moving down the wall when its base is 4 feet from the wall? See the
diagram below where L = 15 and x = 2
Solution: Let b be the distance of the base of the ladder from the house, and let h be the
height of the top of the ladder above the ground. Then the Pythagorean theorem implies
b2 + h2 = 225. Differentiating both sides with respect to time yields
2b
and then
db
dh
+ 2h
=0
dt
dt
and so
2h
dh
db
= −2b
dt
dt
dh
b db
=− · .
dt
h dt
When b = 4, we use the relation b2 + h2 = 225 to find 42 + h2 = 225 and so
√
h2 = 225 − 42 = 209
or
h = 209
We now plug this into
dh
b db
4
=− ·
= −√
· 2 ≈ −0.5534
dt
h dt
209
Thus top of the ladder is moving down the wall at approximately 0.5534 feet per second
Example 6. Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per
minute. It forms a pile in the shape of a right circular cone whose base diameter and height
are equal. How fast is the height of the pile increasing when the pile is 11 feet high? Recall
that the volume of a right circular cone with height h and radius of the base r is given by
V = π3 r2 h.
Solution: In this case r =
Then
Solving for
dh
dt
h
2
since the height and diameter are equal. Therefore V =
dV
3πh2 dh
π
dh
=
·
= h2 ·
dt
12
dt
4
dt
we have
4 · dV
dh
40
dt
=
=
feet per minute
2
dt
πh
121π
when the height is 11 feet.
π 3
h.
12
Example 7. A boat is pulled into a dock by means of a rope attached to a pulley on the dock.
The rope is attached to the front of the boat, which is 11 feet below the level of the pulley.
If the rope is pulled through the pulley at a rate of 12 ft/min, at what rate will the boat be
approaching the dock when 107 ft of rope is out?
Solution: Let x be the horizontal distance of the front of the boat from the dock and let
r be the length of rope that is out. Then by the pythagorean theorem 112 + x2 = r2 .
Differentiating this equation with respect to t we find
2x
dr
dx
r dr
dx
= 2r
and so
= ·
dt
dt
dt
x dt
dr
= −12 ft/min (the negative is
dt
dx
when r = 107 feet. With
because the rope is being pulled in). We are asked to find
dt
this information, from the pythagorean equation above, we know
√
x = 1072 − 112 ≈ 106.4331
From the information given in the question, we know
and therefore,
dx
107
=√
· (−12) ft/min. ≈ −12.0639 ft/min.
dt
1072 − 112
The negative sign indicates that the boat is approaching the dock. Thus, the boat is
approaching the dock at a rate of approximately 12.0639 feet per minute.
Example 8. The altitude (i.e., height) of a triangle is increasing at a rate of 4 cm/minute
while the area of the triangle is increasing at a rate of 4.5 square cm/minute. At what rate is
the base of the triangle changing when the altitude is 7.2 centimeters and the area is 67 square
centimeters. Express answer to four decimal places
1
Solution: The area of a triangle is given by A = bh where b is the base and h is the
2
height. Differentiating both sides of this equation with respect to t yields
dA
1
db
dh
=
h +b
dt
2
dt
dt
We wish to find
db
, and so solving the above equation for it, we have
dt
db
1
dA
dh
=
2
−b
dt
h
dt
dt
dh
dA
Using the quantities
= 4,
= 4.5, h = 7.2 and b = 2A/h = 134
≈ 18.61111111, we
7.2
dt
dt
obtain
db
1
=
((2)(4.5) − (18.61111111)(4)) ≈ −9.0895 cm/min
dt
7.2
Example 9. A baseball diamond has the shape of a square with sides 90 feet long (see figure).
A player runs from second base to third base at a speed of 25 feet per second is 20 feet from
third base. At what rate is the player’s distance from home plate changing?
Solution: Let s represent the distance of the runner to home plate and let x represent the
distance of the runner to 3rd base. Then x2 + 902 = s2 . We differentiate implicitly to find
ds
ds
x dx
dx
= 2s ·
and so
= · .
dt
dt
dt
s dt
√
√
√
We know that x = 20 and so s = 202 + 902 = 10 22 + 92 = 10 85, and we know
dx/dt = −25. Therefore,
√
20
ds
50
10 85
= √ · (−25) = − √ = −
≈ −5.423
dt
17
10 85
85
2x ·
The distance between the runner and home plate is decreasing at approximately 5.4 feet
per second.
Example 10. An airplane flies at an altitude of 5 miles toward a point directly over an
observer (see the figure below). The ground speed of the plane is 460 miles per hour. Find
the rate at which the distance between the observer and plane is changing when the angle of
elevation θ = 70◦ . Round answer to the nearest tenth of a mile per hour.
Solution: With reference to the diagram above, the distance between the observer and
the plane is the hypotenuse of the triangle, we will lable that side h, and we will label the
dx
adjacent side x, we know
= −460 miles per hour and by the Pythagorean theorem,
dt
x2 + 52 = h2 . So we differentiate this equation implicitly to find
d 2
d 2
[x + 25] =
[h ]
dx
dx
and so
⇒
2x
dx
dh
= 2h
dt
dt
dh
x dx
=
. When θ = 70◦ , we have x = 5 cot(70◦ ) and h = 5 csc(70◦ ) therefore
dt
h dt
dh
5 cot(70◦ )
=
(−460) = −460 cos(70◦ ) ≈ −157.3 miles per hour
◦
dt
5 csc(70 )
This means the distance between the jet and observer is decreasing at a rate of 157.3 miles
per hour.
Example 11. Suppose the edges of a cube are increasing at a rate of 5 centimeters per minute.
(a) How fast is the volume changing when each edge is 8 centimeters?
(b) How fast is the volume changing when each edge is 11 centimeters?
Solution: First, the volume of a cube is given by V = x3 where each side has length x
units. Differentiating this implicitly with respect to time yields
dx
dV
= 3x2
dt
dt
(a) Given x = 8 and
dx
= 5, we have
dt
dV
cm3
= 3(8)2 (5) = 960
dt
min
(b) Given x = 11 and
dx
= 5, we have
dt
dV
cm3
= 3(11)2 (5) = 1815
dt
min
Example 12. An airplane fflying at an altitude of 6 miles passes directly over a radar station.
When the airplane is 13 miles from the radar station (s = 13), the radar detects that the
distance s is changing at a rate of 250 miles per hour. What is the speed of the airplane.
(Round answer to nearest mile per hour).
ds
Solution: Using the Pythagorean theorem we note x2 + 62 = s2 , we are given s and
dt
dx
. Then differentiating this implicitly with respect to t, we find
and we look to find
dt
d 2
d
[x + 36] = [s2 ]
dt
dt
and so
⇒
2x
dx
ds
= 2s
dt
dt
s ds
dx
=
.
dt
x dt
√
√
ds
When s = 13, we find that x = 132 − 62 = 133, and we were given
= 250 miles per
dt
hour. Therefore,
dx
13
=√
· 250 ≈ 282 miles per hour
dt
133
That is the plane is travelling approximately 282 miles per hour.