RELATED RATES
PROBLEMS
If a particle is moving along a straight line according
to the equation of motion s  f ( t ) , since the velocity
may be interpreted as a rate of change of distance
with respect to time, thus we have shown that the
velocity of the particle at time “t” is the derivative of
“s” with respect to “t”.
There are many problems in which we are
concerned with the rate of change of two or more
related variables with respect to time, in which it is
not necessary to express each of these variables
directly as function of time. For example, we are given
an equation involving the variables x and y, and that
both x and y are functions of the third variable t,
where t denotes time.
Since the rate of change of x and y with respect to t is
dy
dx
given by and dt , respectively, we differentiate both
dt
sides of the given equation with respect to t by
applying the chain rule.
When two or more variables, all functions of t, are
related by an equation, the relation between their
rates of change may be obtained by differentiating the
equation with respect to t.
A Strategy for Solving Related Rates Problems (p. 205)
Example 1
A 17 ft ladder is leaning against a wall. If the bottom of
the ladder is pulled along the ground away from the wall at
the constant rate of 5 ft/sec, how fast will the top of the
ladder be moving down the wall when it is 8 ft above the
ground?
y
dy
?
dt y 8 ft
x
dx
ft
5
dt
sec
Let x  distance ft  of the bottomof the ladder from the wall at any instant
y  distance ft  of the top of the ladder from the ground at any instant
t  tim e sec since the bottomof the ladderis pulled along the ground
away from the wall
x2  y2  172 Working Equation
dx
dy
2x  2y  0
dt
dt
dx
dx
 2x
x
dy
dt 
dt

dt
2y
y
when y  8  x  172  82  15
dy  155
ft

 9.375
dt
8
sec
Note:
• Values which changes as time changes are denoted by variable.
• The rate is positive if the variable increases as time increases and
is negative if the variable decreases as time increases.
Example 2
A balloon leaving the ground 60 feet from an observer,
rises vertically at the rate 10 ft/sec . How fast is the balloon
receding from the observer after 8 seconds?
h
L
dh
ft
 10
dt
sec
dL
?
dt t 8 sec
Viewer
60 feet
Let h  height ft  of the balloon from the groundat any instant
L  distance ft  of the balloon from the observerat any instant
t  tim esec sincethe balloonstarts to rise from the ground
In the figure : L  h  60
2
2
2
L  h2  3600 Working Equation
 dh 
2h 
dL
 dt 

dt 2 h2  3600
 dh 
h 
dL
dt
 2 
dt
h  3600
dh
ft
Since,
 10
and t  8sec
dt
sec
ft 

h   10
 8sec  80 ft.
 sec 
dL

dt
8010
80  3600
2
dL
800
800


dt
6400  3600
10,000
dL 800
ft

8
dt 100
sec
Example 3
As a man walks across a bridge at the rate of 5 ft/sec ,
a boat passes directly beneath him at 10 ft/sec. If the
bridge is 30 feet above the water, how fast are the man
and the boat separating 3 seconds later?
5
ft
sec
Let t  tim e sec the m an starts
10
ft
sec
to cross the bridge
s distance ft  between the m an
and the boat at any instant
S2  30  L2
ft
5
sec
2
S  900  L2 but L  125t2
30’
S
ft
10
sec
30’
10t
L2  5t   10t 
L 
2
5t   10t 
2
L  125t
2
2
125t 
2
S  900  125t2
2
Working Equation
Differenti ate both sides of the WE wrt time t :
ds
Find:
when t  3 sec
dt
2
900  
1252t
dS

dt 2 900  125t2
125t 
dS

dt
900  125t2
when t  3sec
ds
1253

2
dt
900  1253
dS 1253 25 ft
ft


or 8.33
dt
45
3 sec
sec
Example 4
A man on a wharf of 20 feet above the water pulls in a
rope, to which a boat is attached, at the rate of 4 ft/sec. At
what rate is the boat approaching the wharf when there is
25 feet of rope out?
dR
ft
 4
dt
sec
R
20ft
x
Let t  tim e sec since the boat starts to approach the wharf
x  distance ft  of the boat from the wharf at any instant
R  length ft  of the rope out at any instant
dx
Find
when R  25ft
When R  25ft
dt
dR
ft
R2  x 2  202
and
 4
dt
sec
x 2  R2  400
dx
25 4 
2

x  R  400 (Working Equation)
dt
252  400
254 
2

x  R  400
15
dR
dR
dx
20 ft
2R
R

dx
dt
dt

 2
dt
3 sec
2
dt 2 R  400
R  400
Example 5
Water is flowing into a conical reservoir 20 feet deep
and 10 feet across the top, at the rate of 15 ft3/min . Find
how fast the surface is rising when the water is 8 feet
deep?
10 feet
dV
ft 3
 15
dt
m in
5 feet
20 feet
r
h
Let t  tim e m in sin ce the water flows int o the reservoir
r  radius(ft) of the water surfaceat any instant
h  height(ft) of the water at any instant
dh
Find
when the water is 8 ft. deep
dt
By ratio and proportion
1
1 2
V  Bh   r h
5 r
1
3
3
 r  h
20 h
4
2

1 1 
Thus , V   h
3 4 

 3
h  V  h  Working Equation
48

dV   2  dh   2  dh 
 3h    h  
dt 48   dt  16  dt 
dV
16
dh
1615 15 ft
ft
dt



or 1.194
2
2
dt h8 ft
h
8 
4  min
min
Example 6
Water is flowing into a vertical tank at the rate of 24
ft3/min . If the radius of the tank is 4 feet, how fast is
the surface rising?
4 feet
dV
ft 3
 24
dt
m in
h
Let t  tim e m in since the water flows into the tank
h  height ft  of the water at any instant
 
V  volum e ft 3 of the water at any instant
dh
Find
when the radius of the tank is 4 ft.
dt
From
V  Bh
V   r 2 h   r 2h
But r is constant, r  4 ft
V  4  h  16 h Working Equation
2
dV
dh
 16
dt
dt
dV
dh
24
3 ft
dt



dt r 4 ft 16 16 2 min
Example 7
A triangular trough is 10 feet long, 6 feet across the
top, and 3 feet deep. If water flows in at the rate of 12
ft3/min, find how fast the surface is rising when the
water is 6 inches deep?
6 feet
3 feet
ft 3
12
m in
x
h
Let t  tim e m in sincethe water flows into the trough
h  height ft  of the water at any instant
x  horizontalwidth  ft  of the water at the triangular
end at any instant
 
V  volum e ft 3 of the water at any instant
dh
Find
when the water is 6 inches deep.
dt
From
V  Bh
1 
V   x h 10  5xh
2 
x 6
By ratio and proportion,
  x  2h
h 3
Thus, V  5xh  52hh  10h 2 Working Equation
dV
dh
 20h
dt
dt
dV
dh
 dt 
dt h6 in 20h
12
ft
 1.2
min
  1ft 
206in

  12in 
Example 8
A train, starting at noon, travels at 40 mph going
north. Another train, starting from the same point at
2:00 pm travels east at 50 mph . Find how fast the two
trains are separating at 3:00 pm.
C
3pm
y
B
2pm
80 miles
L
dy
mi
 40
dt
hr
A 12pm
2pm
dx
mi
 50
dt
hr
x
3pm
D
dL
Find
when t  1hr.
dt
dx
dy
Since
 50 mph and
 40 mph
dt
dt
A B  402  80 miles
From the figure: L  x  80  y 
2
2
2
L  x  80  y 
2
2
Working Equation
dx
dy
2x  2(80  y)
dL
dt
 dt 2
dt
2 x  (80  y)2
x  50mph1hr   50 miles
After 1 hr  
y  40mph1hr   40 miles
dx
dy
x  (80  y)
dL
dt
 dt 2
dt
x  (80  y)2
dL (50)(50))  ((80  40)(40)

dt
(50)2  (80  40)2
dL
2,500  4 ,800

dt
2,500  14,400
dL
7,300
7,300


dt
16,900 130
dL
mi
 56.15
dt
hr
Example 9
A billboard 10 feet high is located on the edge of a
building 45 feet tall. A girl 5 feet in height approaches
the building at the rate of 3.4 ft/sec . How fast is the
angle subtended at her eye by the billboard changing
when she is 30 feet from the billboard?
10’
45’



 3. 4
x
ft
sec
5’
d
Find
when x  30 ft.
dt
    
In the figure: 
    
Using :     
tan   tan    
tan   tan 
tan  
1  tan  tan 
50
40
but, tan  
and tan  
x
x
50 40

x
x
tan  
 50  40 
1    
 x  x 
10
tan   2 x
x  2000
x2
10x
 2
x  2000
 10x 
Working Equation
  tan1  2

 x  2000
 10x 
Working Equation
  tan  2

 x  2000
1
d

dt
1
2 
 10x 
1  2
 x  2000
10x
d

dt
x
dx
dx
 10x(2x)
dt
dt
x 2  20002
(x 2  2000)(10)
2
 20,000  20x 2 
2
 2000  100x 2
dx
dt
2
dx

20,000  10x 
d
dt

dt x  2000  100x
2
2
2
2
d 20,000  10 30   3.4 

2
dt 302  2000   100 302
2
d 11,000 3.4 

dt
8,500,000
d
rad
 0.0044
dt
sec
Example 10
A picture 40 cm high is placed on a wall with its
base 30 cm above the level of the eye of an observer. If
the observer is approaching the wall at the rate of 40
cm/sec, how fast is the measure of the angle
subtended at the observer’s eye by the picture
changing when the observer is 1 m from the wall?
  2  1
x
1 x
  cot
 cot
70
30
1
x
1 dx
1 dx


d
30 dt
 70 dt

x2
x2
dt
1 2 1 2
70
30
d
dx / dt  4900  dx / dt  900 





2 
dt
70  4900  x 
30  900  x 2 
dx
substitute
 40cm/sec and x  1m  100cm
dt
d
 40 
4900
900
  40 







dt
70  4900  10000  30  900  10000 
d
 700 
 300 
4

4



dt
14900
10900




d 28 12


 0.07782...
dt 149 109
d
 0.078 rad. / sec.
dt
Example 11
A statue 10ft. high is standing on a base 13ft. high.
If an observer’s eye is 5ft. above the ground, how far
should he stand from the base in order that the
angle between his lines of sight to the top and
bottom of the statue be a maximum?
5’
x
From the figure   2  1
x
1 x
cot 2 
 2  cot
10’
18
18
x
x
cot 1   1  cot1
8
8
x
x
  cot1  cot1
18
8
13’
d
1
1
1
1




 
 
2
2
dx
 x   18 
x   8 
1 
1  
 18 
8 

1
1 


d
8 0
  18 2  
2
dx
x
x
  
  
1   1  
 18  
8 
1
18
x2
1 2
18
1
2
 18 
 2  8 2
 18  1  x
82
18
8
 2 2
2
2
18  x 8  x
964  x2   4324  x2 
964  9x2  4324  4x2
5x 2  1296  576
8 
 2
8 
2
5x 2  720
x 2  144  x  12
x  12
Therefore, the observer must be 12
ft from the base of the statue so
that his line of sight from top to
bottom of the statue is maximum.
EXERCISE A:
1. What number exceeds its square by the maximum amount?
2. The sum of two numbers is “K”. find the minimum value of
the sum of their squares.
3. A rectangular field of given area is to be fenced off along the bank
of a river. If no fence is needed along the river, what are the
dimensions of the rectangle that will require the least amount of
fencing?
4. A Norman window consists of a rectangle surmounted by a
semicircle. What shape gives the most light for a given perimeter?
5. A cylindrical glass jar has a plastic top. If the plastic is half as
expensive as the glass per unit area, find the most economical
proportions for the glass.
6. Find the proportions of the circular cone of maximum volume
inscribed in a sphere.
7. A wall 8 feet high and 24.5 feet from a house. Find the shortest
ladder which will reach from the ground to the house when
leaning over the wall
EXERCISE B:
1. A sign 3 ft high is placed on a wall with its base 2 ft
above the eye level of a woman attempting to read it.
Find how far from the wall the woman should stand to
get the “best view” of the sign; that is, so that the
angle subtended at her eye by the sign is maximum.
2. A man on dock is pulling in at the rate of 2ft/sec a
rowboat by means of a rope. The man’s hands are
20ft. above the level of the point where the rope is
attached to the boat. How fast is the measure of the
angle of depression of the rope changing when there
are 52 ft. of rope out?
3. Find the equations of the normal line and tangent lines
to the graph of the equation y  sec12x  1 at the
point  1 , 1   .
2 3 
4. A picture 5 ft high is placed on a wall with its base 7ft
above the level of the eye of an observer is
approaching the wall at the rate of 3ft/sec. How fast is
the measure of the angle subtended at her eye by the
picture changing when the observer is 10ft. from the
wall?
5. An airplane is flying at a speed of 300mi/hr at an altitude
of 4 mi. If an observer is on the ground, find the time rate
of change of the measure of the observer’s angle of
elevation of the airplane when the airplane is over a point
on the ground 2 mi. from the observer.