Math 1000 Related rate problems:
1.
A boat is being pulled up by a rope attached at a rate of 2 m/sec at a wharf. If the
wharf is 5 m above the water level. When there is 13 m left in the rope, find (a)
the rate of the boat moving toward the wharf. (b) the rate of the angle, between
the rope and the water level, changing. (c) the rate of the triangle area, formed by
the rope, water level and the wharf, changing.
(1) Sketch diagram as shown
(2) Assignment variables
X= distance from the boat to wharf
Z= length of the the rope
(3) Relationships
from P-theroem c 2  a 2  b 2
z 2  52  x 2  25  x 2
(4) Differentiate both sides
d 2 d
d
d
z  25  x 2   25  x 2
dt
dt
dt
dt
dz
dx
dz
dx
2z
 2x
z
x
dt
dt
dt
dt
dz
z
dx
 dt
Therefore
dt
x
dx
(5) solve for
dt
(a) when z = 13, we get x from
dz
 2m / sec(Given)
z 2  25  x2  x  z 2  25  132  25  12 and
dt
dz
z
dx
dx 13(2) 13
 dt =

 m / sec
dt
x
dt
12
6
(b) Follow the similar procedures above
Assign the angle between the rope & water level as θ, then find the
relationship sin θ = 5/z
d
d 5
d
dz
sin  
 5 z 1  5(1) z  2
Differentiate both sides:
dt
dt z
dt
dt
d
 5 dz
adj 12
cos 
 2

but when z = 13 , x = 12 and cos 
dt
z dt
hpt 13
d
 5 dz
5
 2

(2)  0.064rad / sec
Solve for
dt
z cos dt 132  12
13
(c) Since the area of the triangle formed can be written by A = ½ base x height
1
5
Therefore A = x (5)  x
2
2
d
d 5
5 dx
A
x
dt
dt 2
2 dt
dx 13
d
5 13 65 2
 ( from(a )) therefore
A 

m / sec
Since
dt
6
dt
2 6 12
Differentiate both sides :
2. A 10 m ladder is leaning against a straight vertical wall of a house. If the base of
the ladder is pulled along the level ground at a constant rate of 2 m/sec, when the
bottom of the ladder is 5 meter away from the house, (a) how fast is the top of the
ladder falling ? (b) what is the rate of the angle, between the ladder and the house,
changing?
(1) Sketch diagram as shown
(2) Assignment variables
X= distance of the house to base of the ladder
Y = distance from the tip of the ladder to the base of the house
(3) Relationships
from P-theroem c 2  a 2  b 2
102  y 2  x 2
(4) Differentiate both sides
d
d
d
d
100   y 2  x 2   y 2  x 2
dt
dt
dt
dt
dy
dx
dy
dx
0  2 y  2x
y
 x
dt
dt
dt
dt
dx
x
dy
dt

Therefore
dt
y
dy
(5) solve for
dt
(a) when x = 5, we get y from
102  y 2  x 2  y  102  52  75  5 3  8.66 and
dx
 2m / sec(Given)
dt
dx
x
dy
dt  5(2)  1.155m / sec

8.66
dt
y
(b) Follow the similar procedures above
Assign the angle between the rope & water level as θ, then find the
relationship sin θ = x/10
d
d x
1 dx
sin  

Differentiate both sides:
dt
dt 10 10 dt
d
1 dx
dx
adj 8.66
cos

 0.1

but when z = 10 , y = 8.66 and cos  
dt
10 dt
dt
hpt
10
Solve for
3.
d
0.1 dx
0.1


(2)  0.231rad / sec
dt
cos dt 8.66
10
A tank in the shape of an inverted cone has a height of 3.6 m and a radius at top of
1.15 m. Water is flowing into the tank at the rate of 0.5 m3/min. (a) How fast is
the water level rising when it is 1.8 m deep? (b) How long will it take to fill up the
tank?
(1) Sketch diagram as shown
(2) Assignment variables
Assume the height is h and radius is r
(3) Relationships
The volume of water at anytime is V =
1
basearea  height
3
1
1
basearea  height = r 2 h
3
3
(4) Reduce the variables in (3) to only one
V=
We can get the relationship between h & r from the similar triangle of ΔABC
& ΔCDE
h 3.6
1.15

 1.15h  3.6r  r 
h  0.32h , thus
r 1.15
3.6
1
1
V  r 2 h   (.32h) 2 h  0.032h3
3
3
d
(5) Differentiate both sides , use
dt
d
d
d
dh
dh
V  0.032h3  0.032 h3  0.032 (3h 2 )  0.096h 2
dt
dt
dt
dt
dt
dh
(6) Solve for
dt
When h = 1.8 m
dV
d
dh
dh
0.5
dt
V  0.096h 2



 0.513m / min
2
dt
dt
dt 0.096h
0.096(1.8) 2
4. Gravel is being dumped from a conveyor belt at a rate of 30 m3/min, and its
coarseness is such that it forms a pile of the shape of a cone whose base diameter
and height are always equal. How fast is the height of the pile increasing when
the pile is 10 m high?
(1) Sketch diagram as shown
(2) Assignment variables
Assume the height is h and radius is r
(3) Relationships
The volume of water at anytime is V =
1
basearea  height
3
1
1
basearea  height = r 2 h
3
3
(4) Reduce the variables in (3) to only one
Because the radius and the height always the same r = h, thus
1
1
V  r 2 h   (h) 2 h  0.333h3
3
3
d
(5) Differentiate both sides , use
dt
d
d
d
dh
dh
V  0.333h3  0.333 h3  0.333 (3h 2 )  h 2
dt
dt
dt
dt
dt
dh
(6) Solve for
dt
When h = 1.8 m
dV
d
dh
dh
30
V  h 2

 dt2 
 0.096m / min
dt
dt
dt h
 (10) 2
V=
5.
A baseball diamond is a square with side 90 ft. A batter hits the ball and run
toward the first base with a speed of 24 ft/s. (a) At what rate is his distance from
the second base decreasing when he is half way to the first base? (b) At what rate
is his distance from the third base increasing when he is half way to the first base?
(1) Sketch diagram as shown
(2) Assignment variables
Assume x is the distance between the runner and the home base at anytime
And z = distance between the runner and the 2nd base at anytime
(3)`Relationships
from P-theroem c 2  a 2  b 2 , we get:
z 2  (90  x 2  902
(4) Reduce the variables in (3) to only one
Not needed
d
(5) Differentiate both sides , use
dt
d 2 d
d
dx
z  ((90  x) 2  902 )  ((90  x) 2  2(90  x)
dt
dt
dt
dt
dz
dx
2z
 2(90  x)
dt
dt
dz
(6) Solve for
dt
When x = 45 (half of 90)
z  (90  x 2  902  (90  452  902  100.62
dz  2(90  x) dx  2(90  45)


(24)  10.73 ft / sec
(a)
dt
2z
dt
2(100.62)
(b) use the similar process listed above:
z = distance between the runner and the third base at anytime:
z 2  x 2  902
d 2 d 2
d
dx
z  ( x  902 )  ( x 2 )  2 x
dt
dt
dt
dt
dx
dx
2x
x
dz
dx
dz
dt  dt
2z
 2x


dt
dt
dt
2z
z
6.
When x = 45 (half of 90) z  x2  902  452  902  100.62
dx
x
dz
45(24)
 dt 
 10.736 ft / sec
dt
z
100.6
A 6ft tall man is walking away at a speed of 2 ft/sec from a lamp-post that is 20ft
high. When he is 10 ft away from the lamp-post, find (a) how fast is the head of
the the shadow moving? (b) the rate of the length of the shadow changing?
(1) Sketch diagram as shown
(2) Assignment variables
Assume x is the distance between the man and the base of the lamp at anytime
And y = distance between the man’s shadow of the head and the base of the
lamp
(3)`Relationships
We can get the relationship between h & r from the similar triangle of ΔABC & ΔADE
y
15
6x 2

 6 y  15( y  x)  9 y  6 x  y 
 x,
yx 6
9
3
(4) Reduce variable to only one: (not needed)
d
(5) Differentiate both sides , use
dt
d
d 2
dy 2 dx
y
x

dt
dt 3
dt 3 dt
(6) Solve for
dy
dt
When x = 10
dy 2 dx 2
4

 (2)  ft / sec
dt 3 dt 3
3
7.
A plane flying horizontally at an attitude of 3 km and at a speed of 600 km/hr,
passes directly over an observer on the ground. How fast is the distance between
the observer and the plane increasing 3 min later?
(1) Sketch diagram as shown
(2) Assignment variables
Assume the distance flown by the airplane is x
and the distance between the plane & the observer is z
(3)`Relationships
z 2  32  x 2
(4) Reduce the variables (not needed to do)
(5) Differentiate both sides , use


d
dt
d 2 d 2
z 
3  x2
dt
dt
dz d 2 d 2
dz
dx
dz x dx
2z
 3  x → 2z
 2x →

dt dt
dt
dt
dt
dt z dt
dh
(6) Solve for
dt
When t = 3min = 3/60 hr = 1/20 hr, x = 600 (1/20) = 30 km
And z 2  32  x 2 → z  32  x2  32  302  9  900  30.15
dz x dx
30

(600) = 597.01 km/hr
Therefore
=
dt z dt 30.15
8.
Two supply ships leave Hibernia oil rig at the same time, with one ship travelling
west at 12 knots and the other travelling south at 15 knots. How fast is the
distance between them changing after 2 hours?
(1) Sketch diagram as shown
(2) Assignment variables
Assume the distance for west bound ship is x
the distance for south bound ship is y and the distance between 2 ships is z
(3)`Relationships
z 2  x2  y 2
(4) Reduce the variables (not needed to do)
(5) Differentiate both sides , use


d
dt
d 2 d 2
z 
x  y2
dt
dt
dz d 2 d 2
dz
dx
dy
dz x dx y dy
2z
 x  y → 2z
 2 x  2 y +→


dt dt
dt
dt
dt
dt
dt z dt z dt
dz
dt
When t = 2hrs x = 12 (2) = 24 and y = 15 (2) = 30
(6) Solve for
And z 2  x 2  y 2 → z  x 2  y 2  242  302  576  900  38.41s
dz x dx y dy
24
30


(12) 
(15) = 7.50+11.72=19.2 knots
Therefore
=
dt z dt z dt 38.41
38.41
9.
A column of water in the shape of a right circular cylinder is being frozen. As it
freezes its height increases at a rate of 0.008 mm/min and its diameter increases at
rate of 0.004 mm/min. At what rate is the volume changing when the volume is
48 π mm3 and the diameter is 4 mm? given the volume of a cone =
given the volume of a cone = r 2 h (where r= radius of the base, h: height )
(1) Sketch diagram as shown
(2) Assignment variables
Assume the radius of the base is r and the height of the
cylinder is h
(3)`Relationships
z 2  x2  y 2
(4) Reduce the variables (not needed to do)
(5) Differentiate both sides , use

d 2 d 2
z 
x  y2
dt
dt
2 knots

d
dt