MCV4U
Related Rates
Related Rates Day # 1
Often it is required that we find the rate at which one variable is changing, given the rate of
change of a related variable.
Steps for Solving a Related Rate Problem
 Make a sketch
 Introduce the variables
 Identify the quantity to be found
 Define equation that relates the variables
 Implicitly differentiate both sides with respect to time
 Substitute in known quantities and known rates
 Solve the equation
 Write a concluding statement including unit
Related Rate Problem involving a circular model
Eg.1: A pebble is thrown into a lake and causes a circular ripple to spread outward at a rate of 2
m/s. Find the rate of change of the area in terms of π.
a) 3 seconds after the pebble is thrown.
b) When the area of the ripple is 9π m2
Solution:
1. Make a sketch. ie: Draw a circle and label radius, increasing outwards.
2. Let r represent the radius in meters, let A represent the area in m2.
dA
3. Required to find the change in Area with respect to time, therefore
.
dt
4. Equation that relates the variables. A  r 2
dA
dr
 2r 
5. Implicitly differentiate both sides w.r.t. time.
dt
dt
dr
 2m
@ t  3 sec . r  6 m
6. Substitute in known quantities and known rates.
s
dt
dA
dr
 2r 
dt
dt
7. Solve the equation:
dA
 2 62  24
dt
8. Concluding statement. Therefore the area of the circle is increasing at a rate of 24π m2/s
at t = 3 sec.
 A  r and A  9
2
b)
r  3 m
dA
dr
 2r 
dt
dt
dA
 2 32   12
dt
Therefore the area of the circle is increasing at a rate of 12π m2/s when the area of the circle
is 9π m2.
MCV4U
Related Rates
Related Rate Problem involving a Right Triangle Model
Eg.2: A dog runs across a path at a rate of 4 m/s. A culvert is located directly beneath the dog.
A fish swims right underneath the dog at a rate of 3 m/s. in a direction perpendicular to the path
of the dog. Find the rate of change of the distance between the two animals 1 second later.
Solution:
1. Sketch
2. Let x represent the distance traveled by the fish (m)
Let y represent the distance traveled by the dog (m)
Let r represent the distance between the dog and the fish (m)
dr
3. Required to find:
dt
2
2
4. Equation: x  y  r 2
x2  y2  r 2
dx
dy
dr
 2y 
 2r 
dt
dt
dt
dx
dy
dr
x
 y
r
dt
dt
dt
6. Substitute in known quantities and known rates.
@ t = 1 sec. x = 3 m , y = 4 m and r = 5 m (using Pythagorean theorem)
dx
dy
 3 and
 4.
dt
dt
dx
dy
dr
x  y
r
dt
dt
dt
dr
7. Solve the equation: 33  4 4   5 
dt
dr
 5 m/s
dt
8. Concluding Statement: Therefore the rate of change between the two animals after 1
second is 5 m/s.
5. Implicitly Differentiate w.r.t. time: 2 x 
MCV4U
Related Rates
Related Rate Problem involving a Conical Model
Eg.3: Water is pouring into an inverted right circular cone at a rate of π m3/min. The height
and the diameter of the base of the cone are both 10 m. How fast is the water level rising when
the depth of the water is 8 m?
Solution:
1. Sketch an inverted cone and label the diameter of 10 m, the radius of 5 m, the height of
the entire cone of 10 m, then label the radius and height of the cone part way up to show
the depth of the water.
2. Let V represent the volume of water in the cone at time, t.
Let r represent the radius of water in the cone at time, t.
Let h represent the height of water in the cone at time, t.
dh
3. Required to find:
dt
1 2
4. Equation: V  r h .
3
5. Before we differentiate implicitly we need to have our equation in terms of only h. Using
r
5
h
r  .
similar triangles from our sketch, 
h 10
2
1 2
V  r h
3
2
1 h
V h      h
3 2
1
V h   h 3
12
dV 1 2 dh
 h 
dt
4
dt
6. Substitute in known quantities and known rates. When h = 8 m. ,
dV
  m3/min.
dt
dV 1 2 dh
 h 
dt
4
dt
1
dh
2
7. Solve the equation:    8 
4
dt
dh 1

dt 16
8. Concluding Statement: Therefore when the depth of the water is 8 m., the level of water
1
in the cone is rising at a rate of
m/min.
16
MCV4U
Related Rates
Related Rate Problem involving Similar Triangle Model
Eg.4: A student who is 1.6 m tall walks directly away from a lamppost at a rate of 1.2 m/s. A
light is situated 8 m above the ground on the lamppost. Show that the student’s shadow
is lengthening at a rate of 0.3 m/s. when she is 20 m from the base of the lamppost.
Solution:
1. Sketch a small right triangle inside a large right triangle. Label the height of the large
right triangle with 8 m to represent the lamppost. Label the height of the small right
triangle with 1.6 m to represent the student.
2. Let x represent the length of the shadow.
Let y represent the distance the student is from the base of the lamppost.
dy
dx
 1.2 m / s and are required to find:
3. We are given
when y = 20 m.
dt
dt
4. To determine an equation that relates x and y, we use similar triangles.
x y
8

x
1 .6
1 .6 x  1 .6 y  8 x
1 .6 y  6 .4 x
dy
dx
 6.4
5. Implicitly Differentiate both sides w.r.t. time: 1.6
dt
dt
dy
 1.2 m / s
6. Substitute in known quantities. When y = 20 m ,
dt
dx
1.61.2   6.4
dt
7. Solve the equation:
dx
 0 .3 m / s
dt
8. Concluding Statement: Therefore the student’s shadow is lengthening at 0.3 m/s.
MCV4U
Related Rates
Related Rate Problem involving a Rectangular Prism Model
Eg.5: The base of a rectangular tank is 3 m by 4 m and is 10 m high. Water is added at a rate of
8 m3/min. Find the rate of change of water level when the water is 5 m deep.
Solution:
1. Sketch a rectangular prism and label.
2. Let V represent the volume of the tank.
Let l represent the length of the tank.
Let w represent the width of the tank.
Let h represent the height of the tank.
3. We are given the length, width and height of the tank and
dh
when the water is 5 m deep.
dt
V  lwh
Equation that relates the variables.
V  12h
Implicitly Differentiate both sides w.r.t. time.
dV
dh
 12 
dt
dt
dV
 8 m3/min.
Substitute in known quantities.
dt
dV
dh
 12 
dt
dt
dh
Solve the equation: 8  12 
dt
dh 2

dt 3
dV
 8.
dt
Required to find
4.
5.
6.
7.
8. Therefore the rate of change of water in the tank is
Homework: Related Rates Worksheet
Day # 2: Handout Related Rates Problem Set
Length and Width remain constant.
2
m/min.
3