2. Related Rates
Related Rates
In this section all functions are in the variable t for time,
and all the derivatives (#)' are taken with respect to t,
unless otherwise specified.
3. Related Rates
Related Rates
In this section all functions are in the variable t for time,
and all the derivatives (#)' are taken with respect to t,
unless otherwise specified.
Let’s use resizing a rectangle on the computer screen
as an example.
4. Related Rates
Related Rates
In this section all functions are in the variable t for time,
and all the derivatives (#)' are taken with respect to t,
unless otherwise specified.
Let’s use resizing a rectangle on the computer screen
as an example. We resize a rectangle by dragging one
of its corners every which way with various speeds.
5. Related Rates
Related Rates
In this section all functions are in the variable t for time,
and all the derivatives (#)' are taken with respect to t,
unless otherwise specified.
Let’s use resizing a rectangle on the computer screen
as an example. We resize a rectangle by dragging one
of its corners every which way with various speeds.
w
l
6. Related Rates
Related Rates
In this section all functions are in the variable t for time,
and all the derivatives (#)' are taken with respect to t,
unless otherwise specified.
Let’s use resizing a rectangle on the computer screen
as an example. We resize a rectangle by dragging one
of its corners every which way with various speeds.
w
W
drag
l
L
7. Related Rates
Related Rates
In this section all functions are in the variable t for time,
and all the derivatives (#)' are taken with respect to t,
unless otherwise specified.
Let’s use resizing a rectangle on the computer screen
as an example. We resize a rectangle by dragging one
of its corners every which way with various speeds.
w
W
drag
l
L
The unchanged relation in this process is that the
rectangular area A enclosed always satisfy A = L* W.
8. Related Rates
w
W
drag
l
L
Suppose we know the rates of changes L'(t) and W'(t),
that is, the speeds of the mouse in the horizontal
direction and in vertical direction, what is the
rate of changes in area A?
9. Related Rates
w
W
drag
l
L
Suppose we know the rates of changes L'(t) and W'(t),
that is, the speeds of the mouse in the horizontal
direction and in vertical direction, what is the
rate of changes in area A? How are all the rates related
in this context?
10. Related Rates
w
W
drag
l
L
Suppose we know the rates of changes L'(t) and W'(t),
that is, the speeds of the mouse in the horizontal
direction and in vertical direction, what is the
rate of changes in area A? How are all the rates related
in this context? These type of problems are called
related rates problems.
11. Related Rates
w
W
drag
l
L
Suppose we know the rates of changes L'(t) and W'(t),
that is, the speeds of the mouse in the horizontal
direction and in vertical direction, what is the
rate of changes in area A? How are all the rates related
in this context? These type of problems are called
related rates problems.
Note that the rate of change in the area A depends not
only on L'(t) and W'(t), it also depends on the size of
the rectangle, more specifically, the L and W at that
moment.
12. Related Rates
w
W
drag
l
L
Suppose we know the rates of changes L'(t) and W'(t),
that is, the speeds of the mouse in the horizontal
direction and in vertical direction, what is the
rate of changes in area A? How are all the rates related
in this context? These type of problems are called
related rates problems.
Note that the rate of change in the area A depends not
only on L'(t) and W'(t), it also depends on the size of
the rectangle, more specifically, the L and W at that
moment. This is the case in most related problems.
13. Related Rates
Since all the variables are assumed to be formulas in t,
the algebra in computing these rates is based on the
Chain Rules so let’s review the algebra.
14. Related Rates
Since all the variables are assumed to be formulas in t,
the algebra in computing these rates is based on the
Chain Rules so let’s review the algebra.
Example A. Assume all the variables are functions in t.
Expand the following derivatives as much as possible.
a. (xy2) '
b. d(u2/ev)
dt
15. Related Rates
Since all the variables are assumed to be formulas in t,
the algebra in computing these rates is based on the
Chain Rules so let’s review the algebra.
Example A. Assume all the variables are functions in t.
Expand the following derivatives as much as possible.
a. (xy2) ' (Product Rule)
= x' y2 + x(y2)'
b. d(u2/ev)
dt
16. Related Rates
Since all the variables are assumed to be formulas in t,
the algebra in computing these rates is based on the
Chain Rules so let’s review the algebra.
Example A. Assume all the variables are functions in t.
Expand the following derivatives as much as possible.
a. (xy2) ' (Product Rule)
= x' y2 + x(y2)'
= x' y2 + x(2yy') or y(x'y + 2xy')
b. d(u2/ev)
dt
17. Related Rates
Since all the variables are assumed to be formulas in t,
the algebra in computing these rates is based on the
Chain Rules so let’s review the algebra.
Example A. Assume all the variables are functions in t.
Expand the following derivatives as much as possible.
a. (xy2) ' (Product Rule)
= x' y2 + x(y2)'
= x' y2 + x(2yy') or y(x'y + 2xy')
b. d(u2/ev) d(u2e–v )
dt = dt
18. Related Rates
Since all the variables are assumed to be formulas in t,
the algebra in computing these rates is based on the
Chain Rules so let’s review the algebra.
Example A. Assume all the variables are functions in t.
Expand the following derivatives as much as possible.
a. (xy2) ' (Product Rule)
= x' y2 + x(y2)'
= x' y2 + x(2yy') or y(x'y + 2xy')
b. d(u2/ev) d(u2e–v ) (Product Rule)
dt = dt
= d(u2) e–v + u2 d(e–v)
dt dt
19. Related Rates
Since all the variables are assumed to be formulas in t,
the algebra in computing these rates is based on the
Chain Rules so let’s review the algebra.
Example A. Assume all the variables are functions in t.
Expand the following derivatives as much as possible.
a. (xy2) ' (Product Rule)
= x' y2 + x(y2)'
= x' y2 + x(2yy') or y(x'y + 2xy')
b. d(u2/ev) d(u2e–v ) (Product Rule)
dt = dt
= d(u2) e–v + u2 d(e–v)
dt dt
du dv
= 2ue dt – u
–v 2 e–v
dt
20. Related Rates
Since all the variables are assumed to be formulas in t,
the algebra in computing these rates is based on the
Chain Rules so let’s review the algebra.
Example A. Assume all the variables are functions in t.
Expand the following derivatives as much as possible.
a. (xy2) ' (Product Rule)
= x' y2 + x(y2)'
= x' y2 + x(2yy') or y(x'y + 2xy')
b. d(u2/ev) d(u2e–v ) (Product Rule)
dt = dt
= d(u2) e–v + u2 d(e–v)
dt dt
du dv du dv )
= 2ue dt – u
–v 2 e–v
dt or ue ( dt – u dt
–v 2
22. Related Rates
In related–rate problems the rates of changes of some
of the variables will be stated. These rates need to be
translated into derivative notations to be used.
23. Related Rates
In related–rate problems the rates of changes of some
of the variables will be stated. These rates need to be
translated into derivative notations to be used.
So if “the length (L) is decreasing at rate of 5 ft/sec”,
it is to be written in derivative as L'(t) = –5,
24. Related Rates
In related–rate problems the rates of changes of some
of the variables will be stated. These rates need to be
translated into derivative notations to be used.
So if “the length (L) is decreasing at rate of 5 ft/sec”,
it is to be written in derivative as L'(t) = –5,
or that “the area A is increasing at a rate of 10 ft2/sec”
is to be written as dA = 10 or A'(t) = 10.
dt
25. Related Rates
In related–rate problems the rates of changes of some
of the variables will be stated. These rates need to be
translated into derivative notations to be used.
So if “the length (L) is decreasing at rate of 5 ft/sec”,
it is to be written in derivative as L'(t) = –5,
or that “the area A is increasing at a rate of 10 ft2/sec”
is to be written as dA = 10 or A'(t) = 10.
dt
Likewise, any specified speed is to be translated as
rates of change of a suitably chosen distance.
26. Related Rates
In related–rate problems the rates of changes of some
of the variables will be stated. These rates need to be
translated into derivative notations to be used.
So if “the length (L) is decreasing at rate of 5 ft/sec”,
it is to be written in derivative as L'(t) = –5,
or that “the area A is increasing at a rate of 10 ft2/sec”
is to be written as dA = 10 or A'(t) = 10.
dt
Likewise, any specified speed is to be translated as
rates of change of a suitably chosen distance.
Example B. We resize a rectangle by dragging one
of its corners. The length L is increasing at a rate of
5 cm/sec and the width W is increasing at a rate of
7cm/sec. What is rate of change in area A at the
moment when the rectangle is 6 cm x 4 cm?
27. Related Rates
We have that A = L* W, where are the variable,
are all functions in t.
28. Related Rates
We have that A = L* W, where are the variable,
are all functions in t. Take the derivative on both sides,
we have that
A ' = L' * W + L* W'
29. Related Rates
We have that A = L* W, where are the variable,
are all functions in t. Take the derivative on both sides,
we have that
A ' = L' * W + L* W'
We are given that L' = 5, W ' = 7, L = 6, and W = 4.
30. Related Rates
We have that A = L* W, where are the variable,
are all functions in t. Take the derivative on both sides,
we have that
A ' = L' * W + L* W'
We are given that L' = 5, W ' = 7, L = 6, and W = 4.
Hence A ' = 5(4) + 7(6) = 62
31. Related Rates
We have that A = L* W, where are the variable,
are all functions in t. Take the derivative on both sides,
we have that
A ' = L' * W + L* W'
We are given that L' = 5, W ' = 7, L = 6, and W = 4.
Hence A ' = 5(4) + 7(6) = 62 or the rate of change is
62 cm2/sec (cm x cm/sec).
32. Related Rates
We have that A = L* W, where are the variable,
are all functions in t. Take the derivative on both sides,
we have that
A ' = L' * W + L* W'
We are given that L' = 5, W ' = 7, L = 6, and W = 4.
Hence A ' = 5(4) + 7(6) = 62 or the rate of change is
62 cm2/sec (cm x cm/sec).
We may see the rates of
change in areas as shown
in the figure. W
L
A ' = L' W + L W'
33. Related Rates
We have that A = L* W, where are the variable,
are all functions in t. Take the derivative on both sides,
we have that
A ' = L' * W + L* W'
We are given that L' = 5, W ' = 7, L = 6, and W = 4.
Hence A ' = 5(4) + 7(6) = 62 or the rate of change is
62 cm2/sec (cm x cm/sec).
L'
We may see the rates of
change in areas as shown
in the figure. W L' W
L
A ' = L' W + L W'
34. Related Rates
We have that A = L* W, where are the variable,
are all functions in t. Take the derivative on both sides,
we have that
A ' = L' * W + L* W'
We are given that L' = 5, W ' = 7, L = 6, and W = 4.
Hence A ' = 5(4) + 7(6) = 62 or the rate of change is
62 cm2/sec (cm x cm/sec).
L'
We may see the rates of
change in areas as shown
in the figure. W L' W
L
W ’ L’
W' L W'
A ' = L' W + L W'
35. Related Rates
We have that A = L* W, where are the variable,
are all functions in t. Take the derivative on both sides,
we have that
A ' = L' * W + L* W'
We are given that L' = 5, W ' = 7, L = 6, and W = 4.
Hence A ' = 5(4) + 7(6) = 62 or the rate of change is
62 cm2/sec (cm x cm/sec).
L'
We may see the rates of
change in areas as shown
in the figure. W L' W
This is the geometric
version of the Product Rule L
W ’ L’
of derivatives. W' L W'
(W'L' is negligible) A ' = L' W + L W'
36. Related Rates
The Cosine Law
c2 = a2 + b2 – 2ab*cos(C) b c
is an extension of the
Pythagorean Theorem
o) hence it’s to be C
(i.e. C = 90
considered as the general a
Distance Formula in a slanted coordinate system
where the coordinate axes intersect at angle C.
37. Related Rates
The Cosine Law
c2 = a2 + b2 – 2ab*cos(C) b c
is an extension of the
Pythagorean Theorem
o) hence it’s to be C
(i.e. C = 90
considered as the general a
Distance Formula in a slanted coordinate system
where the coordinate axes intersect at angle C.
Example C. Two cars approach an intersection where
the two streets intersect at an angle of π/3.
A is traveling at 20m/sec and B is traveling at 30m/sec.
How fast is their distance shrinking when A is 30m
away from the intersection and B is 20m away from
the intersection?
38. Related Rates
30m/s
Here is a diagram.
B
Let’s label the sides as
x and y as shown. D
y
π/3
o x A
20m/s
40. Related Rates
30m/s
Here is a diagram.
B
Let’s label the sides as
x and y as shown. D
y
By the Cosine Law
π/3
D2 = x2 + y2 – 2xy*cos(π/3) or o x A
20m/s
41. Related Rates
30m/s
Here is a diagram.
B
Let’s label the sides as
x and y as shown. D
y
By the Cosine Law
π/3
D2 = x2 + y2 – 2xy*cos(π/3) or o x A
D2 = x2 + y2 – xy 20m/s
42. Related Rates
30m/s
Here is a diagram.
B
Let’s label the sides as
x and y as shown. D
y
By the Cosine Law
π/3
D2 = x2 + y2 – 2xy*cos(π/3) or o x A
D2 = x2 + y2 – xy 20m/s
We are given the following:
x = 30
at the instance
y = 20
43. Related Rates
30m/s
Here is a diagram.
B
Let’s label the sides as
x and y as shown. D
y
By the Cosine Law
π/3
D2 = x2 + y2 – 2xy*cos(π/3) or o x A
D2 = x2 + y2 – xy 20m/s
We are given the following:
x = 30 dx
at the instance we have dt = –20
y = 20 dy
dt = –30
44. Related Rates
30m/s
Here is a diagram.
B
Let’s label the sides as
x and y as shown. D
y
By the Cosine Law
π/3
D2 = x2 + y2 – 2xy*cos(π/3) or o x A
D2 = x2 + y2 – xy 20m/s
We are given the following:
x = 30 dx
at the instance we have dt = –20
y = 20 dy
dt = –30
The negative sign means that x and y are shrinking.
45. Related Rates
30m/s
Here is a diagram.
B
Let’s label the sides as
x and y as shown. D
y
By the Cosine Law
π/3
D2 = x2 + y2 – 2xy*cos(π/3) or o x A
D2 = x2 + y2 – xy 20m/s
We are given the following:
x = 30 dx
at the instance we have dt = –20
y = 20 dy
dt = –30
The negative sign means that x and y are shrinking.
The question is “what is dD at that instance?”.
dt
46. Related Rates
It’s easier to take the derivatives of
D2 = x2 + y2 – xy (instead of solving for D first).
47. Related Rates
It’s easier to take the derivatives of
D2 = x2 + y2 – xy (instead of solving for D first).
dD dx dy dy dx
2D dt = 2x dt + 2y dt – x dt – y dt
48. Related Rates
It’s easier to take the derivatives of
D2 = x2 + y2 – xy (instead of solving for D first).
dD dx dy dy dx
2D dt = 2x dt + 2y dt – x dt – y dt or
dD
2D dt = (2x – y) dx + (2y – x) dy *
dt dt
49. Related Rates
It’s easier to take the derivatives of
D2 = x2 + y2 – xy (instead of solving for D first).
dD dx dy dy dx
2D dt = 2x dt + 2y dt – x dt – y dt or
dD
2D dt = (2x – y) dx + (2y – x) dy *
dt dt
x = 30
At the instance with D2 = x2 + y2 – xy
y = 20
we have that D2 = 700 or that D = 10√7.
50. Related Rates
It’s easier to take the derivatives of
D2 = x2 + y2 – xy (instead of solving for D first).
dD dx dy dy dx
2D dt = 2x dt + 2y dt – x dt – y dt or
dD
2D dt = (2x – y) dx + (2y – x) dy *
dt dt
x = 30
At the instance with D2 = x2 + y2 – xy
y = 20
we have that D2 = 700 or that D = 10√7.
dx
dt = –20
Substitute all these with dy into *.
dt = –30
51. Related Rates
dD
2D dt = (2x – y) dx + (2y – x) dy
dt dt
D = 10√7, x = 30, y = 20
dx
dt = –20
dy
dt = –30
52. Related Rates
dD
2D dt = (2x – y) dx + (2y – x) dy
dt dt
D = 10√7, x = 30, y = 20
dx
dt = –20
dy
dt = –30
We get
dD
20√7 dt = 40(–20) + 10(–30)
53. Related Rates
dD
2D dt = (2x – y) dx + (2y – x) dy
dt dt
D = 10√7, x = 30, y = 20
dx
dt = –20
dy
dt = –30
We get
dD
20√7 dt = 40(–20) + 10(–30) or that
dD = –1100
dt 20√7
54. Related Rates
dD
2D dt = (2x – y) dx + (2y – x) dy
dt dt
D = 10√7, x = 30, y = 20
dx
dt = –20
dy
dt = –30
We get
dD
20√7 dt = 40(–20) + 10(–30) or that
dD = –1100 = –55 ≈ –20.8 m/sec
dt 20√7 √7
So the distance is decreasing at a rate of 20.8m/sec
56. Related Rates
(Extra Credit: What was the closest distance came
between the two cars?)
Example D. Water is pumped at a rate of 2m3/sec
into a tank in the shape of a circular cone that’s 6m in
diameter at the top and 5m in height. Find the rate of
increase in the radius of the surface at the instance
when the water is filled to the depth of 2m.
3m 3m
x 5m
5x/3
57. Related Rates
(Extra Credit: What was the closest distance came
between the two cars?)
Example D. Water is pumped at a rate of 2m3/sec
into a tank in the shape of a circular cone that’s 6m in
diameter at the top and 5m in height. Find the rate of
increase in the radius of the surface at the instance
when the water is filled to the depth of 2m.
The cross section of a cone 3m 3m
is triangle hence similar
triangles play an important
role in such problems, x 5m
5x/3
58. Related Rates
(Extra Credit: What was the closest distance came
between the two cars?)
Example D. Water is pumped at a rate of 2m3/sec
into a tank in the shape of a circular cone that’s 6m in
diameter at the top and 5m in height. Find the rate of
increase in the radius of the surface at the instance
when the water is filled to the depth of 2m.
The cross section of a cone 3m 3m
is triangle hence similar
triangles play an important
role in such problems, i.e. x 5m
the radius and height at any level
is always of the same ratio as the 5x/3
dimension of the tank.
59. Related Rates
3m 3m
The volume of water at
any given moment is V
πr2h
V= 3 x 5m
h=5x/3
60. Related Rates
3m 3m
The volume of water at
any given moment is V
πr2h and in this case
V= 3 x 5m
5πx3
V= 9 h=5x/3
61. Related Rates
3m 3m
The volume of water at
any given moment is V
πr2h and in this case
V= 3 x 5m
5πx3
V= 9 h=5x/3
We are given that V' = 2
62. Related Rates
3m 3m
The volume of water at
any given moment is V
πr2h and in this case
V= 3 x 5m
5πx3
V= 9 h=5x/3
We are given that V' = 2 and we
want x' at the instance that the height h = 5x/3 = 2,
63. Related Rates
3m 3m
The volume of water at
any given moment is V
πr2h and in this case
V= 3 x 5m
5πx3
V= 9 h=5x/3
We are given that V' = 2 and we
want x' at the instance that the height h = 5x/3 = 2,
or that when x = 6/5.
64. Related Rates
3m 3m
The volume of water at
any given moment is V
πr2h and in this case
V= 3 x 5m
5πx3
V= 9 h=5x/3
We are given that V' = 2 and we
want x' at the instance that the height h = 5x/3 = 2,
or that when x = 6/5.
5πx3
Take the derivative of V = 9 we have
5πx2x'
V' = 3
65. Related Rates
3m 3m
The volume of water at
any given moment is V
πr2h and in this case
V= 3 x 5m
5πx3
V= 9 h=5x/3
We are given that V' = 2 and we
want x' at the instance that the height h = 5x/3 = 2,
or that when x = 6/5.
5πx3
Take the derivative of V = 9 we have
5πx2x'
V' = 3 or x' = 3V '2
5πx
66. Related Rates
3m 3m
The volume of water at
any given moment is V
πr2h and in this case
V= 3 x 5m
5πx3
V= 9 h=5x/3
We are given that V' = 2 and we
want x' at the instance that the height h = 5x/3 = 2,
or that when x = 6/5.
5πx3
Take the derivative of V = 9 we have
5πx2x'
V' = 3 or x' = 3V '2
5πx
set V' = 2 and x = 6/5, we get the rate of change in
the radius is x' = 5/(6π) ≈ 0.265m/sec.
