1
If we are pumping air into a balloon, both the volume and the radius of the balloon are increasing and their rates of
increase are related to each other.
However, it is much easier to measure directly the rate of increase of the volume than the rate of increase of the
radius.
In this section, we will learn how to compute the rate of change of one quantity in terms of that of another quantity.
Objectives:
Identify a mathematical relationship between quantities that are each changing.
Steps for Related Rates Problems:
1.
2.
3.
4.
5.
6.
Draw a picture (sketch).
Write down known information.
Write down what you are looking for.
Write an equation to relate the variables.
Differentiate both sides with respect to t.
Evaluate.
Examples
Consider a sphere of radius 10cm.
If the radius changes 0.1cm (a very small amount) how much does the volume change?
𝑑𝑟 = 0.1 𝑐𝑚
𝑑𝑉 =?
4
V   r3
3
dV  4 r 2 dr
dV  4 10cm   0.1cm
2
dV  40 cm3
The volume would change by approximately 40 𝜋 𝑐𝑚3
Now, suppose that the radius is changing at an instantaneous rate of 0.1 cm/sec.
At what rate is the sphere growing when r = 10 cm?
𝑑𝑟
= 0.1𝑐𝑚/ sec
𝑑𝑡
𝑑𝑉
=?
𝑑𝑡
4
V   r3
3
dV
dr
 4 r 2
dt
dt
dV
cm 
2 
 4 10cm    0.1

dt
 sec 
dV
cm3
 40
dt
sec
The sphere is growing at a rate of 40 𝜋 𝑐𝑚3 /𝑠𝑒𝑐
2
Air is being blown into a sphere at the rate of 6 cubic inches per minute.
How fast is the radius changing when the radius of the sphere is 2 inches?
What we know:
4
v  r 3
3
dv
dr
 4r 2
dt
dt
dr
6  4 (2) 2
dt
dr
6
3


in / min
dt 16 8
dv
 6 in 3 / min
dt
r  2 in
Example: Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm3/s.
How fast is the radius of the balloon increasing when the diameter is 50 cm?
𝑑𝑉
𝑐𝑚3
= 100
𝑑𝑡
𝑠
𝑑𝑟
=?
𝑑𝑡
𝑤ℎ𝑒𝑛 𝑟 = 25 𝑐𝑚
4
V   r3
3
𝑑𝑉 𝑑𝑉 𝑑𝑟
𝑑𝑟
=
= 4𝜋𝑟 2
𝑑𝑡
𝑑𝑟 𝑑𝑡
𝑑𝑡
𝑑𝑟
1 𝑑𝑉
=
𝑑𝑡 4𝜋𝑟 2 𝑑𝑡
→
𝑑𝑟
1
1
=
100 =
𝑑𝑡 4𝜋(25)2
25𝜋
The radius of the balloon is increasing at the rate of 1/(25π) ≈ 0.0127 cm/s.
The edge of a cube is increasing at a rate of 2 inches per minute.
At the instant the edge is 3 inches, how fast is the volume increasing?
What we know:
dv
 2 in 3 / min
dt
s  3 in
v  s3
dv
ds
 3s 2
dt
dt
ds
2  3(3) 2
dt
ds
2 3

in / min
dt 27
3
Water is draining from a cylindrical tank at 3 liters/second. How fast is the surface dropping?
3
cm
dV
L
 3000
 3
dt
sec
sec
𝑑ℎ
=?
𝑑𝑡
V   r 2h
𝑟 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡!
dV
dh
  r2
dt
dt
dh

dt
cm3
sec
 r2
3000
Truck A travels east at 40 mi/hr. Truck B travels north at 30 mi/hr.
How fast is the distance between the trucks changing 6 minutes later?
@ 6 min x= 4 mi & y = 3 mi from initial position
z2 = x 2 + y 2
𝑑 = 𝑣𝑡
@ 6 min z = 5 mi
𝑑𝑧
𝑑𝑥
𝑑𝑦
= 2𝑥
+ 2𝑦
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑧
5
= 4 ∙ 40 + 3 ∙ 30
𝑑𝑡
𝑑𝑧
= 50 𝑚𝑖/ℎ𝑟
𝑑𝑡
2𝑧
Car A is traveling west at 50 mi/h and car B is traveling north at 60 mi/h. Both are headed for the intersection
of the two roads.
At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?
dx / dt = –50 mi/h & dy / dt = –60 mi/h both x & y are decreasing
𝑑𝑧
=?
𝑑𝑡
z2 = x2 + y2 ⇒ 2𝑧
𝑑𝑧
𝑑𝑡
= 2𝑥
𝑑𝑥
𝑑𝑡
+ 2𝑦
𝑑𝑦
𝑑𝑡
When x = 0.3 mi & y = 0.4 mi, z = 0.5 mi.
𝑑𝑧
1
[0.3(−50) + 0.4(−6)0]
=
𝑑𝑡 0.5
𝑑𝑧
= −78 𝑚𝑖/ℎ
𝑑𝑡
4
A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a
rate of 1 ft/sec, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6
ft from the wall?
What we know:
dx
 1 ft / sec
dt
r  10 ft
x  6 ft
x2  y2  r 2
dx
dy
dr
2x  2 y
 2r
dt
dt
dt
dy 

(2  6  1)   2  8    0
dt 

dy
16
 12
dt
dy  12  3


ft / sec
dt
16
4
The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a
rate of 2 cm2/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area
is 100 cm2?
What we know:
dh
 1 cm / min
dt
dA
 2 cm 2 / min
dt
h  10 cm
A  100 cm
1
bh
2
1
100  b  10
2
b  20 cm
1
A  bh
2
dA 1 
dh
db 
 b 
 h

dt
2
dt
dt 
A
2
1
db 
 20  1  10 

2
dt 
db
 1.6 cm / min
dt
A water tank has the shape of an inverted circular cone with base radius 2m and height 4 m. If water is
being pumped into the tank at a rate of 2m3/min, find the rate at which the water level is rising when the
water is 3 m deep.
5
What we know:
1 3
h
12
dv 1
dh
  h2
dt 4
dt
1
2 dh
2   3
4
dt
r 2

h 4
h
r
2
1
v   r 2h
3
rbase  2 m
htotal  4 m
dv
 2 m 3 / min
dt
h3m
v
2
1 h
v   h
3 2
dh 8

m / min
dt 9
A point moves along the curve y   x  3 such that its x-coordinate is increasing at 4 units per second.
2
(a) At the moment x = 1, how fast is the y-coordinate changing? Interpret your answer based on the shape
of the graph and the location of the point.
(b)
What we know:
y   x  3
dx
 4 units / sec
dt
x 1
dy
dx
 2( x  3)
dt
dt
dy
 2(1  3)  4  16 units / sec
dt
2
At the moment x = 1, how fast is the point’s distance from the origin changing?
What we know:

D  ( x  0) 2  ( x  3) 2  0

2
2
dx
D  ( x) 2  ( x  3) 2 
 4 units / sec
dt
D  x 2  ( x  3) 4
x 1
1/ 2


D  x 2  ( x  3) 4
1 / 2
dD 1 2
dx
 x  ( x  3) 4
 2 x  4( x  3) 3 
dt 2
dt
1 / 2
dD 1 2
 (1)  (1  3) 4
 2(1)  4(1  3) 3  4
dt 2
dD 1
1 / 2
1 / 2
 17    30  4  6017   14.552 units / min
dt 2
Calculus Maximus
PROBLEMS:
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