Section 3.10 Related Rates
Sec 3.10_B
1
Useful Formulas in Related Rates Problems
Pythagorean Theorem
a +b = c
2
2
2
c
a
b
Sec 3.10_B
2
Useful Formulas in Related Rates Problems
E
Similar triangles
B
2
A
h
3
4
6
C
D
12
F
If triangle ABC is similar to triangle DEF, then:
2 4
=
6 12
h 3
= , etc
12 4
Sec 3.10_B
3
Distance, Rate and Time
d = rt
Sec 3.10_B
4
Useful Formulas in Related Rates Problems
Volume of a Sphere
Volume of a Cone
4
V = π r3
3
1
V = π r 2h
3
Volume of a right circular cylinder
Area of a Circle of radius r
V = π r 2h
A = π r2
Circumference of a Circle of radius r C = 2π r
Sec 3.10_B
5
Related Rate Problem Strategy
1)
2)
Draw a picture and name the variables and constants. Use
“t” for time and assume all variables are differentiable
functions of t.
Determine what is asked to be found
3) Write down an equation that relates the variables
4) Differentiate with respect to t. Then express the rate you
want to find in terms of the rate and variables whose values
you know.
5. Use known values to find the unknown rate.
Sec 3.10_B
6
Changing Dimensions In a Rectangle
The length l of a rectangle is decreasing at the rate of 2
cm/sec while the width is increasing at the rate of 2 cm/sec.
When l = 12 cm and w =5 cm. find the rates of change of:
a) The area
b) The perimeter
w
l
Sec 3.10_B
7
Rectangle Solution
The length l of a rectangle is decreasing at the rate of 2
cm/sec while the width is increasing at the rate of 2 cm/sec.
When l = 12 cm and w =5 cm. find the rates of change of:
a) The area:
A = lw;
Since l=12 and w=5,
dA
dw
dl
=l
+ w = l (2) + w( −2) = 2l − 2 w
dt
dt
dt
dA
cm
= 2(12) − 2(5) = 14
dt
sec
w
l
Sec 3.10_B
8
Rectangle-Continued
Sec 3.10_B
9
Typed Solution
The length l of a rectangle is decreasing at the rate of 2
cm/sec while the width is increasing at the rate of 2 cm/sec.
When l = 12 cm and w =5 cm. find the rates of change of:
b) The perimeter:
P = 2l + 2 w;
dP
dl
dw
=2 +2
= 2(−2) + 2(2) = 0
dt
dt
dt
w
l
Sec 3.10_B
10
Ice Cube
A cube of ice is melting. The side of the cube is
decreasing at the constant rate of 2 inches per
minute. How fast is the volume decreasing?
s
Let s = the side of the cube
Sec 3.10_B
11
Typed Solution: Ice Cube
A cube of ice is melting. The side of the cube is
decreasing at the constant rate of 2 inches per
minute. How fast is the volume decreasing?
s
Let s = the side of the cube
dV
V =s ,
=?
dt
3
ds
But
= −2
dt
dV
2 ds
= 3s
dt
dt
dV
= 3s 2 ( −2) = −6s 2
dt
The rate of change of the volume is
directly proportional to the square of the
side
Sec 3.10_B
12
Solution:Triangle Problem
The altitude of a triangle is increasing at a rate of 1 cm / min while the area of
2
the triangle is increasing at the rate of 2 cm / min At what rate is the base of
100 cm 2 ?
the triangle changing when the altitude is 10 cm and the area is
Sec 3.10_B
13
Typed Solution:Triangle Problem
2
The altitude of a triangle is increasing at a rate of 1 cm / min while the area of
2
the triangle is increasing at the rate of 2 cm / min . At what rate is the base of
2
the triangle changing when the altitude is 10 cm and the area is 100 cm ?
1
A = bh
2
B
h
A
C
b
dh
dA
db
Given
= 1,
= 2, find
when h = 10 and A = 100
dt
dt
dt
dA 1 ⎡ dh
db ⎤
= ⎢b + h ⎥
dt 2 ⎣ dt
dt ⎦
1⎡
db ⎤
2 = ⎢ 20(1) + 10 ⎥
2⎣
dt ⎦
When h=10 and A=100, using the formula for
the area of a triangle,100 = 1 b(10) = 10b, b = 20
2
Sec 3.10_B
db
1.6cm
=−
dt
min
14
Problem: Street Light: Similar to WW9 # 7
A street light is mounted at the top of a 15 ft tall pole. A
man 6 ft walks away from the pole with a speed of 5 ft/sec
along a straight path. How fast is the tip of his shadow
moving when he is 40 feet from the pole?
15
x
6
y
x
Sec 3.10_B
15
Problem: Street Light Solution
Let x =distance from the pole to the man
15
Let y =distance from the man to the tip of his shadow
6
x
y
Observation: As the man walks from the pole, the length of his
shadow increases and the tip of his shadow can viewed as a
moving object on the end of the shadow and whose distance
from the pole is x + y
We want to find
d ( x + y)
dt
Since we don’t know
x
dy
dt
d ( x + y ) dx dy
=
+
dt
dt dt
we express y in terms of x, since
Sec 3.10_B
dx
is known
dt
16
By similar triangles:
Continued
15 x + y
=
6
y
15
x
15 y = 6 x + 6 y
9 y = 6 x, y =
6
x
9
6
OR y =
y
2
x
3
dy 2 dx
=
dt 3 dt
d ( x + y ) dx dy dx 2 dx 5 dx
=
+
=
+
=
dt
dt dt dt 3 dt 3 dt
Since
dx
=5
dt
d ( x + y) 5
25
= (5) =
ft / sec
dt
3
3
Sec 3.10_B
17
Problem: Street Light: Solution
A street light is mounted at the top of a 15 ft tall pole. A
man 6 ft walks away from the pole with a speed of 5 ft/sec
along a straight path. How fast is the tip of his shadow
moving when he is 40 feet from the pole?
15
x
6
y
x
Sec 3.10_B
18
Problem: Travel
Similar to WW 9, Problem 10
At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35
km/hr and ship B is sailing north at 25 km /hour. How fast is the distance
between the ships changing at 4 PM
Sec 3.10_B
19
Solution
At noon, ship A is 150 km west of ship B. Ship A is sailing east at 35
km/hr and ship B is sailing north at 25 km /hour. How fast is the distance
between the ships changing at 4 PM
B
Let x = distance traveled by A after t hours
past noon
y
Let y = distance traveled by B after t hours
past noon
z
150-x
A
Let x = distance between the ships after t
hours past noon
Given: At noon, ship A is 150 km west of ship B, note
that at any time past noon, ship A is 150 –x miles from
where B departed at 12 noon, dx = 35 and dy = 25
Unknown:
dz
at 4 pm
dt
dt
Sec 3.10_B
dt
20
Continued
B
z
z 2 = (150 − x ) + y 2
2
y
150-x
2z
A
dz
dy
⎛ dx ⎞
= 2 (150 − x ) ⎜ − x ⎟ + 2 y
dt
dt
⎝ dt ⎠
At 4 pm, t=4, x = 4(35) = 140 and y = 4(25) = 100
z 2 = (150 − 140 ) + (100 ) = 10,100
2
2
z = 10,100
dz 1 ⎡
dx
dy ⎤
= ⎢( x − 150 ) + y ⎥
dt z ⎣
dt
dt ⎦
=
−10(35) + 100(25) 215
km
=
= 21.4
hr
10,100
101
Sec 3.10_B
21