§3-10 Related Rates
Homework：3,5,7,8,9,12,14,17,19,25,31,35,38
Related Rates：若問題出現如下敘述：
討論某些量 x 對時間 t 的變化率 (rate of change)的應用問題.
}
dx
at what rate ..................x.......?
則是問 = ? (即 x 對時間 t 的變化率) .
How fast(slow).............x.......?
dt
Steps to solve related rates problems.
1. 引進〝符號〞將題目 (文字) 轉成〝圖〞.
2. 找一方程式〝連結〞此圖.
3.
d
(此方程式).
dt
Example1：
Air is being pumped into a spherical balloon so that its volume increases at a rate
3
of 100 cm /s. How fast is the radius of the balloon increasing when the diameter is
50 cm?
Solution： 已知
dV
dr
= 100 , 求
dt
dt
r = 25 .
4
V = π r3
3
dV
dr
dr
1 cm3
(
= 4π r 2
⇒
=
).
s
dt
dt
dt 25π
Example2：
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/s. How fast is the top of the ladder sliding
down the wall when the ladder is 6 ft from the wall?
Solution：
dx
= 1 (ft s)
dt
dy
求 :
|x =6
.
dt
已知:
⎛ dx ⎞
⎛ dy ⎞
x 2 + y 2 = 100 ⇒ 2 x ⎜ ⎟ + 2 y ⎜ ⎟ = 0. 當x = 6 ⇒ y = 8 .
⎝ dt ⎠
⎝ dt ⎠
dy
3
⇒
= − ( ft s ) .
dt
4
Example3：
A water tank has the shape of an inverted circular cone with base radius 2 m and
height 4 m. If water is being pumped into the tank at a rate of 2 m3 min , find the
rate at which the water level is rising when the water is 3 m deep.
Solution：
2
dV
已知： = 2 (m3 min).
dt
dh
求：
=?
dt h =3
r
4
h
1
1 ⎛h⎞
π
V = π r 2 h = π ⎜ ⎟ h = h3
3
3 ⎝2⎠
12
dV π 2 dh
dh 8
⇒
= h
⇒
=
(m3 min ).
dt
dt
dt 9π
4
2
Example4：
A baseball diamond is a square with side 90 ft. A batter hits the ball and runs
toward first base with a speed of 24 ft/s. At what rate is his distance from third
base increasing when he is half way to first base?
Solution：
dx
= 24 (ft s).
dt
dr
求 :
x = 45 = ?
dt
已知:
dr
dx
= 2x
dt
dt
dr x dx 24 24 5
=
=
=
. ( ∵當 x = 45 , r = 45 5.)
dt r dt
5
5
r 2 = 902 + x 2
⇒
⇒ 2r
Example5：
The minute hand on a watch is 8 mm long and the hour hand is 4 mm long. How
fast is the distance between the tips of the hands changing at one o’clock?
Solution：
已知:
(
11
dθ π
= − 2π = − π rad
h
6
dt 6
分針的角速度 =2π
時針的角速度 =
求 :
ds
dt
θ=
π
π
6
=?
6
s 2 = 64 + 16 − 64 cos θ
−88π
ds
dθ
ds
⇒ 2 s = 64sin θ
⇒
=
(rad h).
dt
dt
dt 3 80 − 32 3
(When θ =
π
, s = 80 − 32 3 ).
6
)