Applications of Derivatives 165
5 m
h
r 18
12
Figure 8.6-2
Differentiate:
dV
dt
=
4
9
πh^2
dh
dt
.
Substituting known values:
− 4 =
4 π
9
(6)^2
dh
dt
⇒− 4 = 16 π
dh
dt
or
dh
dt
=−
1
4 π
ft/min. The water level is
dropping at
1
4 π
ft/min whenh=6 ft.
- (See Figure 8.6-3.)
Step 1: Using the Pythagorean Theorem,
you havex^2 +y^2 =z^2. You also
have
dx
dt
=40 and
dy
dt
=30.
y
x
z
N
S
WE
Figure 8.6-3
Step 2: Differentiate:
2 x
dx
dt
+ 2 y
dy
dt
= 2 z
dz
dt
.
Atx=120, both cars have
traveled 3 hours and thus,
y=3(30)=90. By the
Pythagorean Theorem,
(120)^2 +(90)^2 =z^2 ⇒z=150.
Step 3: Substitute all known values into
the equation:
2(120)(40)+2(90)(30)=2(150)
dz
dt
.
Thus,
dz
dt
=50 mph.
Step 4: The distance between the two cars
is increasing at 50 mph atx=120.
7.(See Figure 8.6-4.)
y
x
9 – x 9 – x
x
Figure 8.6-4
Step 1: Applying the Pythagorean
Theorem, you have
x^2 =y^2 +(9−x)^2 ⇒y^2 =
x^2 −(9−x)^2 =
x^2 −
(
81 − 18 x+x^2
)
=
18 x− 81 = 9 ( 2 x− 9 ),or
y=±
√
9 ( 2 x− 9 )=
± 3
√
( 2 x− 9 )sincey>0,
y= 3
√
( 2 x− 9 ).
The area of the triangle
A=
1
2
(
3
√
2 x− 9
)
( 18 − 2 x)=
(
3
√
2 x− 9
)
( 9 −x)=
3 ( 2 x− 9 )^1 /^2 ( 9 −x).
