Applications of Derivatives 169
dθ
dt
= 0 .197 radian/sec. The angle
of elevation is changing at
0.197 radian/sec, 5 seconds after
liftoff.
- (See Figure 8.6-9.)
h
20
Figure 8.6-9
Sin 20◦=
h
300 t
h=(sin 20◦)300t;
dh
dt
=(sin 20◦)(300)≈ 102 .606 mph. The
plane is gaining altitude at 102.606 mph.
- Vcone=
1
3
πr^2 h
Similar triangles:
4
10
=
r
h
⇒ 5 r= 2 h or
r=
2 h
5
.
Vcone=
1
3
π
(
2 h
5
) 2
h=
4 π
75
h^3 ;
dV
dt
=
4 π
75
(3)h^2
dh
dt
.
Substitute known values:
− 15 =
4 π
25
(5)^2
dh
dt
;
− 15 = 4 π
dh
dt
;
dh
dt
=
− 15
4 π
≈− 1 .19 ft/min.
The water level in the cone is falling at
− 15
4 π
ft/min≈− 1 .19 ft/min when the
water level is 5 feet high.
Vcylinder=πR^2 H=π(6)2H= 36 πH.
dV
dt
= 36 π
dH
dt
;
dH
dt
=
1
36 π
dV
dt
;
dH
dt
=
1
36 π
(15)=
5
12 π
ft/min
≈ 0 .1326 ft/min or 1.592 in/min.
The water level in the cylinder is rising at
5
12 π
ft/min= 0 .133 ft/min.
- Step 1: Letxbe the distance of the foot
of the ladder from the higher wall.
Letybe the height of the point
where the ladder touches the
higher wall. The slope of the
ladder ism=
y− 6
0 − 8
orm=
6 − 0
8 −x
.
Thus,
y− 6
− 8
=
6
8 −x
⇒(y−6)(8−x)
=− 48
⇒ 8 y−xy− 48 + 6 x=− 48
⇒y(8−x)=− 6 x⇒y=
− 6 x
8 −x
.
Step 2: Phythagorean Theorem:
l^2 =x^2 +y^2 =x^2 +
(
− 6 x
8 −x
) 2
Sincel>0,l=
√
x^2 +
(
− 6 x
8 −x
) 2
,
x> 8.
Step 3: √Entery 1 =
{
x∧ 2 +[(− 6 ∗x)/( 8 −x)]∧ 2
}
.
The graph ofy 1 is continuous on
the intervalx>8. Use the
[Minimum] function of the
calculator and obtainx= 14 .604;
y= 19 .731. Thus, the minimum
value oflis 19.731, or the shortest
ladder is approximately
19.731 feet.
- Step 1: Average cost:
C=
C
x
; thus,C(x)
=
2500 + 0. 02 x+ 0. 004 x^2
x
=
2500
x
+ 0. 02 +. 004 x.
