f At time t, let x be the horizontal distance of the plane from the point directly over R, and let u be the distance
between the plane and the station. Then u
2 = x
2 + (4)
2
. So, 2u • D,u = 2x • D,x, u • D,u - x • D,x.
When w = 5, (5)
2 = x
2 + (4)
2
, * = 3, and we are also told that D,u is 300. Substituting in u • D,u =
x • D,x, 5 • 300 = 3 • D,x, D,x = 500 kilometers per hour.
14.17
14.18
Fig. 14-5
Fig. 14-6
A boat passes a fixed buoy at 9 a.m. heading due west at 3 miles per hour. Another boat passes the same buoy at
10a.m. heading due north at 5 miles per hour. How fast is the distance between the boats changing at 11:30
a.m.?
I Refer to Fig. 14-6. Let the time t be measured in hours after 9 a.m. Let x be the number of miles that the
first boat is west of the buoy at time t, and let y be the number of miles that the second boat is north of the buoy at
time ;. Let u be the distance between the boats at time /. For any time ra=l, u
2 = x
2 + y
2 . Then
2u • D,u = 2x • D t x + 2y D,y, u • D,u = x • D,x + y • D,y. We are given that D,x = 3 and D,y=5. So,
u • D,u = 3>x + 5y. At ll:30a.m. the first boat has travelled 2k hours at 3 miles per hour; so, x=
s £.
Similarly, the second boat has travelled at 5 miles per hour for Ik hours since passing the buoy; so, y = ".
Also, «
2 = (f )
2 + (¥)
2 = ¥, « = 15/V2. Substituting in u-D,u = 3x + 5y, (15/V3)- D,u = 3 • f +
5-¥=60, D,w = 4\/2 = 5.64miles per hour.
Water is pouring into an inverted cone at the rate of 3.14 cubic meters per minute. The height of the cone is 10
meters, and the radius of its base is 5 meters. How fast is the water level rising when the water stands 7.5 meters
above the base?
2- D,r= -3.14/7r(1.25)
2 , D,w = -D,y = 3.14/7r(1.25)
2
= 0.64 m/min.
14.19 A particle moves along the curve y=x +2x. At what point(s) on the curve are the x- and ^-coordinates of the
particle changing at the same rate?
D,y = 2x-D,x + 2-D,x = D,x(2x + 2). When D,y = D,x, 2x + 2=\, 2x = -l, x = -|, y = -$.
RELATED RATES
91
Let w be the level of the water above the base, and let r be the radius of the circle that forms the surface of the
water. Let y = 10— w. Then y is the height of the cone-shaped region above the water (see Fig. 14-7). So,
the volume of that cone is V, = \irr
2 y. The total volume of the conical container is V 2 = 5?r(5)
2 • 10 =
250?r/3. Thus, the total volume of the water is V= V 2 -V l = 25077/3 - irr
2 y/3. By similar triangles,
10/5 = y/r, y = 2r. So, V= 25077/3 - irr
2 (2r}/3 = 2507T/3 - 2irr
3 /3. Hence, D,V= -2-nr
2 • D,r. We
aregiventhat D,V=3.14. So, 3.14= -2-rrr
2 • D,r. Thus, D,r = -3.14/2-trr
2 . When the water stands
7.5 meters in the cone, w = 7.5, y = 10- 7.5 = 2.5 r=ky = l.25. So £>/=-3.14/2ir(1.25)
2 . D,y =
Fig. 14-7
between the plane and the station. Then u
2 = x
2 + (4)
2
. So, 2u • D,u = 2x • D,x, u • D,u - x • D,x.
When w = 5, (5)
2 = x
2 + (4)
2
, * = 3, and we are also told that D,u is 300. Substituting in u • D,u =
x • D,x, 5 • 300 = 3 • D,x, D,x = 500 kilometers per hour.
14.17
14.18
Fig. 14-5
Fig. 14-6
A boat passes a fixed buoy at 9 a.m. heading due west at 3 miles per hour. Another boat passes the same buoy at
10a.m. heading due north at 5 miles per hour. How fast is the distance between the boats changing at 11:30
a.m.?
I Refer to Fig. 14-6. Let the time t be measured in hours after 9 a.m. Let x be the number of miles that the
first boat is west of the buoy at time t, and let y be the number of miles that the second boat is north of the buoy at
time ;. Let u be the distance between the boats at time /. For any time ra=l, u
2 = x
2 + y
2 . Then
2u • D,u = 2x • D t x + 2y D,y, u • D,u = x • D,x + y • D,y. We are given that D,x = 3 and D,y=5. So,
u • D,u = 3>x + 5y. At ll:30a.m. the first boat has travelled 2k hours at 3 miles per hour; so, x=
s £.
Similarly, the second boat has travelled at 5 miles per hour for Ik hours since passing the buoy; so, y = ".
Also, «
2 = (f )
2 + (¥)
2 = ¥, « = 15/V2. Substituting in u-D,u = 3x + 5y, (15/V3)- D,u = 3 • f +
5-¥=60, D,w = 4\/2 = 5.64miles per hour.
Water is pouring into an inverted cone at the rate of 3.14 cubic meters per minute. The height of the cone is 10
meters, and the radius of its base is 5 meters. How fast is the water level rising when the water stands 7.5 meters
above the base?
2- D,r= -3.14/7r(1.25)
2 , D,w = -D,y = 3.14/7r(1.25)
2
= 0.64 m/min.
14.19 A particle moves along the curve y=x +2x. At what point(s) on the curve are the x- and ^-coordinates of the
particle changing at the same rate?
D,y = 2x-D,x + 2-D,x = D,x(2x + 2). When D,y = D,x, 2x + 2=\, 2x = -l, x = -|, y = -$.
RELATED RATES
91
Let w be the level of the water above the base, and let r be the radius of the circle that forms the surface of the
water. Let y = 10— w. Then y is the height of the cone-shaped region above the water (see Fig. 14-7). So,
the volume of that cone is V, = \irr
2 y. The total volume of the conical container is V 2 = 5?r(5)
2 • 10 =
250?r/3. Thus, the total volume of the water is V= V 2 -V l = 25077/3 - irr
2 y/3. By similar triangles,
10/5 = y/r, y = 2r. So, V= 25077/3 - irr
2 (2r}/3 = 2507T/3 - 2irr
3 /3. Hence, D,V= -2-nr
2 • D,r. We
aregiventhat D,V=3.14. So, 3.14= -2-rrr
2 • D,r. Thus, D,r = -3.14/2-trr
2 . When the water stands
7.5 meters in the cone, w = 7.5, y = 10- 7.5 = 2.5 r=ky = l.25. So £>/=-3.14/2ir(1.25)
2 . D,y =
Fig. 14-7