DIFFERENTIAL CALCULUS: SOLVED PROBLEMS IN TIME RATES Time Rates If a quantity x is a function of time t, the time rate of change of x is given by dx/dt. When two or more quantities, all functions of t, are related by an equation, the relation between their rates of change may be obtained by differentiating both sides of the equation with respect to t. Basic Time Rates Velocity, v=dtds, where s is the distance. Acceleration, a=dtdv=dt2d2s, where v is velocity and s is the distance.
Solution 01
Volume of water: V=10(3)y=30y
Volume of water:
dtdV=30dtdy
V=r2h=(42)h=16h
12=30dtdy
dtdV=16dtdh
dtdy=04 ft/min
24=16dtdh
Problem 04
dtdh=0477 ft/min
answer
Problem 02 Water flows into a vertical cylindrical tank at 12 ft3/min, the surface rises 6 in/min. Find the radius of the tank.
answer
A triangular trough 10 ft long is 4 f t across the top, and 4 ft deep. If water flows in at the rate of 3 ft3/min, find how fast the surface is rising when the water is 6 in deep. Solution 04
Discharge, Q=dtdV, where V is the volume at any time. Angular Speed, =dtd, where is the angle at any time.
Steps in Solving Time Rates Problem Identify what are changing and what are fixed. Assign variables to those that are changing and appropriate value (constant) to those that are fixed. Create an equation relating all the variables and constants in Step 2.
Solution 02
Volume of water: V=r2h dtdV=r2dtdh
Volume of water:
12=r2(05)
V=21xy(10)=5xy
r=1205=276 ft Differentiate the equation with respect to time.
answer
By similar triangle: yx=44
Problem 03
x=y Problem 01
A rectangular trough is 10 ft long and 3 ft wide. Find how fast the surface rises, if water flows in at the rate of 12 ft3/min.
V=5y2 dtdV=10ydtdy when y = 6 in or 0.5 ft
Water is flowing into a vertical cylindrical tank at the rate of 24 ft3/min. If the radius of the tank is 4 ft, how fast is the surface rising?
3=10(05)dtdy dtdy=06 ft/min
answer
Problem 05 A triangular trough is 10 ft long, 6 f t wide across the top, and 3 ft deep. If water flows in at the rate of 12 ft3/min, find how fast the surface is rising when the water is 6 in deep.
Solution 03
1
DIFFERENTIAL CALCULUS: SOLVED PROBLEMS IN TIME RATES x2+y2=202 Solution 05
shadow move?
2xdtdx+2ydtdy=0 xdtdx+ydtdy=0 xdtdx+y(ï2)=0 xdtdxï2y=0 when x = 16 ft 162+y2=202 y=12 ft 16dtdxï2(12)=0
Solution 08
dtdx=15 ft/sec
answer
6sïx=s16
Volume of water: Problem 7
16s²16x=6s 10s=16x
By similar triangle:
In Problem 6, find the rate of change of the slope of the ladder.
yx=36
Solution 07
x=2y
From the figure in Solution 6 above
V=5(2y)y=10y2
m=xy
dtdV=20ydtdy
dtdm=x2xdtdyïydtdx
when y = 6 in or 0.5 ft
where
V=21xy(10)=5xy
10dtds=16(5) dtds=8 mi/hr
answer
Problem 09 In Problem 08, how fast does the shadow lengthen?
x = 16 ft
12=20(05)dtdy dtdy=12 ft/min
10dtds=16dtdx
answer
y = 12 ft
Problem 06
dx/dt = 1.5 ft/sec
A ladder 20 ft long leans a gainst a vertical wall. If the top slides downward at the rate of 2 ft/sec, find how fast the lower end is moving when it is 16 ft from the wall.
dy/dt = ²2 ft/sec
Solution 06
dtdm=ï25128 per second
Solution 09
dtdm=25128 per second decreasing answer
s6=16s+x
dtdm=16216(ï2)ï12(16) dtdm=256ï50
Problem 08
16s=6x+6s 10s=6x
A man 6 ft tall walks away from a lamp post 16 ft high at the rate of 5 miles per hour. How fast does the end of his
10dtds=6dtdx 10dtds=6(5) dtds=3 mi/hr
2
answer
DIFFERENTIAL CALCULUS: SOLVED PROBLEMS IN TIME RATES Problem 10 A boy on a bike rides north 5 mi, then turns east (Fig. 47). If he rides 10 mi/hr, at what rate does his distance to the starting point S changing 2 hour after he left that point? Solution 11 s2=402t2+502(t²2)2 Solution 13
s=1600t2+2500(t²2)2 dtds=3200t+5000(tï2)21600t2+2500(t²2)2 dtds=4100tï50001600t2+2500(t²2)2
Volume of water:
at 3 PM, t = 3
V=21[2+(2+2x)]y(10)
dtds=4100(3)ï50001600(32)+2500(3²2)2
V=20y+10xy
dtds=5615 mi/hr
From the figure:
answer
yx=21
Solution 10
x=21y
For 5 miles:
Problem 12
5=10t
In Problem 11, how fast the trains are separating after along time?
V=20y+5y2 dtdV=20dtdy+10ydtdy
t=05hr Solution 12
when y = 6 in or 0.5 ft
d2=52+102(t²05)2 After a long time, t =
10=20dtdy+10(05)dtdy
d=25+100(t²05)2 dtds=4100tï50001600t2+2500(t²2)2
dtdy=04 ft/min
dtdd=200(tï05)225+100(t²05)2 dtdd=100(tï05)25+100(t²05)2 when t = 2 hrs
answer
dtds=4100tï50001600t2+2500t2ï10000t+ 10000
Problem 14
dtds=4100tï50004100t2ï10000t+100001t 1t
For the trough in Problem 13, how fas t the water surface is rising when the water is 1 foot deep.
dtdd=100(2ï05)25+100(2²05)2 dtds=4100ï5000t4100ï10000t+10000t2 dtdd=150250=1505101010
Solution 14 dtds=4100ï50004100ï10000+100002
dtdd=310 mi/hr
answer
From the Solution 13 dtds=6403 mi/hr
answer
Problem 11
dtdV=20dtdy+10ydtdy Problem 13
A train starting at noon, travels north at 40 miles per hour. Another train s tarting from the same point at 2 PM travels east at 50 miles per hour. Find, to the nearest mile per hour, how fast the two trains are separating at 3 PM.
When y = 1 ft A trapezoidal trough is 10 ft long, 4 ft wide at the top, 2 ft wide at the bottom and 2 ft deep. If water flows in at 10 ft3/min, find how fast the surface is rising, when the water is 6 in deep.
3
dtdV=20dtdy+10ydtdy 10=20dtdy+10(1)dtdy dtdy=31 ft/min
answer
DIFFERENTIAL CALCULUS: SOLVED PROBLEMS IN TIME RATES Problem 15 A light at eye level stands 20 ft from a house and 15 ft from the path leading from the house to the street. A man walks along the path at 6 ft per sec. How fast does his shadow move along the wall when he is 5 ft from the house?
when x = 30 ² 15 = 15 ft dtdy=152ï180(5) Solution 16
dtdy=²4 ft/sec
By Pythagorean Theorem: x2+y2=s2
The negative sign in the answer indicates that the length of the shadow is shortening.
2xdtdx+2ydtdy=2sdtds
Problem 18
xdtdx+ydtdy=sdtds
Solve Problem 17, if the light is 10 ft above the ground.
Solution 15
answer
From Solution 15, when y = 5 ft dx/dt = 8 ft/sec and x = 15(5)/(20 - 5) = 5 f t, then s = ¥(x2 + y2) = ¥(52 + 52) = 5¥2 ft Thus,
Solution 18
5(8)+5(6)=52dtds
By similar triangle:
dtds=99 ft/sec
answer
From the figure:
Problem 17
yx=20x+15
light is placed on the ground 30 ft from a building. A man 6 ft tall walks from the light toward the building at the rate of 5 ft/sec. Find the rate at which the length of his shadow is changing when he is 15 ft from the building.
20x=xy+15y (20²y)x=15y x=15y20ïy dtdx=(20ïy)2(20ïy)15dtdyï15yïdtdy
3010ïy=x4 y=10ïx120 dtdy=ïx2ï120dtdx dtdy=x2120dtdx when x = 30 ² 15 = 15 ft dtdy=152120(5)
Solution 17
dtdy=38 ft/sec
answer
dtdx=(20ïy)215(20ïy)+15ydtdy Problem 19 dtdx=300(20ïy)2dtdy One city A, is 30 mi north and 55 mi east of another city, B. At noon, a car starts west from A at 40 mi/hr, at 12:10 PM, another car starts east from B at 60 mi/hr. Find, in two ways, when the cars will be nearest together.
when y = 5 ft dtdx=300(20ï5)2(6) dtdx=8 ft/sec
answer
By similar triangle:
Problem 16
y30=x6
In Problem 15, when the man is 5 ft from the house, find the time-rate of change of that portion of his shadow which lies on the ground.
y=x180
dtdy=x2ï180dtdx
4
DIFFERENTIAL CALCULUS: SOLVED PROBLEMS IN TIME RATES x=60t²10+40t²55 s=(65²100t)2+900
x=²(65²100t)
dtds=2(65ï100t)(ï100)2(65²100t)2+900
x2=(65²100t)2
dtds=ï100(65ï100t)(65²100t)2+900
s=x2+302
when ds/dt = 0
s=(65²100t)2+900
ï100(65ï100t)(65²100t)2+900=0
Second condition: (after 1:05 PM but before 1:22:30 PM)
ok
ï100(65ï100t)=0 100t=65 t=065 hr t=39 min time: 12:39 PM
answer
Problem 20 For the condition of Problem 19, draw the appropriate figures for times before 12:39 PM and after that time. Show that in terms of time after noon, the formula for distance between the two cars (one formula associated with each figure) are equivalent.
Solution 19 1st Solution (Specific): The figure to the right shows the position of the cars when they are nearest to each other. 40t+60t²6010=55 100t=65 t=065 hr t=39 min time: 12:39 PM
[60(t²1060)²x]+40t=55 x=60t²10+40t²55 x=²(65²100t)
Solution 20 For time before 12:39 PM, see the figure in the general solution of Solution 20.
x2=(65²100t)2 s=x2+302
x=65²100t
s=(65²100t)2+900
s=x2+302=(65ï100t)2+900
Third condition: (after 1:22:30 PM)
For time after 12:39 PM, there are three conditions that worth noting. Each are thoroughly illustrated below.
40t²[x²60(t²1060)]=55
answer
2nd Solution (General): From the figure shown in the right:
First condition: (after 12:39 PM but before 1:05 PM)
s=x2+302 s=x2+900 x=55²40t²60t+10
where: x = 55 ² 40t ² 60(t ² 10/60)
x=65²100t
x = 65 ² 100t
[60(t²1060)²x]+40t=55
5
ok
DIFFERENTIAL CALCULUS: SOLVED PROBLEMS IN TIME RATES x2=(65²100t)2
Solution 23
s=x2+302 s=(65²100t)2+900
Before 12:45 PM ok
For time before 12:45 PM, see the figure in Solution 22.
Problem 21 x=45²60t For Problem 19, compute the time-rate of change of the distance between the cars at (a) 12:15 PM; (b) 12:30 PM; (c) 1:15 PM Solution 21
s=(45²60t)2+(30+40t)2
s=x2+(30+40t)2
After 12:45 PM
where:
s=x2+(30+40t)2
x = 35 ² 60(t ² 10/60)
From Solution 20,
where:
x = 45 ² 60t
x = 60(t ² 10/60) ² 35 = ²(45 ² 60t)
s=(65²100t)2+900 at any time after noon.
s=(45²60t)2+(30+40t)2
From Solution 19:
dtds=2(45²60t)2+(30+40t)22(45² 60t)(ï60)+2(30+40t)(40)=0
dtds=ï100(65ï100t)(65²100t)2+900 (a) at 12:15 PM, t = 15/60 = 0.25 hr
x2 = (45 ² 60t)2 s=(45²60t)2+(30+40t)2
ï60(45²60t)+40(30+40t)=0
Problem 24
²3(45²60t)+2(30+40t)=0
(ok!)
dtds=ï100[65ï100(025)][65ï100(025)]2+ 900
²135+180t+60+80t=0
For Problem 22, compute the time-rate of change of the distance between the cars at (a) 12:15 PM, (b) 12:45 PM.
dtds=ï80 mi/hr
260t=75
Solution 24
(b) at 12:30 PM, t = 30/60 = 0.5 hr
t=02885 hr =17 min 18 sec
dtds=ï100[65ï100(05)][65ï100(05)]2+90 0
time: 12:17:18 PM
Solution 23 above shows that the distance s at any time after noon is given by
dtds=ï4472 mi/hr
answer
answer
(c) at 1:15 PM, t = 1 + 15/60 = 1.25 hr dtds=ï100[65ï100(125)][65ï100(125)]2+ 900 dtds=8944 mi/hr
answer
answer
Problem 23
s=(45²60t)2+(30+40t)2
For the condition of Problem 22, draw the appropriate figure for times before 12:45 PM and after that time. Show that in terms of time after noon, the formulas for distance between the two cars (one formula associated with each figure) are equivalent.
dtds=2(45²60t)2+(30+40t)22(45² 60t)(ï60)+2(30+40t)(40) 60t)(ï60)+2(30+40t)(40) See Solution 22 dtds=(45²60t)2+(30+40t)2ï60(45² 60t)+40(30+40t)
Problem 22 dtds=(45² 60t)2+(30+40t)2ï2700+3600t+1200+1600t
One city C, is 30 mil es north and 35 miles east from another city, D. At noon, a car starts north from C a t 40 miles per hour, at 12:10 PM, another car starts east from D a t 60 miles per hour. Find when the cars will be nearest together.
dtds=5200tï1500(45²60t)2+(30+40t)2 (a) at 12:15 PM, t = 15/60 = 0.25 hr dtds=5200(025)ï1500[45² 60(025)]2+[30+40(025)]2
Solution 22
dtds=²4 mi/hr
6
answer
DIFFERENTIAL CALCULUS: SOLVED PROBLEMS IN TIME RATES dtdm=ï44135 rad/sec (b) at 12:45 PM, t = 4 5/60 = 0.75 hr dtds=5200(075)ï1500[45² 60(075)]2+[30+40(075)]2 dtds=40 mi/hr
answer
Problem 25 One city E, is 20 mi les north and 20 miles east of another city, F. At noon a car starts south from E at 40 mi/hr, at 12:10 PM, another car starts east from F at 60 mi/hr. Find the rate at which the cars approach each other between 12:10 PM and 12:30 PM. What happens at 12:30 PM?
The cars may/will collide at this time. Answer Problem 26 A kite is 40 ft high with 50 ft cord out. If the kite moves horizontally at 5 miles p er hour directly away from the boy f lying it, how fast is the cord being paid out?
answer
Problem 28 At noon a car drives fro m A (Fig. 48) toward C at 60 miles per hour. Another car starting from B at the same time drives toward A at 30 miles per hour. If AB = 42 miles, find when the cars will be nearest each other.
Solution 26 s2=x2+402 2sdtds=2xdtdx sdtds=xdtdx when s = 50 ft Solution 25
502 = x2 + 402 Solution 28 x = 30 ft By cosine law:
Velocity of approach,
Thus,
v=602+402
50dtds=30(322)
v=7211 mi/hr
answer
[math]s2=(60t)2+(42²30t)2²2(60t)(42² 30t)cos60[math]
dtds=44 ft/sec
answer
At 12:30 PM
Problem 27
Distance traveled by car from E
In Problem 26, find the rate at which the slope of the cord is d ecreasing.
= 40(30/60) = 20 miles
Distance traveled by car from F = 60 [(30 - 10)/60] = 20 miles
s2=3600t2+(1764²2520t+900t2)² (2520tï1800t2) s2=6300t2²5040t+1764 s=6300t2²5040t+1764
Solution 27
dtds=12600tï504026300t2²5040t+1764=0
Slope
12600tï5040=0
m=x40
t=25 hr
dtdm=x2ï40dtdx
t=24 min
From Solution 26, x = 30 ft when s = 50 ft
time: 12:24 PM
dtdm=302ï40(322)
7
answer
DIFFERENTIAL CALCULUS: SOLVED PROBLEMS IN TIME RATES Problem 29
Solution 30
Solution 31
Solve Problem 28 if the car from B leaves at noon but the car from A leaves at 12:07 PM.
Solution 29
By Pythagorean Theorem: s=(100²40t)2+(200²40t)2 Set ds/dt = 0 By cosine law: s2=[60(t²760)]2+(42²30t)2²2[60(t² 760)](42²30t)cos60 s2=(3600t2²840t+49)+(1764² 2520t+900t2)²(²1800t2+2730t²294)
dtds=2(100²40t)2+(200²40t)22(100² 40t)(ï40)+2(200ï40t)(ï40)=0
From the isometric box:
2(100²40t)(ï40)+2(200ï40t)(ï40)=0
s=x2+302
(100²40t)+(200ï40t)=0 300²80t=0
s=x2+900
t=375 hrs
where:
s2=6300t2²6090t+2107 s=6300t2²6090t+2107 x2 = (20t)2 + (40 ² 30t)2
t=3 hrs 45 min dtds=12600tï609026300t2²6090t+2107=0 time: 3:45 PM
answer
x2 = 400t2 + 1600 ² 2400t + 900t2
12600t²6090=0 Minimum distance will occur at t = 3.75,
x2 = 1300t2 ² 2400t + 1600
t=2960 hr smin=[100²40(375)]2+[200²40(375)]2
s=(1300t2²2400t+1600)+900
smin=7071 miles
s=1300t2²2400t+2500
t=29 min time: 12:29 PM
answer
answer Problem 31
dtds=2600t²240021300t2²2400t+2500
An elevated train on a track 30 ft above the ground crosses a street at the rate of 20 ft/sec at the instant that a car, approaching at the rate of 30 ft/sec, is 40 ft up the street. Find how fast the train and the car separating 1 second later.
dtds=1300t²12001300t2²2400t+2500
Problem 30 Two railroad tracks intersect at right angles, at noon there is a train on each track approaching the crossing at 40 mi/hr, one being 100 mi, the other 200 mi distant. Find (a) when they will be nearest together, and (b) what will be their minimum distance apart.
8
after 1 sec, t = 1 dtds=1300(1)²12001300(12)² 2400(1)+2500 dtds=267 ft/sec
answer
DIFFERENTIAL CALCULUS: SOLVED PROBLEMS IN TIME RATES Problem 32
s=(5²480t+11600t2)+4
In Problem 31, find when the train and the car are nearest together.
s=9²480t+11600t2 dtds=ï480+23200t29²480t+11600t2=0
Solution 32 ²480+23200t=0 From Solution 31, t=3145 hr dtds=1300t²12001300t2²2400t+2500 t=1729 min
answer
the train and the car are nearest together if ds/dt = 0
Problem 34
1300t²12001300t2²2400t+2500=0
In Problem 33, find how fast the two will be separating after along time.
1300t²1200=0
From the figure:
Solution 34 t=1213 sec
x=square root(4t)2+302
From the Solution 33, t=0923 sec
answer
x=16t2+900
s=9²480t+11600t2 Problem 33 dtds=ï480+23200t29²480t+11600t2 From a car traveling east at 40 miles per hour, an airplane traveling horizontally north at 100 miles per hour is visible 1 mile east, 2 miles south, and 2 miles up. Find when this two will be nearest together.
dtds=ï240+11600t9²480t+11600t21t1t dtds=ït240+116009t2²t480+11600
s=41x s=4116t2+900 dtds=4132t216t2+900
after a long time, t dtds=ï240+1160092²480+11600
Solution 33
s6=x24
dtds=4t16t2+900 when 4t = 40; t = 10 sec
dtds=1160011600
dtds=4(10)16(102)+900
dtds=11600 dtds=1077 mi/hr
answer
dtds=08 ft/sec
answer
Problem 36
Problem 35 An arc light hangs at the height of 30 ft above the center of a street 60 ft wide. A man 6 ft tall walks along the sidewalk at the rate of 4 ft/sec. How fast is his shadow lengthening when he is 40 ft up the street? From the figure: Solution 35 s=x2+22=x2+4 where: x2 = (1 ² 40t)2 + (2 ² 100t)2 x2 = (1 ² 80t + 1600t2) + (4 ² 400t + 10000t2) x2 = 5 ² 480t + 11600t2
9
In Problem 35, how fast is the tip of the shadow moving? Solution 36
DIFFERENTIAL CALCULUS: SOLVED PROBLEMS IN TIME RATES dtds=100t+100100t2+200t+400
after 6 hrs from start, t = 6 ² 2 = 4 hrs dtds=100(4)+100100(42)+200(4)+400 dtds=945 mi/hr
By cosine law, s2=202+(10t)2²2(20)(10t)cos60 s=100t2ï200t+400 dtds=200tï2002100t2ï200t+400 Triangle LAB,
dtds=100tï100100t2ï200t+400
x6=30x+30
after 6 hrs from start, t = 6 ² 2 = 4 hrs
5x=x+30
dtds=100(4)ï100100(42)ï200(4)+400
x=75 ft
dtds=100(4)ï100100(42)ï200(4)+400 dtds=866 mi/hr
Triangle ABC,
Problem 38
sx+30=4t30
Solve Problem 37, if the ship turns N 30° E.
s75+30=2t15
Solution 38
s=5t dtds=5 ft/sec
answer
answer
Problem 37 A ship sails east 20 mi les and then turns N 30° W. If the ship·s speed is 10 mi/hr, find how fast it will be leaving the starting point 6 hr after the start.
Solution 37
By cosine law, s2=202+(10t)2²2(20)(10t)cos120
s=100t2+200t+400
dtds=200t+2002100t2+200t+400
10
answer