Time Rates

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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.

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.

Differentiate the equation with respect to time.

Problem 01

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?

Solution 01

Volume of water:

V=r2h=(42)h=16h

dtdV=16dtdh

24=16dtdh

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.

Solution 02

Volume of water:

V=r2h

dtdV=r2dtdh

12=r2(05)

r=1205=276 ft answer

Problem 03

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.

Solution 03

Volume of water:

V=10(3)y=30y

dtdV=30dtdy

12=30dtdy

dtdy=04 ft/min answer

Problem 04

A triangular trough 10 ft long is 4 ft 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

Volume of water:

V=21xy(10)=5xy

By similar triangle:

yx=44

x=y

V=5y2

dtdV=10ydtdy

when y = 6 in or 0.5 ft

3=10(05)dtdy


Problem 05

A triangular trough is 10 ft long, 6 ft 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.

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Solution 05

Volume of water:

V=21xy(10)=5xy


yx=36

x=2y

V=5(2y)y=10y2

dtdV=20ydtdy


12=20(05)dtdy


Problem 06

A ladder 20 ft long leans against 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.

Solution 06

x2+y2=202

2xdtdx+2ydtdy=0

xdtdx+ydtdy=0

xdtdx+y(−2)=0

xdtdx−2y=0

when x = 16 ft

162+y2=202

y=12 ft

16dtdx−2(12)=0

dtdx=15 ft/sec answer

Problem 7

In Problem 6, find the rate of change of the slope of the ladder.

Solution 07

From the figure in Solution 6 above

m=xy

dtdm=x2xdtdy−ydtdx

where

x = 16 ft

y = 12 ft

dx/dt = 1.5 ft/sec

dy/dt = –2 ft/sec

dtdm=16216(−2)−12(16)

dtdm=256−50

dtdm=−25128 per second

dtdm=25128 per second decreasing answer

Problem 08

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 shadow move?

Solution 08

6s−x=s16

16s–16x=6s

10s=16x

10dtds=16dtdx

10dtds=16(5)

dtds=8 mi/hr answer

Problem 09

In Problem 08, how fast does the shadow lengthen?

Solution 09

s6=16s+x

16s=6x+6s

10s=6x

10dtds=6dtdx

10dtds=6(5)

dtds=3 mi/hr answer

Problem 10

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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 10

For 5 miles:

5=10t

t=05hr

d2=52+102(t–05)2

d=25+100(t–05)2

dtdd=200(t−05)225+100(t–05)2

dtdd=100(t−05)25+100(t–05)2

when t = 2 hrs

dtdd=100(2−05)25+100(2–05)2

dtdd=150250=1505101010

dtdd=310 mi/hr answer

Problem 11

A train starting at noon, travels north at 40 miles per hour. Another train starting 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.

Solution 11

s2=402t2+502(t–2)2

s=1600t2+2500(t–2)2

dtds=3200t+5000(t−2)21600t2+2500(t–2)2

dtds=4100t−50001600t2+2500(t–2)2

at 3 PM, t = 3

dtds=4100(3)−50001600(32)+2500(3–2)2

dtds=5615 mi/hr answer

Problem 12

In Problem 11, how fast the trains are separating after along time?

Solution 12

After a long time, t = ∞

dtds=4100t−50001600t2+2500(t–2)2

dtds=4100t−50001600t2+2500t2−10000t+10000

dtds=4100t−50004100t2−10000t+100001t1t

dtds=4100−5000t4100−10000t+10000t2

dtds=4100−50004100−10000+100002


Problem 13

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.

Solution 13

Volume of water:

V=21[2+(2+2x)]y(10)

V=20y+10xy

From the figure:

yx=21

x=21y

V=20y+5y2

dtdV=20dtdy+10ydtdy


10=20dtdy+10(05)dtdy


Problem 14

For the trough in Problem 13, how fast the water surface is rising when the water is 1 foot deep.

Solution 14

From the Solution 13

dtdV=20dtdy+10ydtdy

When y = 1 ft

dtdV=20dtdy+10ydtdy

10=20dtdy+10(1)dtdy


Problem 15

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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?

Solution 15

From the figure:

yx=20x+15

20x=xy+15y

(20–y)x=15y

x=15y20−y

dtdx=(20−y)2(20−y)15dtdy−15y−dtdy

dtdx=(20−y)215(20−y)+15ydtdy

dtdx=300(20−y)2dtdy

when y = 5 ft

dtdx=300(20−5)2(6)

dtdx=8 ft/sec answer

Problem 16

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.

Solution 16

By Pythagorean Theorem:

x2+y2=s2

2xdtdx+2ydtdy=2sdtds

xdtdx+ydtdy=sdtds

From Solution 15, when y = 5 ft

dx/dt = 8 ft/sec and

x = 15(5)/(20 - 5) = 5 ft, then

s = √(x2 + y2) = √(52 + 52) = 5√2 ft

Thus,

5(8)+5(6)=52dtds

dtds=99 ft/sec answer

Problem 17

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.

Solution 17


y30=x6

y=x180

dtdy=x2−180dtdx

when x = 30 – 15 = 15 ft

dtdy=152−180(5)

dtdy=–4 ft/sec answer

The negative sign in the answer indicates that the length of the shadow is shortening.

Problem 18

Solve Problem 17, if the light is 10 ft above the ground.

Solution 18


3010−y=x4

y=10−x120

dtdy=−x2−120dtdx

dtdy=x2120dtdx

when x = 30 – 15 = 15 ft

dtdy=152120(5)

dtdy=38 ft/sec answer

Problem 19

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.

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Solution 19

1st Solution (Specific):

The figure to the right shows the position of the cars when they are nearest to each other.

40t+60t–6010=55

100t=65

t=065 hr

t=39 min

time: 12:39 PM answer

2nd Solution (General):

From the figure shown in the right:

s=x2+302

s=x2+900

where:

x = 55 – 40t – 60(t – 10/60)

x = 65 – 100t

s=(65–100t)2+900

dtds=2(65−100t)(−100)2(65–100t)2+900

dtds=−100(65−100t)(65–100t)2+900

when ds/dt = 0

−100(65−100t)(65–100t)2+900=0

−100(65−100t)=0

100t=65

t=065 hr

t=39 min


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 20

For time before 12:39 PM, see the figure in the general solution of Solution 20.

x=65–100t

s=x2+302=(65−100t)2+900

For time after 12:39 PM, there are three conditions that worth noting. Each are thoroughly illustrated below.

First condition: (after 12:39 PM but before 1:05 PM)

[60(t–1060)–x]+40t=55

x=60t–10+40t–55

x=–(65–100t)

x2=(65–100t)2

s=x2+302

s=(65–100t)2+900 ok

Second condition: (after 1:05 PM but before 1:22:30 PM)

[60(t–1060)–x]+40t=55

x=60t–10+40t–55

x=–(65–100t)

x2=(65–100t)2

s=x2+302

s=(65–100t)2+900 ok

Third condition: (after 1:22:30 PM)

40t–[x–60(t–1060)]=55

x=55–40t–60t+10

x=65–100t

x2=(65–100t)2

s=x2+302

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s=(65–100t)2+900 ok

Problem 21

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

From Solution 20,

s=(65–100t)2+900 at any time after noon.

From Solution 19:

dtds=−100(65−100t)(65–100t)2+900

(a) at 12:15 PM, t = 15/60 = 0.25 hr

dtds=−100[65−100(025)][65−100(025)]2+900

dtds=−80 mi/hr answer

(b) at 12:30 PM, t = 30/60 = 0.5 hr

dtds=−100[65−100(05)][65−100(05)]2+900

dtds=−4472 mi/hr answer

(c) at 1:15 PM, t = 1 + 15/60 = 1.25 hr

dtds=−100[65−100(125)][65−100(125)]2+900


Problem 22

One city C, is 30 miles north and 35 miles east from another city, D. At noon, a car starts north from C at 40 miles per hour, at 12:10 PM, another car starts east from D at 60 miles per hour. Find when the cars will be nearest together.

Solution 22

s=x2+(30+40t)2

where:

x = 35 – 60(t – 10/60)

x = 45 – 60t

s=(45–60t)2+(30+40t)2

dtds=2(45–60t)2+(30+40t)22(45–60t)(−60)+2(30+40t)(40)=0

−60(45–60t)+40(30+40t)=0

–3(45–60t)+2(30+40t)=0

–135+180t+60+80t=0

260t=75

t=02885 hr =17 min 18 sec

time: 12:17:18 PM answer

Problem 23

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.

Solution 23

Before 12:45 PM

For time before 12:45 PM, see the figure in Solution 22.

x=45–60t

s=(45–60t)2+(30+40t)2

After 12:45 PM

s=x2+(30+40t)2

where:

x = 60(t – 10/60) – 35 = –(45 – 60t)

x2 = (45 – 60t)2

s=(45–60t)2+(30+40t)2 (ok!)

Problem 24

For Problem 22, compute the time-rate of change of the distance between the cars at (a) 12:15 PM, (b) 12:45 PM.

Solution 24

Solution 23 above shows that the distance s at any time after noon is given by

s=(45–60t)2+(30+40t)2

dtds=2(45–60t)2+(30+40t)22(45–60t)(−60)+2(30+40t)(40) See Solution 22

dtds=(45–60t)2+(30+40t)2−60(45–60t)+40(30+40t)

dtds=(45–60t)2+(30+40t)2−2700+3600t+1200+1600t

dtds=5200t−1500(45–60t)2+(30+40t)2

(a) at 12:15 PM, t = 15/60 = 0.25 hr

dtds=5200(025)−1500[45–60(025)]2+[30+40(025)]2

dtds=–4 mi/hr answer

(b) at 12:45 PM, t = 45/60 = 0.75 hr

dtds=5200(075)−1500[45–60(075)]2+[30+40(075)]2


Problem 25

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One city E, is 20 miles 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?

Solution 25

Velocity of approach,

v=602+402

v=7211 mi/hr answer

At 12:30 PM

Distance traveled by car from E

= 40(30/60)

= 20 miles

Distance traveled by car from F

= 60 [(30 - 10)/60]

= 20 miles

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 per hour directly away from the boy flying it, how fast is the cord being paid out?

Solution 26

s2=x2+402

2sdtds=2xdtdx

sdtds=xdtdx

when s = 50 ft

502 = x2 + 402

x = 30 ft

Thus,

50dtds=30(322)


Problem 27

In Problem 26, find the rate at which the slope of the cord is decreasing.

Solution 27

Slope

m=x40

dtdm=x2−40dtdx

From Solution 26, x = 30 ft when s = 50 ft

dtdm=302−40(322)

dtdm=−44135 rad/sec answer

Problem 28

At noon a car drives from 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 28

By cosine law:

[math]s2=(60t)2+(42–30t)2–2(60t)(42–30t)cos60[math]

s2=3600t2+(1764–2520t+900t2)–(2520t−1800t2)

s2=6300t2–5040t+1764

s=6300t2–5040t+1764

dtds=12600t−504026300t2–5040t+1764=0

12600t−5040=0

t=25 hr

t=24 min


Problem 29

Solve Problem 28 if the car from B leaves at noon but the car from A leaves at 12:07 PM.

Solution 29

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By cosine law:

s2=[60(t–760)]2+(42–30t)2–2[60(t–760)](42–30t)cos60

s2=(3600t2–840t+49)+(1764–2520t+900t2)–(–1800t2+2730t–294)

s2=6300t2–6090t+2107

s=6300t2–6090t+2107

dtds=12600t−609026300t2–6090t+2107=0

12600t–6090=0

t=2960 hr

t=29 min


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.

Solution 30

By Pythagorean Theorem:

s=(100–40t)2+(200–40t)2

Set ds/dt = 0

dtds=2(100–40t)2+(200–40t)22(100–40t)(−40)+2(200−40t)(−40)=0

2(100–40t)(−40)+2(200−40t)(−40)=0

(100–40t)+(200−40t)=0

300–80t=0

t=375 hrs

t=3 hrs 45 min


Minimum distance will occur at t = 3.75,

smin=[100–40(375)]2+[200–40(375)]2

smin=7071 miles answer

Problem 31

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.

Solution 31

From the isometric box:

s=x2+302

s=x2+900

where:

x2 = (20t)2 + (40 – 30t)2

x2 = 400t2 + 1600 – 2400t + 900t2

x2 = 1300t2 – 2400t + 1600

s=(1300t2–2400t+1600)+900

s=1300t2–2400t+2500

dtds=2600t–240021300t2–2400t+2500

dtds=1300t–12001300t2–2400t+2500

after 1 sec, t = 1

dtds=1300(1)–12001300(12)–2400(1)+2500


Problem 32

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In Problem 31, find when the train and the car are nearest together.

Solution 32

From Solution 31,

dtds=1300t–12001300t2–2400t+2500

the train and the car are nearest together if ds/dt = 0

1300t–12001300t2–2400t+2500=0

1300t–1200=0

t=1213 sec

t=0923 sec answer

Problem 33

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.

Solution 33

From the figure:

s=x2+22=x2+4

where:

x2 = (1 – 40t)2 + (2 – 100t)2

x2 = (1 – 80t + 1600t2) + (4 – 400t + 10000t2)

x2 = 5 – 480t + 11600t2

s=(5–480t+11600t2)+4

s=9–480t+11600t2

dtds=−480+23200t29–480t+11600t2=0

–480+23200t=0

t=3145 hr

t=1729 min answer

Problem 34

In Problem 33, find how fast the two will be separating after along time.

Solution 34

From the Solution 33,

s=9–480t+11600t2

dtds=−480+23200t29–480t+11600t2

dtds=−240+11600t9–480t+11600t21t1t

dtds=−t240+116009t2–t480+11600

after a long time, t

dtds=−240+1160092–480+11600

dtds=1160011600

dtds=11600


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?

Solution 35

From the figure:

x=square root(4t)2+302

x=16t2+900

s6=x24

s=41x

s=4116t2+900

dtds=4132t216t2+900

dtds=4t16t2+900

when 4t = 40; t = 10 sec

dtds=4(10)16(102)+900


Problem 36

In Problem 35, how fast is the tip of the shadow moving?

Solution 36

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Triangle LAB,

x6=30x+30

5x=x+30

x=75 ft

Triangle ABC,

sx+30=4t30

s75+30=2t15

s=5t


Problem 37

A ship sails east 20 miles 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)2–2(20)(10t)cos60

s=100t2−200t+400

dtds=200t−2002100t2−200t+400

dtds=100t−100100t2−200t+400

after 6 hrs from start, t = 6 – 2 = 4 hrs

dtds=100(4)−100100(42)−200(4)+400

dtds=100(4)−100100(42)−200(4)+400


Problem 38

Solve Problem 37, if the ship turns N 30° E.

Solution 38

By cosine law,

s2=202+(10t)2–2(20)(10t)cos120

s=100t2+200t+400

dtds=200t+2002100t2+200t+400

dtds=100t+100100t2+200t+400

after 6 hrs from start, t = 6 – 2 = 4 hrs

dtds=100(4)+100100(42)+200(4)+400


Time Rates

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