Calculus Timerates Edited

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, , where is the distance.

Acceleration, , where is velocity and is the distance.

Discharge, , where is the volume at any time.

Angular Speed, , where is the angle at any time.

Steps in Solving Time Rates Problem

1. Identify what are changing and what are fixed.

2. Assign variables to those that are changing and appropriate value (constant) to those that are fixed.

3. Create an equation relating all the variables and constants in Step 2.

4. 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:

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:

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:

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:

By similar triangle:

when y = 6 in or 0.5 ft

answer

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.

Solution 05

Volume of water:



answer

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

when x = 16 ft

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

wherex = 16 fty = 12 ftdx/dt = 1.5 ft/secdy/dt = -2 ft/sec

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

answer

Problem 09

In Problem 08, how fast does the shadow lengthen?

Solution 09

answer

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 10

For 5 miles:

when t = 2 hrs

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

at 3 PM, t = 3

answer

Problem 12

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

Solution 12

After a long time, t = ∞

answer

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:

From the figure:


answer

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

When y = 1 ft

answer

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?

Solution 15

From the figure:

when y = 5 ft

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:

From Solution 15, when y = 5 ftdx/dt = 8 ft/sec andx = 15(5)/(20 - 5) = 5 ft, thens = √(x2 + y2) = √(52 + 52) = 5√2 ft

Thus,

answer

Problem 17

A 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


when x = 30 - 15 = 15 ft

answerThe 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


when x = 30 - 15 = 15 ft

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.

Solution 19

1st Solution (Specific):

The figure to the right shows the

position of the cars when they are nearest to each other.

time: 12:39 PM answer

2nd Solution (General):

From the figure shown in the right:

where:x = 55 - 40t - 60(t - 10/60)x = 65 - 100t

when ds/dt = 0


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.

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)

ok

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

ok

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

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,

at any time after noon.

From Solution 19:

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

answer

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

answer

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

answer

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

where:x = 35 - 60(t - 10/60)x = 45 - 60t

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

After 12:45 PM

where:x = 60(t - 10/60) - 35 = -(45 - 60t)x2 = (45 - 60t)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

See Solution 22

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

answer

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

answer

Problem 25

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,

answer

At 12:30 PMDistance 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

when s = 50 ft502 = x2 + 402

x = 30 ft

Thus,

answer

Problem 27

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

Solution 27

Slope

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

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:


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

By cosine law:


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:

Set ds/dt = 0


Minimum distance will occur at t = 3.75,

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:

where:x2 = (20t)2 + (40 - 30t)2

x2 = 400t2 + 1600 - 2400t + 900t2

x2 = 1300t2 - 2400t + 1600

after 1 sec, t = 1

answer

Problem 32

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

Solution 32

From Solution 31,

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

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:

where:x2 = (1 - 40t)2 + (2 - 100t)2

x2 = (1 - 80t + 1600t2) + (4 - 400t + 10000t2)x2 = 5 - 480t + 11600t2

answer

Problem 34

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

Solution 34

From the Solution 33,

after a long time,

answer

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:

when 4t = 40; t = 10 sec

answer

Problem 36

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

Solution 36

Triangle LAB,

Triangle ABC,

answer

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,

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

answer

Problem 38

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

Solution 38

By cosine law,

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

answer

Calculus Timerates Edited

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