problem solving rates

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Chapter 9: Radicals

9.10 Rate Word Problems: Work and Time

If it takes Felicia 4 hours to paint a room and her daughter Katy 12 hours to paint the same room, then working together, they could paint the room in 3 hours. The equation used to solve problems of this type is one of reciprocals. It is derived as follows:

[latex]\text{rate}\times \text{time}=\text{work done}[/latex]

For this problem:

[latex]\begin{array}{rrrl} \text{Felicia's rate: }&F_{\text{rate}}\times 4 \text{ h}&=&1\text{ room} \\ \\ \text{Katy's rate: }&K_{\text{rate}}\times 12 \text{ h}&=&1\text{ room} \\ \\ \text{Isolating for their rates: }&F&=&\dfrac{1}{4}\text{ h and }K = \dfrac{1}{12}\text{ h} \end{array}[/latex]

To make this into a solvable equation, find the total time [latex](T)[/latex] needed for Felicia and Katy to paint the room. This time is the sum of the rates of Felicia and Katy, or:

[latex]\begin{array}{rcrl} \text{Total time: } &T \left(\dfrac{1}{4}\text{ h}+\dfrac{1}{12}\text{ h}\right)&=&1\text{ room} \\ \\ \text{This can also be written as: }&\dfrac{1}{4}\text{ h}+\dfrac{1}{12}\text{ h}&=&\dfrac{1 \text{ room}}{T} \\ \\ \text{Solving this yields:}&0.25+0.083&=&\dfrac{1 \text{ room}}{T} \\ \\ &0.333&=&\dfrac{1 \text{ room}}{T} \\ \\ &t&=&\dfrac{1}{0.333}\text{ or }\dfrac{3\text{ h}}{\text{room}} \end{array}[/latex]

Example 9.10.1

Karl can clean a room in 3 hours. If his little sister Kyra helps, they can clean it in 2.4 hours. How long would it take Kyra to do the job alone?

The equation to solve is:

[latex]\begin{array}{rrrrl} \dfrac{1}{3}\text{ h}&+&\dfrac{1}{K}&=&\dfrac{1}{2.4}\text{ h} \\ \\ &&\dfrac{1}{K}&=&\dfrac{1}{2.4}\text{ h}-\dfrac{1}{3}\text{ h}\\ \\ &&\dfrac{1}{K}&=&0.0833\text{ or }K=12\text{ h} \end{array}[/latex]

Example 9.10.2

Doug takes twice as long as Becky to complete a project. Together they can complete the project in 10 hours. How long will it take each of them to complete the project alone?

[latex]\begin{array}{rrl} \dfrac{1}{R}+\dfrac{1}{2R}&=&\dfrac{1}{10}\text{ h,} \\ \text{where Doug's rate (} \dfrac{1}{D}\text{)}& =& \dfrac{1}{2}\times \text{ Becky's (}\dfrac{1}{R}\text{) rate.} \\ \\ \text{Sum the rates: }\dfrac{1}{R}+\dfrac{1}{2R}&=&\dfrac{2}{2R} + \dfrac{1}{2R} = \dfrac{3}{2R} \\ \\ \text{Solve for R: }\dfrac{3}{2R}&=&\dfrac{1}{10}\text{ h} \\ \text{which means }\dfrac{1}{R}&=&\dfrac{1}{10}\times\dfrac{2}{3}\text{ h} \\ \text{so }\dfrac{1}{R}& =& \dfrac{2}{30} \\ \text{ or }R &= &\dfrac{30}{2} \end{array}[/latex]

This means that the time it takes Becky to complete the project alone is [latex]15\text{ h}[/latex].

Since it takes Doug twice as long as Becky, the time for Doug is [latex]30\text{ h}[/latex].

Example 9.10.3

Joey can build a large shed in 10 days less than Cosmo can. If they built it together, it would take them 12 days. How long would it take each of them working alone?

[latex]\begin{array}{rl} \text{The equation to solve:}& \dfrac{1}{(C-10)}+\dfrac{1}{C}=\dfrac{1}{12}, \text{ where }J=C-10 \\ \\ \text{Multiply each term by the LCD:}&(C-10)(C)(12) \\ \\ \text{This leaves}&12C+12(C-10)=C(C-10) \\ \\ \text{Multiplying this out:}&12C+12C-120=C^2-10C \\ \\ \text{Which simplifies to}&C^2-34C+120=0 \\ \\ \text{Which will factor to}& (C-30)(C-4) = 0 \end{array}[/latex]

Cosmo can build the large shed in either 30 days or 4 days. Joey, therefore, can build the shed in 20 days or −6 days (rejected).

The solution is Cosmo takes 30 days to build and Joey takes 20 days.

Example 9.10.4

Clark can complete a job in one hour less than his apprentice. Together, they do the job in 1 hour and 12 minutes. How long would it take each of them working alone?

[latex]\begin{array}{rl} \text{Convert everything to hours:} & 1\text{ h }12\text{ min}=\dfrac{72}{60} \text{ h}=\dfrac{6}{5}\text{ h}\\ \\ \text{The equation to solve is} & \dfrac{1}{A}+\dfrac{1}{A-1}=\dfrac{1}{\dfrac{6}{5}}=\dfrac{5}{6}\\ \\ \text{Therefore the equation is} & \dfrac{1}{A}+\dfrac{1}{A-1}=\dfrac{5}{6} \\ \\ \begin{array}{r} \text{To remove the fractions, } \\ \text{multiply each term by the LCD} \end{array} & (A)(A-1)(6)\\ \\ \text{This leaves} & 6(A)+6(A-1)=5(A)(A-1) \\ \\ \text{Multiplying this out gives} & 6A-6+6A=5A^2-5A \\ \\ \text{Which simplifies to} & 5A^2-17A +6=0 \\ \\ \text{This will factor to} & (5A-2)(A-3)=0 \end{array}[/latex]

The apprentice can do the job in either [latex]\dfrac{2}{5}[/latex] h (reject) or 3 h. Clark takes 2 h.

Example 9.10.5

A sink can be filled by a pipe in 5 minutes, but it takes 7 minutes to drain a full sink. If both the pipe and the drain are open, how long will it take to fill the sink?

The 7 minutes to drain will be subtracted.

[latex]\begin{array}{rl} \text{The equation to solve is} & \dfrac{1}{5}-\dfrac{1}{7}=\dfrac{1}{X} \\ \\ \begin{array}{r} \text{To remove the fractions,} \\ \text{multiply each term by the LCD}\end{array} & (5)(7)(X)\\ \\ \text{This leaves } & (7)(X)-(5)(X)=(5)(7)\\ \\ \text{Multiplying this out gives} & 7X-5X=35\\ \\ \text{Which simplifies to} & 2X=35\text{ or }X=\dfrac{35}{2}\text{ or }17.5 \end{array}[/latex]

17.5 min or 17 min 30 sec is the solution

For Questions 1 to 8, write the formula defining the relation. Do Not Solve!!

Bill’s father can paint a room in 2 hours less than it would take Bill to paint it. Working together, they can complete the job in 2 hours and 24 minutes. How much time would each require working alone?
Of two inlet pipes, the smaller pipe takes four hours longer than the larger pipe to fill a pool. When both pipes are open, the pool is filled in three hours and forty-five minutes. If only the larger pipe is open, how many hours are required to fill the pool?
Jack can wash and wax the family car in one hour less than it would take Bob. The two working together can complete the job in 1.2 hours. How much time would each require if they worked alone?
If Yousef can do a piece of work alone in 6 days, and Bridgit can do it alone in 4 days, how long will it take the two to complete the job working together?
Working alone, it takes John 8 hours longer than Carlos to do a job. Working together, they can do the job in 3 hours. How long would it take each to do the job working alone?
Working alone, Maryam can do a piece of work in 3 days that Noor can do in 4 days and Elana can do in 5 days. How long will it take them to do it working together?
Raj can do a piece of work in 4 days and Rubi can do it in half the time. How long would it take them to do the work together?
A cistern can be filled by one pipe in 20 minutes and by another in 30 minutes. How long would it take both pipes together to fill the tank?

For Questions 9 to 20, find and solve the equation describing the relationship.

If an apprentice can do a piece of work in 24 days, and apprentice and instructor together can do it in 6 days, how long would it take the instructor to do the work alone?
A carpenter and his assistant can do a piece of work in 3.75 days. If the carpenter himself could do the work alone in 5 days, how long would the assistant take to do the work alone?
If Sam can do a certain job in 3 days, while it would take Fred 6 days to do the same job, how long would it take them, working together, to complete the job?
Tim can finish a certain job in 10 hours. It takes his wife JoAnn only 8 hours to do the same job. If they work together, how long will it take them to complete the job?
Two people working together can complete a job in 6 hours. If one of them works twice as fast as the other, how long would it take the slower person, working alone, to do the job?
If two people working together can do a job in 3 hours, how long would it take the faster person to do the same job if one of them is 3 times as fast as the other?
A water tank can be filled by an inlet pipe in 8 hours. It takes twice that long for the outlet pipe to empty the tank. How long would it take to fill the tank if both pipes were open?
A sink can be filled from the faucet in 5 minutes. It takes only 3 minutes to empty the sink when the drain is open. If the sink is full and both the faucet and the drain are open, how long will it take to empty the sink?
It takes 10 hours to fill a pool with the inlet pipe. It can be emptied in 15 hours with the outlet pipe. If the pool is half full to begin with, how long will it take to fill it from there if both pipes are open?
A sink is ¼ full when both the faucet and the drain are opened. The faucet alone can fill the sink in 6 minutes, while it takes 8 minutes to empty it with the drain. How long will it take to fill the remaining ¾ of the sink?
A sink has two faucets: one for hot water and one for cold water. The sink can be filled by a cold-water faucet in 3.5 minutes. If both faucets are open, the sink is filled in 2.1 minutes. How long does it take to fill the sink with just the hot-water faucet open?
A water tank is being filled by two inlet pipes. Pipe A can fill the tank in 4.5 hours, while both pipes together can fill the tank in 2 hours. How long does it take to fill the tank using only pipe B?

Answer Key 9.10

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How to Solve Rate Problems - Grade 7 Math Questions With Detailed Solutions

How to solve questions on rates in math? Grade 7 math questions are presented along with detailed Solutions and explanations included.

What are rates in math and where are they needed? The rate is a ratio of two quantities having different units. Where are they needed? Example 1: Car A travels 150 kilometers in 3 hours. Car B travels 220 kilometers in 4 hours. We assume that both car travels at constant speeds. Which of the two cars travels faster? Solution Car A travels 150 kilometers in 3 hours. In one hour it travels $ \dfrac{150 \,\, \text{kilometers}}{3 \,\, \text{hours}} = \dfrac{50 \,\, \text{km}}{1 \,\, \text{hour}} $ = 50 km / hour Car B travels 220 kilometers in 4 hours. In one hour it travels $ \dfrac{220 \,\, \text{kilometers}}{4 \,\, \text{hours}} = \dfrac{55 \,\, \text{km}}{1 \,\, \text{hour}} $ = 55 km / hour The quantities 50 km / hour and 55 km / hour are called unit rates because the denominator is one unit of time: 1 hour. In this case the unit rates can be used to find out which car travels faster because we now know how many kilometers are traveled by each car in one hour and we can therefore compare the speed (or rates) and say that car B travels faster.

Example 2: A car travels 150 kilometers in 3 hours. We assume that the car travels at a constant speed. How many hours are needed for this car to travel 250 kilometers at the same speed? Let t be the number of hours needed to travel 250 kilometers. Since the car travels at a constant rate (speed), we can write that the unit rate is the same whatever values for distance and time we use. Hence we write $ \dfrac{150 \,\, \text{km}}{3\,\,\text{hour}} = \dfrac{250 \,\, \text{km}}{\text{t}} $ , t in hours The above equation in t has the form. $ \dfrac{a}{b} = \dfrac{c}{d} $ Multiply both terms of the above by the product of the denominators $b \times d$. $ b \times d \times \dfrac{a}{b} = b \times d \times \dfrac{c}{d} $ Simplify $ \cancel{b}\times d \times\dfrac{a}{\cancel{b}} = b \times \cancel{d} \times \dfrac{c}{\cancel{d}} $ to obtain $ a \times d = b \times c $ Hence the equations $ \dfrac{a}{b} = \dfrac{c}{d} $ and $ a \times d = b \times c $ are equivalent and have the same solution. This method of changing an equation from fractions on each side to products on each side is called "cross muliply" method which we will use to solve our problems. We now go back to our equation $ \dfrac{150 \,\, \text{km}}{3\,\,\text{hour}} = \dfrac{250 \,\, \text{km}}{\text{t}} $ and use the "cross multiply" method to write it as follows. $ 150 \,\, \text{km} \times t = 250 \text{km}\times 3 \text{hours} $ Since we need to find t, we then isolate it by dividing both sides of the above equation by $ 150 \,\, \text{km} $. $ \dfrac{150 \,\, \text{km} \times t}{150 \,\, \text{km}} = \dfrac{250 \text{km}\times 3 \text{hours}}{150 \,\, \text{km}} $ Simplify. $ \dfrac{\cancel{150 \,\, \text{km}} \times t}{\cancel{150 \,\, \text{km}}} = \dfrac{250 \cancel{\text{km}}\times 3 \text{hours}}{150 \,\, \cancel{\text{km}}} $ $ t = \dfrac{250 \times 3}{150} \, \, \text{hours} = 5 \,\, \text{hours}$

The exercises below with solutions and explanations are all about solving rate problems.

Solve the following rate problems.

The distance between two cities on the map is 15 centimeters. The scales on the map is 5 centimeters to 15 kilometers. What is the real distance, in kilometers, between the two cities?
A car consumes 10 gallons of fuel to travel a distance of 220 miles. Assuming a constant rate of consumption, how many gallons are needed to travel 330 miles?
Ten tickets to a cinema theater costs $66. Wha is the cost of 22 tickets to the same cinema theater?
Cans of soda are packaged in boxes containing the same number of cans. There are 36 cans in 4 boxes. a) How many cans are there in 7 boxes? b) How many boxes are needed to package 99 cans of soda?
Joe bought 4 kilograms of apples at the cost of $15. How much would he pay for 11 kilograms of the same apples in the same shop?
It takes a pump 10 minutes to move 55 gallons of water up a hill. Using the same pump under the same condition; a) how much water is moved in 22 minutes? b) how long does it take to move 165 gallons of water?
A container with 324 liters of water, leaks 3 liters every 5 hours. How long does it take for the container to become empty?
Twenty one cans of tomato paste of the same size have a weight of 7300 grams. What is the weight of 5 cans?
An empty container is being filled with water at the rate of 5 liters every 45 seconds and leaks water at the rate of one liter every 180 seconds. What is the quantity of water in the container after one hour?

Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers Home Page

GMAT Math : Rate Problems

Study concepts, example questions & explanations for gmat math, all gmat math resources, example questions, example question #1 : rate problems.

A group of students are making posters to advertise for a bake sale. 12 large signs and 60 small signs are needed. It takes 10 minutes to paint a small sign and 30 minutes to paint a large sign. How many students will be needed to paint all of the signs in 2 hours or less?

$problem solving rates$

Example Question #226 : Problem Solving Questions

Jason is driving across the country. For the first 3 hours, he travels 60 mph. For the next 2 hours he travels 72 mph. Assuming that he has not stopped, what is his average traveling speed in miles per hour?

$problem solving rates$

In the first three hours, he travels 180 miles.

$problem solving rates$

In the next two hours, he travels 144 miles.

$problem solving rates$

for a total of 324 miles.

$problem solving rates$

Divide by the total number of hours to obtain the average traveling speed.

$problem solving rates$

Tom runs a 100m race in a certain amount of time. If John runs the same race, he takes 2 seconds longer. If John ran at 8m/s, approximately how fast did Tom run?

$problem solving rates$

Tom runs a 100m race in a certain amount of time. If John runs the same race, he takes 2 seconds longer. If John ran at 8m/s, how fast did Tom run?

$problem solving rates$

Jim and Julia, a married couple, work in the same building.

One morning, both left at 9:00, but in different cars. Jim arrived at 10:10; Julia arrived 10 minutes later. If Jim's average speed was 54 miles per hour, what was Julia's average speed (nearest whole number)?

$problem solving rates$

or, rounded, 47 miles per hour.

Example Question #229 : Problem Solving Questions

Andy and his wife Donna both work at the same building.

One morning, Andy left home at 8:00; Donna left 5 minutes later. Each arrived at their common destination at 8:50. Andy drove at an average speed of 45 miles per hour; what was Donna's average speed, to the nearest mile per hour?

$problem solving rates$

Example Question #230 : Problem Solving Questions

Kenny and Marie, a married couple, work in the same building.

One morning, both left at 9:00, but in different cars. Kenny arrived at 10:10; Marie arrived 10 minutes later. If Kenny's average speed was 6 miles per hour faster than Marie's, how far is their work place from their home (nearest whole mile)?

$problem solving rates$

Jerry took a car trip of 320 miles. The trip took a total of six hours and forty minutes; for the first four hours, his average speed was 60 miles per hour. What was his average speed for the remaining time?

$problem solving rates$

If Sally drives q miles in 3 hours, her rate is 3/q miles per hour. Plug this rate into the distance equation and solve for the time:

$\dpi{100} \small Distance = rate\times time$

A cat runs at a rate of 12 miles per hour. How far does he run in 10 minutes?

$\dpi{100} 1\ mile$

None of the other answers are correct.

We need to convert hours into minutes and multiply this by the 10 minute time interval:

$\small \frac{12\ miles}{1\ hour}x\frac{1\ hour}{60\ min}x\frac{10\ min}{1}=\frac{120\ miles}{60}=2\ miles$

Example Question #4 : Rate Problems

In order to qualify for the next heat, the race-car driver needs to average 60 miles per hour for two laps of a one mile race-track. The driver only averages 40 miles per hour on the first lap. What must be the driver's average speed for the second lap in order to average 60 miles per hour for both laps?

90 miles per hour

80 miles per hour

120 miles per hour

240 miles per hour

100 miles per hour

If the driver needs to drive two laps, each one mile long, at an average rate of 60 miles per hour. To find the average speed, we need to add the speed for each lap together then divide by the number of laps. The equation would be as follows:

$problem solving rates$

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Study Guides > Calculus Volume 1

Related rates, learning objectives.

Express changing quantities in terms of derivatives.
Find relationships among the derivatives in a given problem.
Use the chain rule to find the rate of change of one quantity that depends on the rate of change of other quantities.

We have seen that for quantities that are changing over time, the rates at which these quantities change are given by derivatives. If two related quantities are changing over time, the rates at which the quantities change are related. For example, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon are increasing. In this section, we consider several problems in which two or more related quantities are changing and we study how to determine the relationship between the rates of change of these quantities.

Setting up Related-Rates Problems

In many real-world applications, related quantities are changing with respect to time. For example, if we consider the balloon example again, we can say that the rate of change in the volume, [latex]V[/latex], is related to the rate of change in the radius, [latex]r[/latex]. In this case, we say that [latex]\frac{dV}{dt}[/latex] and [latex]\frac{dr}{dt}[/latex] are related rates because [latex]V[/latex] is related to [latex]r[/latex]. Here we study several examples of related quantities that are changing with respect to time and we look at how to calculate one rate of change given another rate of change.

Inflating a Balloon

A spherical balloon is being filled with air at the constant rate of [latex]2 \, \text{cm}^3 / \text{sec} [/latex] ((Figure)). How fast is the radius increasing when the radius is [latex]3\, \text{cm}[/latex]?

Three balloons are shown at Times 1, 2, and 3. These balloons increase in volume and radius as time increases.

The volume of a sphere of radius [latex]r[/latex] centimeters is

Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, [latex]t[/latex] seconds after beginning to fill the balloon with air, the volume of air in the balloon is

Differentiating both sides of this equation with respect to time and applying the chain rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equation

The balloon is being filled with air at the constant rate of 2 cm 3 /sec, so [latex]V^{\prime}(t)=2 \, \text{cm}^3 / \sec[/latex]. Therefore,

which implies

When the radius [latex]r=3 \, \text{cm}[/latex],

[latex]\frac{1}{72\pi} \, \text{cm/sec}[/latex], or approximately 0.0044 cm/sec

[latex]\frac{dr}{dt}=\frac{1}{2\pi r^2}[/latex]

Before looking at other examples, let’s outline the problem-solving strategy we will be using to solve related-rates problems.

Problem-Solving Strategy: Solving a Related-Rates Problem

Assign symbols to all variables involved in the problem. Draw a figure if applicable.
State, in terms of the variables, the information that is given and the rate to be determined.
Find an equation relating the variables introduced in step 1.
Using the chain rule, differentiate both sides of the equation found in step 3 with respect to the independent variable. This new equation will relate the derivatives.
Substitute all known values into the equation from step 4, then solve for the unknown rate of change.

Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, if the value for a changing quantity is substituted into an equation before both sides of the equation are differentiated, then that quantity will behave as a constant and its derivative will not appear in the new equation found in step 4. We examine this potential error in the following example.

Examples of the Process

Let’s now implement the strategy just described to solve several related-rates problems. The first example involves a plane flying overhead. The relationship we are studying is between the speed of the plane and the rate at which the distance between the plane and a person on the ground is changing.

An Airplane Flying at a Constant Elevation

An airplane is flying overhead at a constant elevation of [latex]4000[/latex] ft. A man is viewing the plane from a position [latex]3000[/latex] ft from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at the rate of [latex]600[/latex] ft/sec, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?

Step 1. Draw a picture, introducing variables to represent the different quantities involved.

A right triangle is made with a person on the ground, an airplane in the air, and a radio tower at the right angle on the ground. The hypotenuse is s, the distance on the ground between the person and the radio tower is x, and the side opposite the person (that is, the height from the ground to the airplane) is 4000 ft.

As shown, [latex]x[/latex] denotes the distance between the man and the position on the ground directly below the airplane. The variable [latex]s[/latex] denotes the distance between the man and the plane. Note that both [latex]x[/latex] and [latex]s[/latex] are functions of time. We do not introduce a variable for the height of the plane because it remains at a constant elevation of [latex]4000[/latex] ft. Since an object’s height above the ground is measured as the shortest distance between the object and the ground, the line segment of length [latex]4000[/latex] ft is perpendicular to the line segment of length [latex]x[/latex] feet, creating a right triangle.

Step 2. Since [latex]x[/latex] denotes the horizontal distance between the man and the point on the ground below the plane, [latex]dx/dt[/latex] represents the speed of the plane. We are told the speed of the plane is 600 ft/sec. Therefore, [latex]\frac{dx}{dt}=600[/latex] ft/sec. Since we are asked to find the rate of change in the distance between the man and the plane when the plane is directly above the radio tower, we need to find [latex]ds/dt[/latex] when [latex]x=3000[/latex] ft.

Step 3. From the figure, we can use the Pythagorean theorem to write an equation relating [latex]x[/latex] and [latex]s[/latex]:

Step 4. Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

Step 5. Find the rate at which the distance between the man and the plane is increasing when the plane is directly over the radio tower. That is, find [latex]\frac{ds}{dt}[/latex] when [latex]x=3000[/latex] ft. Since the speed of the plane is [latex]600[/latex] ft/sec, we know that [latex]\frac{dx}{dt}=600[/latex] ft/sec. We are not given an explicit value for [latex]s[/latex]; however, since we are trying to find [latex]\frac{ds}{dt}[/latex] when [latex]x=3000[/latex] ft, we can use the Pythagorean theorem to determine the distance [latex]s[/latex] when [latex]x=3000[/latex] and the height is [latex]4000[/latex] ft. Solving the equation

for [latex]s[/latex], we have [latex]s=5000[/latex] ft at the time of interest. Using these values, we conclude that [latex]ds/dt[/latex] is a solution of the equation

Note : When solving related-rates problems, it is important not to substitute values for the variables too soon. For example, in step 3, we related the variable quantities [latex]x(t)[/latex] and [latex]s(t)[/latex] by the equation

Since the plane remains at a constant height, it is not necessary to introduce a variable for the height, and we are allowed to use the constant 4000 to denote that quantity. However, the other two quantities are changing. If we mistakenly substituted [latex]x(t)=3000[/latex] into the equation before differentiating, our equation would have been

After differentiating, our equation would become

As a result, we would incorrectly conclude that [latex]\frac{ds}{dt}=0[/latex].

What is the speed of the plane if the distance between the person and the plane is increasing at the rate of [latex]300[/latex] ft/sec?

[latex]500[/latex] ft/sec

[latex]\frac{ds}{dt}=300[/latex] ft/sec

We now return to the problem involving the rocket launch from the beginning of the chapter.

Chapter Opener: A Rocket Launch

Step 1. Draw a picture introducing the variables.

A right triangle is formed with a camera at one of the nonright angles and a rocket at the other nonright angle. The angle with the camera has measure θ. The distance from the rocket to the ground is h; note that this is the side opposite the angle with measure θ. The side adjacent to the angle with measure θ is 5000 ft.

Let [latex]h[/latex] denote the height of the rocket above the launch pad and [latex]\theta[/latex] be the angle between the camera lens and the ground.

Step 2. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is [latex]1000[/latex] ft off the ground. That is, we need to find [latex]\frac{d\theta}{dt}[/latex] when [latex]h=1000[/latex] ft. At that time, we know the velocity of the rocket is [latex]\frac{dh}{dt}=600[/latex] ft/sec.

Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: [latex]h[/latex] and [latex]\theta[/latex]. How can we create such an equation? Using the fact that we have drawn a right triangle, it is natural to think about trigonometric functions. Recall that [latex]\tan \theta[/latex] is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have

This gives us the equation

Step 4. Differentiating this equation with respect to time [latex]t[/latex], we obtain

Step 5. We want to find [latex]\frac{d\theta}{dt}[/latex] when [latex]h=1000[/latex] ft. At this time, we know that [latex]\frac{dh}{dt}=600[/latex] ft/sec. We need to determine [latex]\sec^2 \theta[/latex]. Recall that [latex]\sec \theta[/latex] is the ratio of the length of the hypotenuse to the length of the adjacent side. We know the length of the adjacent side is [latex]5000[/latex] ft. To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is [latex]5000[/latex] ft, the length of the other leg is [latex]h=1000[/latex] ft, and the length of the hypotenuse is [latex]c[/latex] feet as shown in the following figure.

A right triangle has one angle with measure θ. The hypotenuse is c, the side length opposite the angle with measure θ is 1000, and the side adjacent to the angle with measure θ is 5000.

We see that

and we conclude that the hypotenuse is

Therefore, when [latex]h=1000[/latex], we have

Recall from step 4 that the equation relating [latex]\frac{d\theta}{dt}[/latex] to our known values is

When [latex]h=1000[/latex] ft, we know that [latex]\frac{dh}{dt}=600[/latex] ft/sec and [latex]\sec^2 \theta =\frac{26}{25}[/latex]. Substituting these values into the previous equation, we arrive at the equation

Therefore, [latex]\frac{d\theta}{dt}=\frac{3}{26}[/latex] rad/sec.

What rate of change is necessary for the elevation angle of the camera if the camera is placed on the ground at a distance of [latex]4000[/latex] ft from the launch pad and the velocity of the rocket is [latex]500[/latex] ft/sec when the rocket is [latex]2000[/latex] ft off the ground?

[latex]\frac{1}{10}[/latex] rad/sec

Find [latex]\frac{d\theta}{dt}[/latex] when [latex]h=2000[/latex] ft. At that time, [latex]\frac{dh}{dt}=500[/latex] ft/sec.

In the next example, we consider water draining from a cone-shaped funnel. We compare the rate at which the level of water in the cone is decreasing with the rate at which the volume of water is decreasing.

Water Draining from a Funnel

Water is draining from the bottom of a cone-shaped funnel at the rate of [latex]0.03 \, \text{ft}^3 /\text{sec}[/latex]. The height of the funnel is [latex]2[/latex] ft and the radius at the top of the funnel is [latex]1[/latex] ft. At what rate is the height of the water in the funnel changing when the height of the water is [latex]\frac{1}{2}[/latex] ft?

Step 1: Draw a picture introducing the variables.

A funnel is shown with height 2 and radius 1 at its top. The funnel has water to height h, at which point the radius is r.

Let [latex]h[/latex] denote the height of the water in the funnel, [latex]r[/latex] denote the radius of the water at its surface, and [latex]V[/latex] denote the volume of the water.

Step 2: We need to determine [latex]\frac{dh}{dt}[/latex] when [latex]h=\frac{1}{2}[/latex] ft. We know that [latex]\frac{dV}{dt}=-0.03 \text{ft}^3 / \text{sec}[/latex].

Step 3: The volume of water in the cone is

From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, [latex]\frac{r}{h}=\frac{1}{2}[/latex] or [latex]r=\frac{h}{2}[/latex]. Using this fact, the equation for volume can be simplified to

Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time [latex]t[/latex], we obtain

Step 5: We want to find [latex]\frac{dh}{dt}[/latex] when [latex]h=\frac{1}{2}[/latex] ft. Since water is leaving at the rate of [latex]0.03 \, \text{ft}^3 / \text{sec}[/latex], we know that [latex]\frac{dV}{dt}=-0.03 \, \text{ft}^3 / \text{sec}[/latex]. Therefore,

It follows that

At what rate is the height of the water changing when the height of the water is [latex]\frac{1}{4}[/latex] ft?

[latex]-0.61[/latex] ft/sec

We need to find [latex]\frac{dh}{dt}[/latex] when [latex]h=\frac{1}{4}[/latex].

Key Concepts

To solve a related rates problem, first draw a picture that illustrates the relationship between the two or more related quantities that are changing with respect to time.
In terms of the quantities, state the information given and the rate to be found.
Find an equation relating the quantities.
Use differentiation, applying the chain rule as necessary, to find an equation that relates the rates.
Be sure not to substitute a variable quantity for one of the variables until after finding an equation relating the rates.

For the following exercises, find the quantities for the given equation.

1. Find [latex]\frac{dy}{dt}[/latex] at [latex]x=1[/latex] and [latex]y=x^2+3[/latex] if [latex]\frac{dx}{dt}=4[/latex].

2. Find [latex]\frac{dx}{dt}[/latex] at [latex]x=-2[/latex] and [latex]y=2x^2+1[/latex] if [latex]\frac{dy}{dt}=-1[/latex].

3. Find [latex]\frac{dz}{dt}[/latex] at [latex](x,y)=(1,3)[/latex] and [latex]z^2=x^2+y^2[/latex] if [latex]\frac{dx}{dt}=4[/latex] and [latex]\frac{dy}{dt}=3[/latex].

[latex]\frac{13}{\sqrt{10}}[/latex]

For the following exercises, sketch the situation if necessary and used related rates to solve for the quantities.

4. [T] If two electrical resistors are connected in parallel, the total resistance (measured in ohms, denoted by the Greek capital letter omega, [latex]\Omega[/latex]) is given by the equation [latex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/latex]. If [latex]R_1[/latex] is increasing at a rate of [latex]0.5 \Omega / \text{min}[/latex] and [latex]R_2[/latex] decreases at a rate of [latex]1.1 \Omega / \text{min}[/latex], at what rate does the total resistance change when [latex]R_1=20 \Omega[/latex] and [latex]R_2=50 \Omega[/latex]?

5. A 10 ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall?

A right triangle is formed by a ladder leaning up against a brick wall. The ladder forms the hypotenuse and is 10 ft long.

[latex]2\sqrt{3}[/latex] ft/sec

6. A 25 ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially [latex]20[/latex] ft away from the wall, how fast does the ladder move up the wall [latex]5[/latex] sec after we start pushing?

7. Two airplanes are flying in the air at the same height: airplane [latex]A[/latex] is flying east at 250 mi/h and airplane [latex]B[/latex] is flying north at [latex]300[/latex] mi/h. If they are both heading to the same airport, located 30 miles east of airplane [latex]A[/latex] and 40 miles north of airplane [latex]B[/latex], at what rate is the distance between the airplanes changing?

A right triangle is formed by two airplanes A and B moving perpendicularly to each other. The hypotenuse is the distance between planes A and B. The other sides are extensions of each plane’s path until they meet.

The distance is decreasing at [latex]390[/latex] mi/h.

8. You and a friend are riding your bikes to a restaurant that you think is east; your friend thinks the restaurant is north. You both leave from the same point, with you riding at 16 mph east and your friend riding [latex]12[/latex] mph north. After you traveled [latex]4[/latex] mi, at what rate is the distance between you changing?

9. Two buses are driving along parallel freeways that are [latex]5[/latex] mi apart, one heading east and the other heading west. Assuming that each bus drives a constant [latex]55[/latex] mph, find the rate at which the distance between the buses is changing when they are [latex]13[/latex] mi apart, heading toward each other.

The distance between them shrinks at a rate of [latex]\frac{1320}{13}\approx 101.5[/latex] mph.

10. A 6-ft-tall person walks away from a 10 ft lamppost at a constant rate of [latex]3[/latex] ft/sec. What is the rate that the tip of the shadow moves away from the pole when the person is [latex]10[/latex] ft away from the pole?

A lamppost is shown that is 10 ft high. To its right, there is a person who is 6 ft tall. There is a line from the top of the lamppost that touches the top of the person’s head and then continues to the ground. The length from the end of this line to where the lamppost touches the ground is 10 + x. The distance from the lamppost to the person on the ground is 10, and the distance from the person to the end of the line is x.

11. Using the previous problem, what is the rate at which the tip of the shadow moves away from the person when the person is 10 ft from the pole?

[latex]\frac{9}{2}[/latex] ft/sec

12. A 5-ft-tall person walks toward a wall at a rate of 2 ft/sec. A spotlight is located on the ground 40 ft from the wall. How fast does the height of the person’s shadow on the wall change when the person is 10 ft from the wall?

13. Using the previous problem, what is the rate at which the shadow changes when the person is 10 ft from the wall, if the person is walking away from the wall at a rate of 2 ft/sec?

It grows at a rate [latex]\frac{4}{9}[/latex] ft/sec

14. A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/sec. You are running on the ground starting directly under the helicopter at a rate of 10 ft/sec. Find the rate of change of the distance between the helicopter and yourself after 5 sec.

15. Using the previous problem, what is the rate at which the distance between you and the helicopter is changing when the helicopter has risen to a height of 60 ft in the air, assuming that, initially, it was 30 ft above you?

The distance is increasing at [latex]\frac{135\sqrt{26}}{26}[/latex] ft/sec

For the following exercises, draw and label diagrams to help solve the related-rates problems.

16. The side of a cube increases at a rate of [latex]\frac{1}{2}[/latex] m/sec. Find the rate at which the volume of the cube increases when the side of the cube is 4 m.

17. The volume of a cube decreases at a rate of 10 m/sec. Find the rate at which the side of the cube changes when the side of the cube is 2 m.

[latex]-\frac{5}{6}[/latex] m/sec

18. The radius of a circle increases at a rate of 2 m/sec. Find the rate at which the area of the circle increases when the radius is 5 m.

19. The radius of a sphere decreases at a rate of 3 m/sec. Find the rate at which the surface area decreases when the radius is 10 m.

[latex]240\pi \, \text{m}^2[/latex]/sec

20. The radius of a sphere increases at a rate of 1 m/sec. Find the rate at which the volume increases when the radius is 20 m.

21. The radius of a sphere is increasing at a rate of 9 cm/sec. Find the radius of the sphere when the volume and the radius of the sphere are increasing at the same numerical rate.

[latex]\frac{1}{2\sqrt{\pi}}[/latex] cm

22. The base of a triangle is shrinking at a rate of 1 cm/min and the height of the triangle is increasing at a rate of 5 cm/min. Find the rate at which the area of the triangle changes when the height is 22 cm and the base is 10 cm.

23. A triangle has two constant sides of length 3 ft and 5 ft. The angle between these two sides is increasing at a rate of 0.1 rad/sec. Find the rate at which the area of the triangle is changing when the angle between the two sides is [latex]\pi /6[/latex].

The area is increasing at a rate [latex]\frac{(3\sqrt{3})}{8} \, \text{ft}^{2} / \text{sec}[/latex].

24. A triangle has a height that is increasing at a rate of 2 cm/sec and its area is increasing at a rate of 4 [latex]\text{cm}^2 / \text{sec}[/latex]. Find the rate at which the base of the triangle is changing when the height of the triangle is 4 cm and the area is 20 [latex]\text{cm}^2[/latex].

For the following exercises, consider a right cone that is leaking water. The dimensions of the conical tank are a height of 16 ft and a radius of 5 ft.

25. How fast does the depth of the water change when the water is 10 ft high if the cone leaks water at a rate of 10 [latex]\text{ft}^3[/latex]/min?

The depth of the water decreases at [latex]\frac{128}{125\pi}[/latex] ft/min.

26. Find the rate at which the surface area of the water changes when the water is 10 ft high if the cone leaks water at a rate of 10 [latex]\text{ft}^3[/latex]/min.

27. If the water level is decreasing at a rate of 3 in./min when the depth of the water is 8 ft, determine the rate at which water is leaking out of the cone.

The volume is decreasing at a rate of [latex]\frac{(25\pi )}{16}{\text{ft}}^{3}\text{/min}.[/latex]

28. A vertical cylinder is leaking water at a rate of 1 [latex]\text{ft}^3[/latex]/sec. If the cylinder has a height of 10 ft and a radius of 1 ft, at what rate is the height of the water changing when the height is 6 ft?

29. A cylinder is leaking water but you are unable to determine at what rate. The cylinder has a height of 2 m and a radius of 2 m. Find the rate at which the water is leaking out of the cylinder if the rate at which the height is decreasing is 10 cm/min when the height is 1 m.

The water flows out at rate [latex]\frac{2\pi}{5} \, \text{m}^3[/latex]/min.

30. A trough has ends shaped like isosceles triangles, with width 3 m and height 4 m, and the trough is 10 m long. Water is being pumped into the trough at a rate of [latex]5 \, \text{m}^3[/latex]/min. At what rate does the height of the water change when the water is 1 m deep?

A trough is shown with ends shaped like isosceles triangles. These triangles have width 3 and height 4. The trough is made up of rectangles that are of length 10. There is some water in the trough.

31. A tank is shaped like an upside-down square pyramid, with base of 4 m by 4 m and a height of 12 m (see the following figure). How fast does the height increase when the water is 2 m deep if water is being pumped in at a rate of [latex]\frac{2}{3} \, \text{m}^3[/latex]/sec?

An upside-down square pyramid is shown with square side lengths 4 and height 12. There is an unspecified amount of water inside the shape.

[latex]\frac{3}{2}[/latex] m/sec

For the following problems, consider a pool shaped like the bottom half of a sphere, that is being filled at a rate of 25 [latex]\text{ft}^3[/latex]/min. The radius of the pool is 10 ft.

32. Find the rate at which the depth of the water is changing when the water has a depth of 5 ft.

33. Find the rate at which the depth of the water is changing when the water has a depth of 1 ft.

[latex]\frac{25}{19\pi}[/latex] ft/min

34. If the height is increasing at a rate of 1 in/sec when the depth of the water is 2 ft, find the rate at which water is being pumped in.

35. Gravel is being unloaded from a truck and falls into a pile shaped like a cone at a rate of 10 [latex]\text{ft}^3[/latex]/min. The radius of the cone base is three times the height of the cone. Find the rate at which the height of the gravel changes when the pile has a height of 5 ft.

[latex]\frac{2}{45\pi}[/latex] ft/min

36. Using a similar setup from the preceding problem, find the rate at which the gravel is being unloaded if the pile is 5 ft high and the height is increasing at a rate of 4 in/min.

For the following exercises, draw the situations and solve the related-rate problems.

37. You are stationary on the ground and are watching a bird fly horizontally at a rate of 10 m/sec. The bird is located 40 m above your head. How fast does the angle of elevation change when the horizontal distance between you and the bird is 9 m?

The angle decreases at [latex]\frac{400}{1681}[/latex] rad/sec.

38. You stand 40 ft from a bottle rocket on the ground and watch as it takes off vertically into the air at a rate of 20 ft/sec. Find the rate at which the angle of elevation changes when the rocket is 30 ft in the air.

39. A lighthouse, [latex]L[/latex], is on an island 4 mi away from the closest point, [latex]P[/latex], on the beach (see the following image). If the lighthouse light rotates clockwise at a constant rate of 10 revolutions/min, how fast does the beam of light move across the beach 2 mi away from the closest point on the beach?

A right triangle is formed by a lighthouse L, a point P on the shore that is perpendicular to the line from the lighthouse to the shore, and a point 2 miles to the right of the point P. The distance from P to L is 4 miles.

[latex]100\pi[/latex] mi/min

40. Using the same setup as the previous problem, determine at what rate the beam of light moves across the beach 1 mi away from the closest point on the beach.

41. You are walking to a bus stop at a right-angle corner. You move north at a rate of 2 m/sec and are 20 m south of the intersection. The bus travels west at a rate of 10 m/sec away from the intersection – you have missed the bus! What is the rate at which the angle between you and the bus is changing when you are 20 m south of the intersection and the bus is 10 m west of the intersection?

The angle is changing at a rate of [latex]\frac{21}{25}[/latex] rad/sec.

For the following exercises, refer to the figure of baseball diamond, which has sides of 90 ft.

A baseball field is shown, with the bases labeled Home, 1st, 2nd, and 3rd making a square with side lengths 90 ft.

42. [T] A batter hits a ball toward third base at 75 ft/sec and runs toward first base at a rate of 24 ft/sec. At what rate does the distance between the ball and the batter change when 2 sec have passed?

43. [T] A batter hits a ball toward second base at 80 ft/sec and runs toward first base at a rate of 30 ft/sec. At what rate does the distance between the ball and the batter change when the runner has covered one-third of the distance to first base? ( Hint : Recall the law of cosines.)

The distance is increasing at a rate of 62.50 ft/sec.

44. [T] A batter hits the ball and runs toward first base at a speed of 22 ft/sec. At what rate does the distance between the runner and second base change when the runner has run 30 ft?

45. [T] Runners start at first and second base. When the baseball is hit, the runner at first base runs at a speed of 18 ft/sec toward second base and the runner at second base runs at a speed of 20 ft/sec toward third base. How fast is the distance between runners changing 1 sec after the ball is hit?

The distance is decreasing at a rate of 11.99 ft/sec.

46. [T] A building that is 225 feet tall casts a shadow of various lengths [latex]x[/latex] as the day goes by. An angle of elevation [latex]\theta[/latex] is formed by lines from the top and bottom of the building to the tip of the shadow, as seen in the following figure. Find the rate of change of the angle of elevation [latex]\frac{d\theta}{dx}[/latex] when [latex]x=272[/latex] feet.

A building is shown with height 225 ft. A triangle is made with the building height as the opposite side from the angle θ. The adjacent side has length x.

47. [T] A pole stands 75 feet tall. An angle [latex]\theta[/latex] is formed when wires of various lengths of [latex]x[/latex] feet are attached from the ground to the top of the pole, as shown in the following figure. Find the rate of change of the angle [latex]\frac{d\theta}{dx}[/latex] when a wire of length 90 feet is attached.

A flagpole is shown with height 75 ft. A triangle is made with the flagpole height as the opposite side from the angle θ. The hypotenuse has length x.

Answer: -0.0168 radians per foot

48. [T] A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by [latex]\theta = \tan^{-1}(\frac{x}{2000})[/latex], where [latex]x[/latex] is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5000 feet apart.

A rocket is shown with in the air with the distance from its nose to the ground being x. A triangle is made with the rocket height as the opposite side from the angle θ. The adjacent side has length 2000.

49. [T] A local movie theater with a 30-foot-high screen that is 10 feet above a person’s eye level when seated has a viewing angle [latex]\theta[/latex] (in radians) given by [latex]\theta = \cot^{-1}(\frac{x}{40})- \cot^{-1}(\frac{x}{10})[/latex],

where [latex]x[/latex] is the distance in feet away from the movie screen that the person is sitting, as shown in the following figure.

A person is shown with a right triangle coming from their eye (the right angle being on the opposite side from the eye), with height 10 and base x. There is a line drawn from the eye to the top of the screen, which makes an angle θ with the triangle’s hypotenuse. The screen has a height of 30.

Find [latex]\frac{d\theta}{dx}[/latex].
Evaluate [latex]\frac{d\theta}{dx}[/latex] for [latex]x=5,10,15[/latex], and 20.
Interpret the results in b.
Evaluate [latex]\frac{d\theta}{dx}[/latex] for [latex]x=25,30,35[/latex], and 40
Interpret the results in d. At what distance [latex]x[/latex] should the person sit to maximize his or her viewing angle?

Answer: a. [latex]\frac{d\theta}{dx}=\frac{10}{100+x^2}-\frac{40}{1600+x^2}[/latex] b. [latex]\frac{18}{325}, \, \frac{9}{340}, \, \frac{42}{4745}, \, 0[/latex] c. As a person moves farther away from the screen, the viewing angle is increasing, which implies that as he or she moves farther away, his or her screen vision is widening. d. [latex]-\frac{54}{12905}, \, -\frac{3}{500}, \, -\frac{198}{29945}, \, -\frac{9}{1360}[/latex] e. As the person moves beyond 20 feet from the screen, the viewing angle is decreasing. The optimal distance the person should sit for maximizing the viewing angle is 20 feet.

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Module 4: Applications of Derivatives

Related-rates problem-solving, learning outcomes.

Express changing quantities in terms of derivatives.
Find relationships among the derivatives in a given problem.
Use the chain rule to find the rate of change of one quantity that depends on the rate of change of other quantities.

Setting up Related-Rates Problems

Example: Inflating a Balloon

A spherical balloon is being filled with air at the constant rate of [latex]2 \, \frac{\text{cm}^3}{\text{sec}}[/latex] (Figure 1). How fast is the radius increasing when the radius is [latex]3\, \text{cm}[/latex]?

Figure 1. As the balloon is being filled with air, both the radius and the volume are increasing with respect to time.

The volume of a sphere of radius [latex]r[/latex] centimeters is

The balloon is being filled with air at the constant rate of 2 cm 3 /sec, so [latex]V^{\prime}(t)=2 \, \text{cm}^3 / \sec[/latex]. Therefore,

which implies

When the radius [latex]r=3 \, \text{cm}[/latex],

Watch the following video to see the worked solution to Example: Inflating a Balloon.

What is the instantaneous rate of change of the radius when [latex]r=6 \, \text{cm}[/latex]?

[latex]\frac{dr}{dt}=\dfrac{1}{2\pi r^2}[/latex]

[latex]\dfrac{1}{72\pi} \, \text{cm/sec}[/latex], or approximately 0.0044 cm/sec

Before looking at other examples, let’s outline the problem-solving strategy we will be using to solve related-rates problems.

Problem-Solving Strategy: Solving a Related-Rates Problem

Assign symbols to all variables involved in the problem. Draw a figure if applicable.
State, in terms of the variables, the information that is given and the rate to be determined.
Find an equation relating the variables introduced in step 1.
Using the chain rule, differentiate both sides of the equation found in step 3 with respect to the independent variable. This new equation will relate the derivatives.
Substitute all known values into the equation from step 4, then solve for the unknown rate of change.

We are able to solve related-rates problems using a similar approach to implicit differentiation. In the example below, we are required to take derivatives of different variables with respect to time [latex]{t}[/latex], ie. [latex]{s}[/latex] and [latex]{x}[/latex]. When this happens, we can attach a [latex]\frac{ds}{dt}[/latex] or a [latex]\frac{dx}{dt}[/latex] to the derivative, just as we did in implicit differentiation.

Examples of the Process

Example: An Airplane Flying at a Constant Elevation

Step 1. Draw a picture, introducing variables to represent the different quantities involved.

Figure 2. An airplane is flying at a constant height of 4000 ft. The distance between the person and the airplane and the person and the place on the ground directly below the airplane are changing. We denote those quantities with the variables [latex]s[/latex] and [latex]x[/latex], respectively.

Step 3. From Figure 2, we can use the Pythagorean theorem to write an equation relating [latex]x[/latex] and [latex]s[/latex]:

[latex][x(t)]^2+4000^2=[s(t)]^2[/latex].

Step 4. Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

[latex]x\frac{dx}{dt}=s\frac{ds}{dt}[/latex].

for [latex]s[/latex], we have [latex]s=5000[/latex] ft at the time of interest. Using these values, we conclude that [latex]ds/dt[/latex] is a solution of the equation

[latex]\frac{ds}{dt}=\frac{3000 \cdot 600}{5000}=360[/latex] ft/sec.

After differentiating, our equation would become

As a result, we would incorrectly conclude that [latex]\frac{ds}{dt}=0[/latex].

Watch the following video to see the worked solution to Example: An Airplane Flying at a Constant Elevation.

What is the speed of the plane if the distance between the person and the plane is increasing at the rate of [latex]300[/latex] ft/sec?

[latex]\frac{ds}{dt}=300[/latex] ft/sec

[latex]500[/latex] ft/sec

Find [latex]\frac{d\theta}{dt}[/latex] when [latex]h=2000[/latex] ft. At that time, [latex]\frac{dh}{dt}=500[/latex] ft/sec.

[latex]\frac{1}{10}[/latex] rad/sec

Example: Water Draining from a Funnel

Step 1: Draw a picture introducing the variables.

Figure 3. Water is draining from a funnel of height 2 ft and radius 1 ft. The height of the water and the radius of water are changing over time. We denote these quantities with the variables [latex]h[/latex] and [latex]r,[/latex] respectively.

Let [latex]h[/latex] denote the height of the water in the funnel, [latex]r[/latex] denote the radius of the water at its surface, and [latex]V[/latex] denote the volume of the water.

Step 2: We need to determine [latex]\frac{dh}{dt}[/latex] when [latex]h=\frac{1}{2}[/latex] ft. We know that [latex]\frac{dV}{dt}=-0.03 \text{ft}^3 / \text{sec}[/latex].

Step 3: The volume of water in the cone is

From Figure 3, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, [latex]\frac{r}{h}=\frac{1}{2}[/latex] or [latex]r=\frac{h}{2}[/latex]. Using this fact, the equation for volume can be simplified to

[latex]V=\frac{1}{3}\pi (\frac{h}{2})^2 h=\frac{\pi}{12}h^3[/latex]

Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time [latex]t[/latex], we obtain

[latex]\frac{dV}{dt}=\frac{\pi}{4}h^2 \frac{dh}{dt}[/latex]

It follows that

At what rate is the height of the water changing when the height of the water is [latex]\frac{1}{4}[/latex] ft?

[latex]-0.61[/latex] ft/sec

We need to find [latex]\frac{dh}{dt}[/latex] when [latex]h=\frac{1}{4}[/latex].

4.1 Related Rates. Authored by : Ryan Melton. License : CC BY: Attribution
Calculus Volume 1. Authored by : Gilbert Strang, Edwin (Jed) Herman. Provided by : OpenStax. Located at : https://openstax.org/details/books/calculus-volume-1 . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike . License Terms : Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction

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3.9: Related Rates

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This page is a draft and is under active development.

Gilbert Strang & Edwin “Jed” Herman

Learning Objectives

Express changing quantities in terms of derivatives.
Find relationships among the derivatives in a given problem.
Use the Chain Rule to find the rate of change of one quantity that depends on the rate of change of other quantities.
Apply the modified version of Polya's Problem Solving Strategy to solve related rate problems.

Setting up Related-Rates Problems

In many real-world applications, related quantities are changing with respect to time. For example, if we consider the balloon example again, we can say that the rate of change in the volume, $V$, is related to the rate of change in the radius, $r$. In this case, we say that $\frac{dV}{dt}$ and $\frac{dr}{dt}$ are related rates because $V$ is related to $r$. Here we study several examples of related quantities that are changing with respect to time and we look at how to calculate one rate of change given another rate of change.

Example $\PageIndex{1}$: Inflating a Balloon

A spherical balloon is being filled with air at the constant rate of $2\,\text{cm}^3\text{/sec}$ (Figure $\PageIndex{1}$). How fast is the radius increasing when the radius is $3$ cm?

$Three balloons are shown at Times 1, 2, and 3. These balloons increase in volume and radius as time increases.$

The volume of a sphere of radius $r$ centimeters is

\[V=\frac{4}{3} \pi r^3\,\text{cm}^3.\nonumber\]

Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, $t$ seconds after beginning to fill the balloon with air, the volume of air in the balloon is

\[V(t)=\frac{4}{3} \pi \big[r(t)\big]^3\text{cm}^3.\nonumber\]

Differentiating both sides of this equation with respect to time and applying the Chain Rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equation

\[V^{\prime}(t)=4 \pi \big[r(t)\big]^2 r^{\prime}(t).\nonumber\]

The balloon is being filled with air at the constant rate of $2 \,\text{cm}^3\text{/sec}$, so $V^{\prime}(t)=2\,\text{cm}^3\text{/sec}$. Therefore,

\[2\,\text{cm}^3\text{/sec}=\Big(4 \pi \big[r(t)\big]^2\;\text{cm}^2\Big) \cdot \Big(r^{\prime}(t)\;\text{cm/s}\Big),\nonumber \]

which implies

\[r^{\prime}(t)=\dfrac{1}{2 \pi \big[r(t)\big]^2}\;\text{cm/sec}.\nonumber\]

When the radius $r=3$ cm,

\[r^{\prime}(t)=\dfrac{1}{18 \pi }\;\text{cm/sec}.\nonumber\]

Exercise $\PageIndex{1}$

What is the instantaneous rate of change of the radius when $r=6$ cm?

$\frac{dr}{dt}=\frac{1}{2 \pi r^2}$

$\frac{1}{72 \pi }$ cm/sec, or approximately 0.0044 cm/sec

While memorizing procedures is the absolute worst way to enjoy mathematics, applications (also known as "word problems") tend to confound students so much that it is beneficial to develop a procedure for solving. Let’s outline a modified problem-solving strategy developed by Polya.

In 1945 George Polya published the book How To Solve It which quickly became his most prized publication. It sold over one million copies and has been translated into 17 languages. In this book he identifies basic principles of problem solving. I have modified these problem-solving strategies to best fit calculus (we will see these again in a section titled, Optimization Problems).

Polya's Modified Problem-Solving Process

Read the given problem.
Understand the given problem. This might require you to rephrase the problem in terms that you can understand or, more commonly, to draw a picture of the situation.
Label unknowns. All problems presented in mathematics have unknown quantities. In this step, you take the time to label these unknowns using variables. This step is often combined with the previous step (especially when drawing pictures).
List the Givens and Wants . This step is dedicated to making a table with one column dedicated to given information and another column dedicated to the wanted or desired information. Normally, we only list the given and desired rates in this table.
Create a Master Equation involving the variables within the given and wanted rate information. This is a critical step in all problem-solving processes involving Polya's method. Warning: The Master Equation should be a relation between two variables (those listed in the Givens and Wants table). If your Master Equation has more than two variables, you need to find a relation between the excess variables to reduce the number of final variables in your Master Equation to two.
Find the Rate Equation from the Master Equation. This almost always requires implicit differentiation.
Substitute in any constants given within the problem.
Solve the resulting equation for the desired "wanted" piece of information.

Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, if the value for a changing quantity is substituted into an equation before both sides of the equation are differentiated, then that quantity will behave as a constant and its derivative will not appear in the new equation found during the rate equation step. We examine this potential error in the following example.

Examples of Polya's Problem-Solving Process

Example $\PageIndex{2}$: An Airplane Flying at a Constant Elevation

An airplane is flying overhead at a constant elevation of $4000$ ft. A man is viewing the plane from a position $3000$ ft from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at the rate of $600$ ft/sec, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?

Read. This step is only listed so you can have a "win." All of us can read a problem, but understanding requires a bit more effort.

Understand and Label. Draw a picture, introducing variables to represent the different quantities involved.

$A right triangle is made with a person on the ground, an airplane in the air, and a radio tower at the right angle on the ground. The hypotenuse is s, the distance on the ground between the person and the radio tower is x, and the side opposite the person (that is, the height from the ground to the airplane) is 4000 ft.$

As shown, $x$ denotes the distance between the man and the position on the ground directly below the airplane. The variable $s$ denotes the distance between the man and the plane. Note that both $x$ and $s$ are functions of time. We do not introduce a variable for the height of the plane because it remains at a constant elevation of $4000$ ft. Since an object’s height above the ground is measured as the shortest distance between the object and the ground, the line segment of length 4000 ft is perpendicular to the line segment of length $x$ feet, creating a right triangle.

Givens and Wants.

Since $x$ denotes the horizontal distance between the man and the point on the ground below the plane, $dx/dt$ represents the speed of the plane. We are told the speed of the plane is $600$ ft/sec. Therefore, $\frac{dx}{dt}=600$ ft/sec. Since we are asked to find the rate of change in the distance between the man and the plane when the plane is directly above the radio tower, we need to find $ds/dt$ when $x=3000$ ft.

As was mentioned in Polya's modified problem-solving process, we only list the given and desired rate information. That is, don't bother listing the facts that the height of the plane is always $4000$ feet and the distance from the tower to our position is always $3000$ feet.

Master Equation. The variables within our table above are $x$ and $s$. From the figure, we can use the Pythagorean Theorem to write an equation relating these two variables:

$[x(t)]^2+4000^2=[s(t)]^2.$

Rate Equation. Differentiating this equation with respect to time and using the fact that the derivative of a constant is zero, we arrive at the equation

\[x\frac{dx}{dt}=s\frac{ds}{dt}.\nonumber \]

Substitute. When we "stop time" to do the problem, we know that $\frac{dx}{dt}=600$ ft/sec and $x=3000$ ft. Thus,

\[ (3000)(600) = s \frac{ds}{dt}. \nonumber \]

Solve. You can see that the only thing we are missing is $s$. However, we can use the Pythagorean Theorem to determine the distance $s$ when $x=3000$ ft and the height is $4000$ ft. Solving the equation

$3000^2+4000^2=s^2$

for $s$, we have $s=5000$ ft at the time of interest. Using these values, we conclude that $ds/dt$

is a solution of the equation

$(3000)(600)=(5000) \cdot \dfrac{ds}{dt}$.

$\dfrac{ds}{dt}=\dfrac{3000 \cdot 600}{5000}=360\,\text{ft/sec}.$

Note: When solving related-rates problems, it is important not to substitute values for the variables too soon. For example, in the Master Equation step, we related the variable quantities $x(t)$ and $s(t)$ by the equation

Since the plane remains at a constant height, it is not necessary to introduce a variable for the height, and we are allowed to use the constant 4000 to denote that quantity. However, the other two quantities are changing. If we mistakenly substituted $x(t)=3000$ into the equation before differentiating, our equation would have been

$3000^2+4000^2=[s(t)]^2.$

After differentiating, our equation would become

$0=s(t)\dfrac{ds}{dt}.$

As a result, we would incorrectly conclude that $\frac{ds}{dt}=0.$

Exercise $\PageIndex{2}$

What is the speed of the plane if the distance between the person and the plane is increasing at the rate of $300$ ft/sec?

$\dfrac{ds}{dt}=300$ ft/sec

$500$ ft/sec

We now return to the problem involving the rocket launch from the beginning of the chapter.

Example $\PageIndex{3}$: Chapter Opener - A Rocket Launch

A rocket is launched so that it rises vertically. A camera is positioned $5000$ ft from the launch pad. When the rocket is $1000$ ft above the launch pad, its velocity is $600$ ft/sec.

$A photo of a rocket lifting off.$

Find the necessary rate of change of the camera’s angle as a function of time so that it stays focused on the rocket.

Read. Again, this is a freebie step. Make sure you read the problem thoroughly!

Understand and Label. Draw a picture introducing the variables.

$A right triangle is formed with a camera at one of the nonright angles and a rocket at the other nonright angle. The angle with the camera has measure θ. The distance from the rocket to the ground is h; note that this is the side opposite the angle with measure θ. The side adjacent to the angle with measure θ is 5000 ft.$

Let $h$ denote the height of the rocket above the launch pad and $ \theta $ be the angle between the camera lens and the ground.

Givens and Wants. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is 1000 ft off the ground. That is, we need to find $\frac{d \theta }{dt}$ when $h=1000$ ft. At that time, we know the velocity of the rocket is $\frac{dh}{dt}=600$ ft/sec.

Master Equation. Using the fact that we have drawn a right triangle, it is natural to think about trigonometric functions. Recall that $\tan{(\theta)}$ is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have

$\tan{(\theta)} = \dfrac{h}{5000}$.

This gives us the equation

$h=5000\tan{(\theta)}.$

Rate Equation. Differentiating this equation with respect to time $t$, we obtain

$\dfrac{dh}{dt} = 5000\sec^2{(\theta)} \dfrac{d\theta}{dt}$.

Substitute. Letting $h = 1000$ and $\dfrac{dh}{dt} = 600$, we get

\[ 600 = 5000 \sec^2{(\theta)} \dfrac{d\theta}{dt}. \nonumber \]

Solve. We want to find $\frac{d\theta}{dt}$ when $h=1000$ ft. We need to determine $\sec^2{(\theta)}$. Recall that $\sec{(\theta)}$ is the ratio of the length of the hypotenuse to the length of the adjacent side. We know the length of the adjacent side is $5000$ ft. To determine the length of the hypotenuse, we use the Pythagorean Theorem, where the length of one leg is $5000$ ft, the length of the other leg is $h=1000$ ft, and the length of the hypotenuse is $c$ feet as shown in the following figure.

$A right triangle has one angle with measure θ. The hypotenuse is c, the side length opposite the angle with measure θ is 1000, and the side adjacent to the angle with measure θ is 5000.$

We see that

$1000^2+5000^2=c^2$

and we conclude that the hypotenuse is

$c=1000\sqrt{26}\,\text{ft}.$

Therefore, when $h=1000,$ we have

$\sec^2{(\theta)} = \left(\dfrac{1000\sqrt{26}}{5000}\right)^2=\dfrac{26}{25}.$

Doing a final substitution into our formula from the Substitute step, we arrive at

$600=5000\left(\frac{26}{25}\right)\dfrac{d\theta}{dt}$.

Therefore, $\dfrac{d \theta }{dt}=\dfrac{3}{26}$ rad/sec.

Exercise $\PageIndex{3}$

What rate of change is necessary for the elevation angle of the camera if the camera is placed on the ground at a distance of $4000$ ft from the launch pad and the velocity of the rocket is $500$ ft/sec when the rocket is $2000$ ft off the ground?

Find $\frac{d \theta }{dt}$ when $h=2000$ ft. At that time, $\frac{dh}{dt}=500$ ft/sec.

$\frac{1}{10}$ rad/sec

Example $\PageIndex{4}$: Water Draining from a Funnel

Water is draining from the bottom of a cone-shaped funnel at the rate of $0.03\,\text{ft}^3\text{/sec}$. The height of the funnel is $2$ ft and the radius at the top of the funnel is $1$ ft. At what rate is the height of the water in the funnel changing when the height of the water is $\frac{1}{2}$ ft?

This time, we will shorthand our steps to mirror the pacing you should get used to.

Read. Done.

Understand and Label. See the picture below.

$A funnel is shown with height 2 and radius 1 at its top. The funnel has water to height h, at which point the radius is r.$

Let $h$ denote the height of the water in the funnel, $r$ denote the radius of the water at its surface, and $V$ denote the volume of the water.

Master Equation.

$V=\frac{1}{3} \pi r^2h.$

Since the Master Equation involves more than two variables, we want to find a relation to rewrite $r$ in terms of either $h$ or $V$.

From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the same. Therefore, $\frac{r}{h}=\frac{1}{2}$ or $r=\frac{h}{2}.$ Using this fact, the equation for volume can be simplified to

$V=\frac{1}{3} \pi \left(\frac{h}{2}\right)^2h=\frac{ \pi }{12}h^3$.

Rate Equation.

\[\frac{dV}{dt}=\frac{\pi}{4}h^2\frac{dh}{dt}.\nonumber \]

Substitute.

\[−0.03 = \frac{\pi}{4}\cdot \left(\frac{1}{2}\right)^2 \cdot \dfrac{dh}{dt},\nonumber \]

Solve. The previous step implies

\[−0.03=\frac{\pi}{16}\dfrac{dh}{dt}.\nonumber \]

It follows that

\[\dfrac{dh}{dt}=−\frac{0.48}{\pi} \approx −0.153\,\text{ft/sec}.\nonumber \]

Exercise $\PageIndex{4}$

At what rate is the height of the water changing when the height of the water is $\frac{1}{4}$ ft?

We need to find $\frac{dh}{dt}$ when $h=\frac{1}{4}.$

$−0.61$ ft/sec

Key Concepts

To solve a related rates problem, first draw a picture that illustrates the relationship between the two or more related quantities that are changing with respect to time.
In terms of the quantities, state the information given and the rate to be found.
Find an equation relating the quantities.
Use differentiation, applying the Chain Rule as necessary, to find an equation that relates the rates.
Be sure not to substitute a variable quantity for one of the variables until after finding an equation relating the rates.

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Unit Rate Worksheets with Word Problems

Help students of grade 5 through high school to heighten their logical reasoning with this batch of meticulously drafted unit rate worksheets. Over 60 plus well-researched word problems based on unit rates, unitary method and comparing unit rates are featured here! Answer keys are provided below every worksheet. Browse through some of these worksheets for free.

Unit Rate: Graphs

In this printable practice set, 7th grade and 8th grade students need to carefully observe the graphs where the x coordinate is 1. Find the corresponding y coordinate to determine the unit rate and note down your answers.

Download the set

Rate and Unit Rate

Read the word problems in these printable high school worksheets. Express the phrases in the form of rates. Then, find the unit rate by simplifying the fractions.

Unit Rate Word Problems: Standard

Assist young learners in grade 6 and grade 7 to improve their analytical skills with this set of diligently prepared unit-rate word problem with factual scenarios. Determine the unit rate in each problem.

Unit Rate Word Problems: Unitary Method

Read the phrases provided in this set of 5th grade pdf worksheets. Determine the unit rate. Then apply the unitary method to solve each problem.

Unit Rate Word Problems: Comparing Unit Rates

Read the word problems. Determine the unit rates to make comparisons. Then, draw conclusions based on the questions provided. Use the download option to access the entire set of worksheets.

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Word Problems Involving Rates and Ratios

The ratio is to compare two numbers. Rate is one type of ratio and is used to measure the variety of one thing or quantity in comparison to other. Word problems involving comparing rates deal with distances, time, rates, wind or water current, money, and age.

A step-by-step guide to solving rates and ratios word problems

To solve the word problems involving rates and ratios, follow these steps: Step 1: Find the known ratio and the unknown ratio. Step 2: Write the proportion. Step 3: Use cross-multiply and solve. Step 4: Plug the result into the unknown ratio to check the answers.

Word Problems Involving Rates and Ratios – Examples 1

If 11 apple pies cost $88, what will 8 apple pies cost? Solution: Write as a rate. $\frac{88÷11}{11÷11}=\frac{8}{1}$ Write a proportion to know the cost of 8 apple pies. $\frac{8}{1}=\frac{x}{6}→8×6=1×x→x=48$

Word Problems Involving Rates and Ratios – Examples 2

If 6 cookbooks cost $120, how much would a dozen cookbooks cost? Solution: Write as a rate. $\frac{120÷6}{6÷6}=\frac{20}{1}$ Write a proportion to know the cost of 12 cookbooks. $\frac{20}{1}=\frac{x}{12}→20×12=1×x→x=240$

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East Asian societies have the world’s lowest birth rates—and are learning that ‘throwing a bit of money’ at the problem isn’t solving anything

Governments across Asia—in Singapore and Beijing, Tokyo and Seoul—are facing a crisis : plummeting birth rates.

For several decades now, people in East Asian economies have had fewer and fewer children. Last year, South Korea beat its own record for having the world’s lowest birth rate, reporting 0.72 births per woman for 2023, down from 0.78 in 2022. Singapore reported 0.97 births per woman, the first time the rate has fallen below one. Japan has one of the world’s oldest populations, with a median age of 49.5. Hong Kong, Taiwan, and mainland China are all reporting falling birth rates as well.

All of these economies have fertility rates far below 2.1, the “replacement rate” which allows for a stable population. They haven’t reported a rate above 2.1 for years, if not decades.

A low birth rate leads to a shrinking population, and a smaller workforce to produce the goods and services that lead to economic growth. Slower economic activity results in drops in fiscal revenue, giving fewer resources to a government that now needs to provide welfare for a growing elderly population.

Academics often point to the cost of childcare, poor work-life balance, a lack of support for new parents (particularly mothers), and the stresses of modern society as reasons for falling birth rates. “In all the cosmopolitan cities, the fertility rate tends to be much lower because [people have] a lot of choices. The higher the development, [the] more urbanized, the more education that women get, the smaller the family size,” Paul Cheung, director of the Asia Competitiveness Institute at the Lee Kuan Yew School of Public Policy, says.

Faced with this looming crisis, Asian governments have turned to a straightforward solution: Give prospective parents money if they have kids. The connection is simple to understand. If a major barrier to having children is the cost of childcare, then alleviating that cost with extra cash should change someone’s economic calculus.

Except it hasn’t worked. Even Singapore, which Cheung suggests had a “way more generous [policy] than all the Asian countries,” has not succeeded in arresting the decline in fertility.

“Low birth rates are a reflection of big institutional, cultural, structural problems,” said Stuart Gietel-Basten, a professor of social science and public policy at the Hong Kong University of Science and Technology. “Throwing a bit of money at it is not going to fix it.”

What are governments currently doing to stop falling birth rates?

Cheung, before his stint as an academic, was the director of Singapore’s population planning unit between 1987 and 1994. He helped start Singapore’s pronatalist policy, offering a relatively more generous set of incentives to encourage more births. The government even organized events to help single Singaporeans to meet.

Singapore’s government officially inaugurated its baby bonus scheme in 2001. The most current payout is 11,000 Singapore dollars ($8,263) for each first and second child and 13,000 Singapore dollars ($9,766) each for the third and subsequent child.

Other governments are also trying to dole out incentives. Japan increased its lump-sum childbirth benefit to 500,000 yen ($3,400) in April last year. Starting this October, the government will also offer 15,000 yen ($102) a month to households after the birth of a first and second child until the age of 2, and then continue providing 10,000 yen ($68) till high school. The government will offer more money to families with more than two children.

South Korea has increased its incentives, too. The government gives 2 million Korean won ($1,519) to parents when a baby is born, which increases to 3 million won ($2,279) for the second child. Parents will also get an allowance of up to 18 million won ($13,674) in total for the first two years of the child’s life.

Hong Kong, on the other hand, is offering a one-off cash allowance of HKD 20,000 ($2,557).

Singapore’s birth rate is declining at a slower pace than that of other Asian economies, only falling below 1.0 last year. (By comparison, Hong Kong’s fertility rate first fell below 1.0 in 2001, and hovered around that level before falling back below 1.0 again in 2020). Singapore’s population is still stable, but that may be the result of the country’s more liberal immigration policies, compared with Japan and South Korea.

All these measures seem to do is “delay the population decline a little into the future,” Cheung says.

‘I feel sorry for the government’

The scary thought for demographers may now be that there’s no easy fix for falling fertility. Even Nordic countries, whose more generous pro-child policies were credited with keeping birth rates relatively high, have seen fertility collapse after the COVID pandemic.

“The strange thing with fertility is nobody really knows what’s going on,” Anna Rotkirch, a research director at Family Federation of Finland’s Population Research Institute, told the Financial Times earlier this year. The demographer, who advised former prime minister Sanna Marin on population policy, now thinks fertility decline is “not primarily driven by economics or family policies. It’s something cultural, psychological, biological, cognitive.”

Research from Singapore implies that a drop in fertility could be the result of something more fundamental in how people live in modern society. Tan Poh Lin, a research fellow at the National University of Singapore, found rates of sexual intercourse among married heterosexual couples in Singapore—a “high-stress” society—were lower than the ideal frequency to conceive, generally considered to be five or six times every 30 days. There were “strong negative effects of both stress and fatigue, especially during weekdays,” she writes. Other surveys in Japan and South Korea report similar findings.

But if monetary incentives or better social welfare programs and work-life balance, as in northern European countries, are not getting birth rates to what they should be, then what can Asian economies do to raise them?

“I feel sorry for the government because it’s the only organization or institution doing anything,” Gietel-Basten said. “In reality, everyone has to take responsibility for this. Companies have to change their attitude and recognize children are a social good, and that parents should be supported and not penalized,” he says. “But that costs money.”

Some companies in Asia have made high-profile offers to support employees having children. In February, a South Korean construction firm, the Booyoung Group, offered a bonus worth 100 million won ($76,000) to encourage female employees to have children. China’s Trip.com Group also offered some employees a 10,000 yuan ($1,391) annual bonus for households for every child under the age of five.

But there’s no quick solution, says Gietel-Basten. Instead, he suggests that governments focus on other economic well-being issues—like youth unemployment, job security, and a sense that work is being valued—and hope that indirectly improves fertility rates.

In mainland China, “there’s not even jobs for the young people who are alive now,” he says. “Why do you want to have more children?”

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Mathematics > Optimization and Control

Title: solving the waste bin location problem with uncertain waste generation rate: a bi-objective robust optimization approach.

Abstract: An efficient Municipal solid waste (MSW) system is critical to modern cities in order to enhance sustainability and livability of urban life. With this aim, the planning phase of the MSW system should be carefully addressed by decision makers. However, planning success is dependent on many sources of uncertainty that can affect key parameters of the system, e.g., the waste generation rate in an urban area. With this in mind, this paper contributes with a robust optimization model to design the network of collection points (i.e., location and storage capacity), which are the first points of contact with the MSW system. A central feature of the model is a bi-objective function that aims at simultaneously minimizing the network costs of collection points and the required collection frequency to gather the accumulated waste (as a proxy of the collection cost). The value of the model is demonstrated by comparing its solutions with those obtained from its deterministic counterpart over a set of realistic instances considering different scenarios defined by different waste generation rates. The results show that the robust model finds competitive solutions in almost all cases investigated. An additional benefit of the model is that it allows the user to explore trade-offs between the two objectives.

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Rate problems (video)
Rate problems. In this math lesson, we learn to find unit rates and use them to solve problems. We first calculate the rate for one unit, like cars washed per day or cost per battery. Then, we multiply the unit rate by the desired quantity to find the answer.
Rate problems
Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: https://www.khanacademy.org/math/cc-sixth-grade-math/x0267d782:...
17.5: Solving Rate Problems
Summary. Sometimes we can find and use more than one unit rate to solve a problem. Suppose a grocery store is having a sale on shredded cheese. A small bag that holds 8 ounces is sold for $2.
Art of Problem Solving
A rate is a type of ratio where something of one unit is compared with something else of another unit. Rates are applied in many real-world scenarios like unit conversions, speed/velocity, and per-item price. Unit cancellation is a common strategy used in rate and conversion problems where if there are two instances of a unit written -- one on ...
Rate Problems
The key to solving rate problems is to figure out the context of the problem and then identify a formula that relates all of the information in the problem. In this problem, our context is distance: we have two objects traveling at different rates and in opposite directions. We can relate all of the information using the simple formula distance ...
Unit Rate in Math
Learn to define what a unit rate in math is and see examples of solving rate problems. Discover what unit rate word problems are and how to set them up. Updated: 11/21/2023
9.10 Rate Word Problems: Work and Time
The equation used to solve problems of this type is one of reciprocals. It is derived as follows: rate ×time = work done rate × time = work done. For this problem: Felicia's rate: F rate × 4 h = 1 room Katy's rate: Krate × 12 h = 1 room Isolating for their rates: F = 1 4 h and K = 1 12 h Felicia's rate: F rate × 4 h = 1 room Katy's rate: K ...
How to Solve Rate Problems
This method of changing an equation from fractions on each side to products on each side is called "cross muliply" method which we will use to solve our problems. We now go back to our equation 150 km 3 hour = 250 km t 150 km 3 hour = 250 km t and use the "cross multiply" method to write it as follows. 150 km × t = 250km × 3hours 150 km × t ...
Rate Problems
100 miles per hour. Correct answer: 80 miles per hour. Explanation: If the driver needs to drive two laps, each one mile long, at an average rate of 60 miles per hour. To find the average speed, we need to add the speed for each lap together then divide by the number of laps. The equation would be as follows:
Study Guide
Problem-Solving Strategy: Solving a Related-Rates Problem. Assign symbols to all variables involved in the problem. Draw a figure if applicable. State, in terms of the variables, the information that is given and the rate to be determined. Find an equation relating the variables introduced in step 1.
Related-Rates Problem-Solving
Problem-Solving Strategy: Solving a Related-Rates Problem. Assign symbols to all variables involved in the problem. Draw a figure if applicable. State, in terms of the variables, the information that is given and the rate to be determined. Find an equation relating the variables introduced in step 1.
4.1: Related Rates
is a solution of the equation. (3000)(600) = (5000) ⋅ ds dt. Therefore, ds dt = 3000 ⋅ 600 5000 = 360ft/sec. Note: When solving related-rates problems, it is important not to substitute values for the variables too soon. For example, in step 3, we related the variable quantities x(t) and s(t) by the equation.
3.9: Related Rates
for s, we have s = 5000 ft at the time of interest. Using these values, we conclude that ds / dt. is a solution of the equation. (3000)(600) = (5000) ⋅ ds dt. Therefore, ds dt = 3000 ⋅ 600 5000 = 360ft/sec. Note: When solving related-rates problems, it is important not to substitute values for the variables too soon.
Rates and Unit Rates Worksheets with Word Problems
Unit Rate Worksheets with Word Problems. Help students of grade 5 through high school to heighten their logical reasoning with this batch of meticulously drafted unit rate worksheets. Over 60 plus well-researched word problems based on unit rates, unitary method and comparing unit rates are featured here! Answer keys are provided below every ...
Unit Rates
Welcome to Unit Rates with Mr. J! Need help with how to solve unit rate problems? You're in the right place!Whether you're just starting out, or need a quick...
Solving Related Rates Problems in Calculus
Solution. Let r and A be the radius and area of the circle, respectively. Take note that the variable t is in minutes. The rate of change of the oil film is given by the derivative dA/dt, where. Learn to solve different kinds of related rate problems in calculus. This article is a full guide that shows the step-by-step procedure for solving ...
Word Problems Involving Rates and Ratios
Word problems involving comparing rates deal with distances, time, rates, wind or water current, money, and age. A step-by-step guide to solving rates and ratios word problems. To solve the word problems involving rates and ratios, follow these steps: Step 1: Find the known ratio and the unknown ratio. Step 2: Write the proportion.
finding rate Worksheets
Mastering this skill will help students in comparing different scenarios, understanding rates and ratios, and solving practical everyday problems. ... Problem Solving Skills This worksheet serves as an effective tool to enhance problem-solving abilities. With a total of six problems, students would be able to efficiently analyze and solve ...
Solving Unit Rate Word Problems
Welcome to Solving Unit Rate Word Problems with Mr. J! Need help with unit rates? You're in the right place!Whether you're just starting out, or need a quick...
East Asia's baby bonuses aren't solving falling birth rates
Singapore's birth rate fell below 1.0 for the first time in 2023. ... rates—and are learning that 'throwing a bit of money' at the problem isn't solving anything ... problems," said ...
What is the three-body problem in '3 Body Problem'?
Solving the three-body problem might seem like the most pressing way to move through the game, but it's not the ultimate mission — nor is it even really possible. Jack and Jin advance to Level 3 ...
[2403.19476] Solving the waste bin location problem with uncertain
An efficient Municipal solid waste (MSW) system is critical to modern cities in order to enhance sustainability and livability of urban life. With this aim, the planning phase of the MSW system should be carefully addressed by decision makers. However, planning success is dependent on many sources of uncertainty that can affect key parameters of the system, e.g., the waste generation rate in ...
Accelerating preconditioned ADMM via degenerate proximal ...
In this paper, we aim to accelerate a preconditioned alternating direction method of multipliers (pADMM), whose proximal terms are convex quadratic functions, for solving linearly constrained convex optimization problems. To achieve this, we first reformulate the pADMM into a form of proximal point method (PPM) with a positive semidefinite preconditioner which can be degenerate due to the lack ...

9.10 Rate Word Problems: Work and Time

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Study Guides > Calculus Volume 1

Setting up Related-Rates Problems

Inflating a Balloon

Problem-Solving Strategy: Solving a Related-Rates Problem

Examples of the Process

An Airplane Flying at a Constant Elevation

Chapter Opener: A Rocket Launch

Water Draining from a Funnel

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Module 4: Applications of Derivatives

Setting up Related-Rates Problems

Example: Inflating a Balloon

Problem-Solving Strategy: Solving a Related-Rates Problem

Examples of the Process

Example: An Airplane Flying at a Constant Elevation

Example: Water Draining from a Funnel

3.9: Related Rates

Learning Objectives

Setting up Related-Rates Problems

Example \(\PageIndex{1}\): Inflating a Balloon

Exercise \(\PageIndex{1}\)

Polya's Modified Problem-Solving Process

Examples of Polya's Problem-Solving Process

Example \(\PageIndex{2}\): An Airplane Flying at a Constant Elevation

Exercise \(\PageIndex{2}\)

Example \(\PageIndex{3}\): Chapter Opener - A Rocket Launch

Exercise \(\PageIndex{3}\)

Example \(\PageIndex{4}\): Water Draining from a Funnel

Exercise \(\PageIndex{4}\)

Key Concepts

Unit Rate Worksheets with Word Problems

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A step-by-step guide to solving rates and ratios word problems

Word Problems Involving Rates and Ratios – Examples 1

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finding rate Worksheets

East Asian societies have the world’s lowest birth rates—and are learning that ‘throwing a bit of money’ at the problem isn’t solving anything

What are governments currently doing to stop falling birth rates?

‘I feel sorry for the government’

Latest in Leadership

Larry Summers, a ‘friend of China,’ warns Beijing that its actions are making it hard for people like him to push for stronger ties

Boeing’s issues are prompting Hong Kong’s flagship airline Cathay Pacific to consider an ‘ABC’ future: Airbus, Boeing…and China’s COMAC

HKEX’s first woman CEO says Hong Kong’s exchange could name and shame listed companies that stick with single-gender boards

China’s President Xi meets with U.S. CEOs from FedEx, Blackstone and Qualcomm and tells them the Chinese economy hasn’t peaked

Worsening tensions between China and the U.S. will make Hong Kong and its ‘unique status’ even more important, says the city’s top financial official

Free taxis, education subsidies, and gratis egg freezing—how Chinese companies are crafting next-gen benefits for mothers amid the country’s baby bust

Ozempic maker Novo Nordisk facing pressure as study finds $1,000 appetite suppressant can be made for just $5

A woman purchased a vacant Hawaiian lot for about $22,000. She was surprised to see a $500,000 home was built on it by mistake

Home Depot’s largest acquisition ever is an $18.25 billion bet on the housing market’s severe shortage of new homes

For New Yorkers, ‘there are no consequences for not paying your property taxes’ as delinquencies spike to $880 million

The U.S. and its allies want to bring the entire chip supply chain in-house—and that could create an OPEC-style cartel for the digital age

Here’s what’s open (and closed) on Good Friday 2024

Mathematics > Optimization and Control

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