how to solve work problems in calculus

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"Work" Word Problems

Painting & Pipes Tubs & Man-Hours Unequal Times Etc.

"Work" problems usually involve situations such as two people working together to paint a house. You are usually told how long each person takes to paint a similarly-sized house, and you are asked how long it will take the two of them to paint the house when they work together.

Many of these problems are not terribly realistic — since when can two laser printers work together on printing one report? — but it's the technique that they want you to learn, not the applicability to "real life".

The method of solution for "work" problems is not obvious, so don't feel bad if you're totally lost at the moment. There is a "trick" to doing work problems: you have to think of the problem in terms of how much each person / machine / whatever does in a given unit of time . For instance:

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Suppose one painter can paint the entire house in twelve hours, and the second painter takes eight hours to paint a similarly-sized house. How long would it take the two painters together to paint the house?

To find out how much they can do together per hour , I make the necessary assumption that their labors are additive (in other words, that they never get in each other's way in any manner), and I add together what they can do individually per hour . So, per hour, their labors are:

But the exercise didn't ask me how much they can do per hour; it asked me how long they'll take to finish one whole job, working togets. So now I'll pick the variable " t " to stand for how long they take (that is, the time they take) to do the job together. Then they can do:

This gives me an expression for their combined hourly rate. I already had a numerical expression for their combined hourly rate. So, setting these two expressions equal, I get:

I can solve by flipping the equation; I get:

An hour has sixty minutes, so 0.8 of an hour has forty-eight minutes. Then:

They can complete the job together in 4 hours and 48 minutes.

The important thing to understand about the above example is that the key was in converting how long each person took to complete the task into a rate.

hours to complete job:

first painter: 12

second painter: 8

together: t

Since the unit for completion was "hours", I converted each time to an hourly rate; that is, I restated everything in terms of how much of the entire task could be completed per hour. To do this, I simply inverted each value for "hours to complete job":

completed per hour:

Then, assuming that their per-hour rates were additive, I added the portion that each could do per hour, summed them, and set this equal to the "together" rate:

adding their labor:

As you can see in the above example, "work" problems commonly create rational equations . But the equations themselves are usually pretty simple to solve.

One pipe can fill a pool 1.25 times as fast as a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

My first step is to list the times taken by each pipe to fill the pool, and how long the two pipes take together. In this case, I know the "together" time, but not the individual times. One of the pipes' times is expressed in terms of the other pipe's time, so I'll pick a variable to stand for one of these times.

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Since the faster pipe's time to completion is defined in terms of the second pipe's time, I'll pick a variable for the slower pipe's time, and then use this to create an expression for the faster pipe's time:

slow pipe: s

together: 5

Next, I'll convert all of the completion times to per-hour rates:

Then I make the necessary assumption that the pipes' contributions are additive (which is reasonable, in this case), add the two pipes' contributions, and set this equal to the combined per-hour rate:

multiplying through by 20 s (being the lowest common denominator of all the fractional terms):

20 + 25 = 4 s

45/4 = 11.25 = s

They asked me for the time of the slower pipe, so I don't need to find the time for the faster pipe. My answer is:

The slower pipe takes 11.25 hours.

Note: I could have picked a variable for the faster pipe, and then defined the time for the slower pipe in terms of this variable. If you're not sure how you'd do this, then think about it in terms of nicer numbers: If someone goes twice as fast as you, then you take twice as long as he does; if he goes three times as fast as you, then you take three times as long as him. In this case, if he goes 1.25 times as fast, then you take 1.25 times as long. So the variables could have been " f  " for the number of hours the faster pipe takes, and then the number of hours for the slower pipe would have been " 1.25 f  ".

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Math Work Problems - Two Persons

In these lessons, we will learn how to solve work problems that involve two persons who may work at different rates.

Related Pages Work Problems Solving Work Word Problems Using Algebra More Algebra Lessons

Work Problems are word problems that involve different people doing work together but at different rates . If the people were working at the same rate then we can use the Inversely Proportional Method instead.

How To Solve Work Problems: Two Persons, Unknown Time

We will learn how to solve math work problems that involve two persons. We will also learn how to solve work problems with unknown time.

The following diagram shows the formula for Work Problems that involve two persons. Scroll down the page for more examples and solutions on solving algebra work problems.

This formula can be extended for more than two persons .

"Work" Problems: Two Persons

Example: Peter can mow the lawn in 40 minutes and John can mow the lawn in 60 minutes. How long will it take for them to mow the lawn together?

Solution: Step 1: Assign variables : Let x = time to mow lawn together.

Step 3: Solve the equation The LCM of 40 and 60 is 120 Multiply both sides with 120

Answer: The time taken for both of them to mow the lawn together is 24 minutes.

Work Problems With One Unknown Time

• Catherine can paint a house in 15 hours. Dan can paint a house in 30 hours. How long will it take them working together.
• Evan can clean a room in 3 hours. If his sister, Faith helps, it takes them two and two-fifths hours. How long will it take Faith working alone?

Variations Of GMAT Combined Work Problems

• Working at a constant rate, Joe can paint a fence in 4 hours. Working at a constant rate, his brother can paint the same fence in 2 hours. How long will it take them to paint the fence if they both work together at their respective constant rates?
• Working alone at a constant rate, machine A takes 2 hours to build a care. Working alone at a constant rate, machine B takes 3 hours to build the same car. If they work together for 1 hour at their respective constant rates and then machine B breaks down, how much additional time will it take machine A to finish the car by itself?
• Working alone at a constant rate, Carla can wash a load of dishes in 42 minutes. If Carla works together with Dan and they both work at constant rates, it takes them 28 minutes to wash the same load of dishes. Working at a constant rate, how long would it take Dan to wash the load of dishes by himself?

How To Solve “Working Together” Problems?

Example: It takes Andy 40 minutes to do a particular job alone. It takes Brenda 50 minutes to do the same job alone. How long would it take them if they worked together?

Word Problem: Work, Rates, Time To Complete A Task

We are given that a person can complete a task alone in 32 hours and with another person they can finish the task in 19 hours. We want to know how long it would take the second person working alone.

Example: Latisha and Ricky work for a computer software company. Together they can write a particular computer program in 19 hours. Latisha van write the program by herself in 32 hours. How long will it take Ricky to write the program alone?

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Work Calculus – Definition, Definite Integral, and Applications

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Work definition in calculus

Practice questions.

Work , in physics, indicates the amount of force acting through a given distance. In this article, we’ll highlight the mathematical definition of work and learn how we can define work in terms of definite integrals. We’ll also get to see how essential integral calculus is in physics and engineering in our discussion.

Work in calculus is simply equal to the area under the curve of $\boldsymbol{F(x)}$, where $\boldsymbol{F(x)}$ is simply the function representing the force acted upon from $\boldsymbol{x =a}$ to $\boldsymbol{x = b}$.

We’ll show you how to calculate for an object’s work by evaluating definite integrals, so keep your notes on integral properties and antiderivative formulas handy. For now, let’s do a quick recap of what work is and learn how definite integrals come into place!

What is work in calculus?

Work in calculus reflects the area under the curve of $F(x)$. In our introductory physics class, work is simply the product of the force acting upon the object over a given distance (or displacement) .

\begin{aligned}W &= Fd\end{aligned}

Keep in mind that in SI metric system, force is measured in Newtons ($N$), the distance is in meters ($m$), and work is in terms of $N \cdot m$ or in joules ($J$). For example, if a person exerts a force of $10 \phantom{x}N$ to lift an object $40 \phantom{x}m$ off the floor, the amount of work done is as shown below.

\begin{aligned}W &= F\cdot d\\&= (10 \phantom{x}N) \cdot (40 \phantom{x}m)\\&= 400 \phantom{x}N\cdot m\\&= 400 \phantom{x}J\end{aligned}

Now, let’s extend the definition of work by considering the fact that the force applied can be variable.

In the earlier discussion, we’ve shown how work can be calculated when given a constant value for the force exerted. However, there are instances, when force is not constant . Suppose that we’re observing an object that is moving along the positive horizontal direction from $x=a$ to $x =b$.

Let’s say $F(x)$ is the function representing the amount of force exerted over the interval, $[a, b]$. If $F(x)$ is continuous throughout the interval, we can approximate the amount of work done by partitioning the interval into subintervals with the following endpoints: $\{x_0, x_1, x_2, …, x_n\}$ and a uniform width of $\Delta x$. Hence, we can estimate the total amount of work as shown below:

\begin{aligned}W &\approx \sum_{i =1}^{n} f(c_i)\Delta x_i \end{aligned}

To find a better approximation, as we have learned before, can be acquired by evaluating the quantity as $n\rightarrow \infty$.

This means that work exerted on a moving object is simply the area under the curve of the function for force , $F(x)$ from $x=a$ to $x =b$.

How to solve work problems in calculus?

Now that we understand how we can define work in terms of the definite integral of the force function, $F(x)$, let’s break down the steps to solve problems involving work:

• Identify the function that represents the force, $F(x)$, exerted from $x=a$ to $x= b$.
• Set up the expression for work in terms of $F(x)$: $W = \int_{a}^{b} F(x)\phantom{x}dx$.
• Evaluate the definite integral by applying key integral properties and formulas.

For example, if we want to calculate the amount of work done given $F(x) = 2x + 1$ Newtons over the interval, $[2, 8]$. Hence, we can calculate the amount of work done as shown below.

\begin{aligned}W &= \int_{2}^{8} (2x +1)\phantom{x}dx\\&= \int_{2}^{8}2x \phantom{x}dx + \int_{2}^{8}1 \phantom{x}dx\\&= 2\left[\dfrac{x^2}{2}\right]_{2}^{8} – [x]_{2}^{8}\\&= [(8^2 -2^2) – (8 -2)]\\&= 54 \end{aligned}

This means that the work done to move the object is equal to $54$ Newton-meters.

A particle is moving along the $x$-axis and located $x$ feet from the origin. If a force of $x^2 + x + 4$ pounds is being acted upon the article, calculate the total amount of work done to move the particle from $x = 2$ to $x = 6$.

We can see that the force acting upon the particle is equal to $F(x) = x^2 + x +4$. Hence, use this expression to calculate the amount of work done to move the particle over the interval, $[2, 6]$.

\begin{aligned} W&= \int_{2}^{6} (x^2 + x + 4) \phantom{x}dx\end{aligned}

Evaluate the definite integral to find $W$.

\begin{aligned} W&= \int_{2}^{6} x^2\phantom{x}dx +\int_{2}^{6} x\phantom{x}dx +\int_{2}^{6} 4 \phantom{x}dx\\ &= \left[\dfrac{x^3}{3} \right ]_{2}^{6} + \left[\dfrac{x^2}{2} \right ]_{2}^{6} + [4x]_{2}^{6}\\&= \left[\left(\dfrac{6^3}{3} – \dfrac{2^3}{3} \right ) +\left(\dfrac{6^2}{2} – \dfrac{2^2}{2} \right )+ (4\cdot 6 – 4\cdot 2) \right ]\\&= \dfrac{208}{3} + 16 + 16\\&= \dfrac{304}{3}\end{aligned}

This means that the total amount of work done upon the particle is $\dfrac{304}{3}\phantom{x}\text{ft-lb}$.

We need a force of $30 \phantom{x}\text{N}$ to hold a spring that has been stretched from its natural length of $0.20 \phantom{x}\text{m}$ to a length of $0.45 \phantom{x}\text{m}$. Calculate the amount of work done if we stretch the spring from $0.25 \phantom{x}\text{m}$ to $0.48 \phantom{x}\text{m}$.

Hint : Recall that Hooke’s Law states that the force needed to maintain the stretch of $x$ units is $F(x) = kx$, where $k$ is the spring constant.

Calculate the sprint constant by using the given values: $F(x) = 30 \phantom{x}\text{N}$ and $x = 0.45 -0.20 = 0.25 \phantom{x}\text{m}$.

\begin{aligned} k&= \dfrac{F(x)}{x}\\&= \dfrac{30}{0.25}\\&= 120\end{aligned}

Using $k = 120$, we have $F(x) = 120x$. Let’s use this expression for the force needed to maintain a stretch from $0.25 \phantom{x}\text{m}$ to $0.48 \phantom{x}\text{m}$. Hence, we have the following expression for the amount of work needed:

\begin{aligned}W &= \int_{0.25}^{0.48} 120x \phantom{x}dx \end{aligned}

Evaluate the definite integral to find the exact value of work in $\text{N}\cdot \text{m}$.

\begin{aligned}W &= 120\int_{0.25}^{0.48} x \phantom{x}dx\\&= 120\left[\dfrac{x^2}{2} \right ]_{0.25}^{0.48}\\&= 60\left[x^2 \right ]_{0.25}^{0.48}\\&= 60[(0.48)^2 – (0.25)^2]\\&= 10.074 \phantom{x}\text{N}\cdot \text{m}\\&= 10.074 \phantom{x}\text{J} \end{aligned}

This means that the total amount of work done is $10.074 \phantom{x}\text{N}\cdot \text{m}$ or $10.074 \phantom{x}\text{J}$.

1. A particle is moving along the $x$-axis and located $x$ feet from the origin. If a force of $2x^2 + 4x$ pounds is being acted upon the article, calculate the total amount of work done to move the particle from $x = 1$ to $x = 5$.

2. We need a force of $40 \phantom{x}\text{N}$ to hold a spring that has been stretched from its natural length of $0.10 \phantom{x}\text{m}$ to a length of $0.35 \phantom{x}\text{m}$. Calculate the amount of work done if we stretch the spring from $0.15 \phantom{x}\text{m}$ to $0.60 \phantom{x}\text{m}$.

1. $\dfrac{392}{3} \phantom{x}\text{ft-lb}$ 2. $27 \phantom{x}\text{N}\cdot \text{m}$ or $27 \phantom{x}\text{J}$

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The Ultimate Guide to Solving Calculus Word Problems

Many math-phobic readers cringe at the sight of calculus word problems. There’s so much information given, so many problem types, and so many possible strategies to apply.

Where do you even start!?

That’s why I’ve put together this ultimate guide to solving calculus word problems. Calculus is sometimes regarded as the most challenging of the math disciplines. But, with this guide, it doesn’t have to be! 

Here I will share step-by-step tips and basic approaches to calculus word problems that you can use to solve any problem. I will also walk you through a set of sample problems for each problem type you will encounter in your studies of calculus!

It’s time to demystify these difficult problems!

Step-By-Step Tips and Basic Approaches to Calculus Word Problems

Calculus word problems will look different depending on the course you are in and the overall plan of the program you are studying in. You could be facing limits, continuity, derivatives, integral calculus, differential equations, and more! Calculus is a big complex topic and each curriculum covers it to a different extent.

But, regardless of the calculus word problems you are solving, there are a few basic approaches you should consider taking. These step-by-step tips will help you solve any calculus word problems that you encounter!

• Read the problem carefully:  Reading (and re-reading) the problem is important. This helps you look for key information, as well as important equations and quantities. Highlight them to help them stand out!
• Reflect on any new mathematical principles have you learned:  Are you studying differential calculus or integral calculus? Knowing the difference between the two will help you select an appropriate problem solving strategy.
• Select a strategy for the problem type : Read the problem again for any prompts that direct you toward a specific strategy or any related theorems. For example, optimization problems ask you to find the maximum of some quantity. Rate of change problems usually ask about how quickly a quantity changes. Sample problems of each of these are provided below!

Calculus Word Problems Sample Problems

As you review this collection of basic sample problems, consider the step-by-step tips outlined above. Focus your thinking on how you can use these tips to help you progress with the problem!

Basic Problem: Rate of Change

When it comes to traditional calculus first-course content, rate of change calculus word problems are quite common. The following sample problem will show you how to apply derivatives to solve a rate of change problem.

A boat is traveling along a straight path on the surface of the water in a lake. The boat’s position at time t is given by the function \(s(t)=2t^3−5t^2+3t+10\) , where \(s(t)\) is measured in meters and t is measured in seconds. Determine the boat’s instantaneous velocity at \(t=3\) seconds.

Reading the problem carefully, we can see that we are given a position function and are asked to find instantaneous velocity. The instantaneous velocity at a specific time t is given by the derivative of the position function given. We can apply the power rule to find the derivative:

$$ \begin{split} v(t)=&\frac{ds}{dt} \\ \\=&\frac{d}{dt}(2t^3−5t^2+3t+10) \\ \\ =&6t^2−10t+3 \end{split}$$

To find the instantaneous velocity a time t = 3 seconds, we substitute t = 3 into the derivative function:

$$ \begin{split} v(3)=&6(3)^2−10(3)+3 \\ \\ = &27 \end{split} $$

Therefore, the instantaneous rate of change of the boat on the surface of the water at t = 3 seconds is 27 m/s.

Difficult Calculus Word Problems: Optimization Problems

When I was a math student, I remember some of the most challenging advanced problems in differential calculus being optimization problems. From factoring to applying the first derivative test , there were so many steps that I struggled to even start the problem!

In the video below, I will show you step-by-step solutions that can be applied to solve optimization calculus word problems!

Integral Calculus

One of the tricks to integral calculus problems is recognizing that you are being asked to go backwards from a function given. Remembering that the integral is the opposite of the derivative will help you get started. Consider the following problem:

A particle moves along a straight line. Its velocity at time t is given by the function \(v(t)=3t^2 – 2t + 5\) , where v(t) is measured in meters per second and t is measured in seconds. Find the displacement of the particle on the interval \([0, 4]\).

Since velocity is the derivative of displacement, we need to integrate the velocity function v(t)  in order to find a displacement function for the particle.

$$ \begin{split} s(t)=& \int v(t) dt \\ \\ =&\int 3t^2 – 2t + 5 dt\\ \end{split} $$

From here, we can integrate each term in the expression for v(t) and apply the power rule for integration:

$$ \begin{split} s(t)=& \int 3t^2 dt – \int 2t dt+\int 5 dt \\ \\ =& t^3 -t^2+5t+C\\ \end{split} $$

Lastly, we can find the displacement over the interval [0, 4] by evaluating \(s(4) – s(0)\):

$$ \begin{split} &s(4)-s(0) \\ \\ =& (4)^3 -(4)^2+5(4)+C – (0^3 -0^2+5(0)+C) \\ \\ =& 68\end{split} $$

Therefore, the displacement of the particle over the time interval [0, 4] is 68 meters.

Reflecting on Calculus Word Problems

As a math student, I always found word problems to be the hardest mathematical problems. I always felt like something was missing to get me started on a solution.

As a math teacher of many years, I now have my own set of step-by-step tips that I can share with you when you feel the same way! And when you use helpful strategies like these, calculus isn’t really that hard !

Which of these problems do you find the most challenging? I hope these basic approaches help you with solving real-world calculus word problems!

Did you find this guide to calculus word problems helpful? Share this post and subscribe to Math By The Pixel on YouTube for more helpful calculus content!

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Older Adults and Balance Problems

Have you ever felt dizzy, lightheaded, or as if the room were spinning around you? These can be troublesome sensations. If the feeling happens often, it could be a sign of a balance problem.

On this page:

Causes of balance problems

Symptoms of balance disorders, treatments for balance problems and disorders, coping with a balance disorder.

Many older adults experience problems with balance and dizziness. Problems can be caused by certain medications, balance disorders, or other medical conditions. Balance problems are one reason older people fall. Maintaining good balance as you age and learning about fall prevention can help you get around, stay independent, and carry out daily activities.

People are more likely to have problems with balance as they grow older. In some cases, you can help reduce your risk for certain balance problems, but problems often can start suddenly and without obvious cause.

Balance problems can be caused by certain medications or medical conditions. The list below covers some common causes of balance problems.

• Medications. Check with your doctor if you notice balance problems while taking certain medications. Ask if other medications can be used instead, if the dosage can be safely reduced, or if there are other ways to reduce unwanted side effects.
• Inner ear problems. A part of the inner ear called the labyrinth is responsible for balance. When the labyrinth becomes inflamed, a condition called labyrinthitis occurs, causing vertigo and imbalance. Certain ear diseases and infections can lead to labyrinthitis.
• Alcohol. Alcohol in the blood can also cause dizziness and balance problems by affecting how the inner ear works.
• Other medical conditions. Certain conditions, such as diabetes, heart disease, stroke , or problems with your vision, thyroid, nerves, or blood vessels can cause dizziness and other balance problems.

Visit the NIH National Institute on Deafness and Other Communication Disorders website for more information on specific balance disorders .

If you have a balance disorder, you might experience symptoms such as:

• Dizziness or vertigo (a spinning sensation)
• Falling or feeling as if you are going to fall
• Staggering when you try to walk
• Lightheadedness, faintness, or a floating sensation
• Blurred vision
• Confusion or disorientation

Other symptoms might include nausea and vomiting; diarrhea; changes in heart rate and blood pressure and feelings of fear, anxiety, or panic. Symptoms may come and go over short periods or last for a long time and can lead to fatigue and depression.

Exercises that involve moving the head and body in certain ways can help treat some balance disorders. Patient-specific exercises are developed by a physical therapist or other professional who understands balance and its relationship with other systems in the body.

Balance problems due to high blood pressure may be managed by eating less salt (sodium), maintaining a healthy weight , and exercising . Balance problems due to low blood pressure may be managed by drinking plenty of fluids such as water; avoiding alcohol ; and being cautious regarding your body’s posture and movement, such as never standing up too quickly. Consult with your doctor about making any changes in your diet or activity level.

Some people with a balance disorder may not be able to fully relieve their dizziness and will need to find ways to cope with it. A vestibular rehabilitation therapist can help develop an individualized treatment plan.

Chronic balance problems can affect all aspects of your life, including your relationships, work performance, and your ability to carry out daily activities. Support groups provide the opportunity to learn from other people with similar experiences and challenges.

If you have trouble with your balance, talk to your doctor about whether it’s safe to drive, and about ways to lower your risk of falling during daily activities, such as walking up or down stairs, using the bathroom, or exercising. To reduce your risk of injury from dizziness, do not walk in the dark. Avoid high heels and, instead, wear nonskid, rubber-soled, low-heeled shoes. Don’t walk on stairs or floors in socks or in shoes and slippers with smooth soles. If necessary, use a cane or walker. Make changes to add safety features at your home and workplace, such as adding handrails.

Read about this topic in Spanish . Lea sobre este tema en español .

You may also be interested in

• Learning more about falls and falls prevention
• Find out more about ways to prevent falls in certain rooms
• Watching a video on balance exercises

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For more information on balance problems.

MedlinePlus National Library of Medicine       www.medlineplus.gov

Mayo Clinic www.mayoclinic.org/patient-care-and-health-information

National Institute on Deafness and Other Communication Disorders 800-241-1044 800-241-1055 (TTY) [email protected] www.nidcd.nih.gov

This content is provided by the NIH National Institute on Aging (NIA). NIA scientists and other experts review this content to ensure it is accurate and up to date.

Content reviewed: September 12, 2022

nia.nih.gov

An official website of the National Institutes of Health

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