Working related rates (also called rate of change) problems involves two main steps; translating the word problem into an equation (or set of equations), then manipulating those equations to get the answer. In these problems, watch your units. This page covers the basics of related rates problems.
If you want a complete lecture on this topic, we recommend this video.
video by Prof Leonard 

Before we get started, this first panel will help you review word problems by giving some general advice and techniques. Even if you already know how to work word problems, you may find the information in this panel helpful.
Word problems are what math students dread the most. I completely understand. I had a lot of difficulty too. However, I found a technique so that I was not only able to successfully work word problems but, eventually, I came to like them and am now able to teach them.
I will tell you up front that figuring out how to work word problems is not easy and takes some independent work on your part to master them. But once you do, you will find them enjoyable and, since so many students struggle with them, most teachers give pretty easy problems, even on exams. So you should be able to breeze your way through them.
First, what doesn't work. Most books try to lump all word problems together and give you general guidelines on how to work them. I have NEVER found that helpful. It wasn't until I was able separate out the different types of word problems, that I came to understand how to work them. Since there are different types of word problems, there are different ways to work them.
Here is what you need to do.
1. Find plenty of problems with worked out solutions. Here are some suggested resources.
 Get a good book with examples and worked out solutions of the type of word problems you are studying. We have posted several suggestions on the books page.
 Check out the solution manual for problems in your textbook.
 17calculus practice problems
2. Once you have a good selection of worked out solutions, go through them carefully and pick up patterns on how they set up the problems, solve them and give the final answer. Pick the ones that are similar to ones in your textbook that you are working on for your class.
3. Key   Group the problems into categories that make sense to you. Some examples might be problems with triangles, problems with right circular cylinders, problems asking you to find areas or volumes. A single problem can go into multiple categories based on configuration or type of question or any other category that makes sense to you.
4. Work the problems yourself before looking at the solutions. Then compare your solutions with the book. Determine what you did wrong and what you need to learn in order to work the problems correctly. At first, this will be slow and painful but once your brain catches on, it will start to be fun. Be patient with yourself, work hard and don't give up. [ In the case of videos, stop the video after the presenter has given the problem statement and work it yourself before watching them solve it. ]
5. Important   Once you have finished a problem, write down the meaning of your answer in words and then reread the problem statement to make sure that your answer is what the problem asked for, including units. For example, I worked a problem about a skydiver and the problem asked for the time it takes for the skydiver to hit the ground after he opens his parachute. When I finished the problem, I had calculated the time that it takes for the skydiver to hit the ground since he jumped out of the plane. When I checked my answer in the back of the book, I was confused until I realized that the number I had was not what the problem asked for. If I had written what my answer means in words and then looked back at the problem, I would have realized right away what I needed to do to finish the problem. Doing this will save you from losing points on homework and exams and it takes only a few seconds.
6. Make sure you understand every single step and, when looking at the solution, figure out why they do things the way they do. If you made a mistake, try to understand what your mistake was and what you need to understand in order to not make the same mistake again. [ Also remember that no textbook or video is always 100% correct. If you can not figure out your mistake, find someone to ask and see if the solution manual is incorrect. ]
7. Pick up patterns and general ideas from each group of problems by working the same type of problems all together. Don't jump around to different types. Stay with one type for several problems. I won't tell you exactly how many. You need to determine that by how difficult the problems are, how well you think you understand the current type, how much time you have and how well you want to do on your homework and exams. Sometimes you can go on after working 5 of the same type, sometimes it takes 10 or more.
8. Find a friend to work with and go over the problems with them AFTER you have worked them on your own. Remember, at exam time you will be on your own. So don't rely on someone else too much. If you know more than the other person, explain your work to them. Communicating your work to someone else helps you understand it better. If you know less, ask lots of questions and ask them to explain their solution to you.
9. Do NOT do shortcuts. Shortcuts are good AFTER you have learned the material, not while you are learning the material. Do it the long way for a while until you are know it really well.
10. 2nd Key   Do not just look at the solutions or watch someone else work the problems. You need to get out a pencil and paper and work them yourself. You are going to get frustrated. You are going to want to quit, but don't quit. Use that feeling to motivate yourself and show yourself that you can do it. It feels great to master something that is difficult. If you have never pushed through something difficult before, try it now. It is not easy but it is worth it. I know because I went through this same process myself.
11. Finally, do not skip ANYTHING and NEVER GIVE UP. Make sure you understand every single step in every single problem. Here's why: Chances are, if you skip something, it will show up on an exam precisely because the part you don't understand is probably the most difficult part of the problem and teachers expect you to skip it. So they put it on exams to see if you understand the difficult parts.
So far, I have found that implementing these ideas as the best way to figure out how to work word problems. There are tons of general guidelines in books (most likely in your textbook too) that never really helped me. Give this technique a try. Remember, you are now in charge of your own learning. No one is going to help you from here on out. You need to do it.
Getting Started
As mentioned in the word problems panel above, the main key to working word problems is to work similar problems, one right after another. The related rates word problems on this site are set up in the following categories. If these are not in patterns that work for you, rework the categories into groups that make sense to you.
1. Distances
2. Areas of Geometric Shapes
3. Volumes of Geometric Shapes
Once you have the equation(s) you need, solving them is very easy, IF you know the chain rule and implicit differentiation. The idea of related rates problems is that you are trying to find something that is changing. You always take the derivative of these equations with respect to time, so we usually use the variable \(t\). The first step is to get used to taking the derivative like this. Here is an example.
Calculate the rate of change of \(y\) in the equation \(yx^2=3\).
\(\displaystyle{ \frac{dy}{dt} = 2x \frac{dx}{dt} }\)
Calculate the rate of change of \(y\) in the equation \(yx^2=3\).
Solution 

Calculate the rate of change of \(y\) in the equation \(yx^2=3\).
This is asking for the rate of change of \(y\), which, in equation form is \(dy/dt\). The words rate of change tell you that you need to take the derivative of the entire equation with respect to \(t\).
\(\begin{array}{rcl}
\displaystyle{\frac{d}{dt}[yx^2]} & = & \displaystyle{\frac{d}{dt}[3]} \\
\displaystyle{\frac{dy}{dt}  \frac{d}{dt}[x^2]} & = & 0 \\
\displaystyle{\frac{dy}{dt}  2x \frac{dx}{dt}} & = & 0 \\
\displaystyle{\frac{dy}{dt}} & = & \displaystyle{2x \frac{dx}{dt}} \\
\end{array}\)
Final Answer: \(\displaystyle{ \frac{dy}{dt} = 2x \frac{dx}{dt} }\)
Notice we used the chain rule on both terms on the left side of the equal sign.
Final Answer 

\(\displaystyle{ \frac{dy}{dt} = 2x \frac{dx}{dt} }\) 
With related rates problems, you almost always end up with more than one equation. If you do, you have two options related to WHEN you take the derivative.
1. You can take the derivative of each equation separately and then put the results together.
2. You can combine all the equations into one and then take the derivative of the one equation.
Your choice will depend on the number of equations, how complex the equations are and what the final equation looks like. We recommend that you try option 2 when you are first learning. Most of the initial practice problems you will be given will be simple enough that combining them can be done easily. Later on, try to do the first technique and see if you can figure it out. The derivatives are usually easier to evaluate but the algebra can get messy.
What To Do With Constants In Related Rates Problems
What do you do with constants that are given in the problem? First of all, you never want to just go in and plug in all your constants before you take the derivative.
Safe Answer   Wait and plug in your constants only after you have the derivative. So, you would label all distances with variables, take the derivative with respect to t and then plug in all your given constants. This is what you need to do when you first start learning to work related rates problems. After you have some experience, you can go on to the more experienced technique.
Experienced Answer   Once you learn the basics of related rates problems, you will have a feel for which constants you can plug in right away and which ones you can't. The difference you need to look for is
 if the variable is NOT changing, then you can substitute the constant in before taking the derivative;
 but, if the variable is changing over time, then you must wait until after you take the derivative before you can substitute the constant into the equation.
Strategies
The best way to learn to work these problems is to first get some practice with taking the derivative of given equations with respect to t. Don't spend too much time on this first part. Work only a few problems, to get a feel for it. Then move on.
Next, work similar problems in groups until you get a feel for them. This is where you should spend the majority of your time. Do not go on to the next type until you have mastered each type. You need to work similar problems to begin to see patterns and develop techniques in your mind. If you move on too quickly, you will never see those patterns and will continue to struggle with related rates and other types of word problems.
Note: Keep in mind that the problems on this site are organized in a way that makes sense to us. If they don't make sense to you, rearrange them. For example, you might want to group problems based on a certain geometric shape, like triangles or circles.
Okay, the discussion so far has been pretty theoretical. The best and only way to learn to work these types of problems is to work lots of practice problems.
These problems are a great way to start working related rates problems. Most of the information is given to you and you do not need to worry about the configuration of the problem at this point. You just need to apply the derivative to an equation which gives you some practice before going on to more complicated problems.
Practice
These related rates problems don't fit into one of the area, distance or volume categories and may give you the equation you need.
If \(x^2 + y^2 = 25\) and \(dx/dt = 7\), find \(dy/dt\) when \(x=3\).
Problem Statement 

If \(x^2 + y^2 = 25\) and \(dx/dt = 7\), find \(dy/dt\) when \(x=3\).
Solution 

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If \(z^2 = x^2 + y^2\), \(dx/dt = 4\) and \(dy/dt = 5\). Find \(dz/dt\) when \(x=8\) and \(y=15\).
Problem Statement 

If \(z^2 = x^2 + y^2\), \(dx/dt = 4\) and \(dy/dt = 5\). Find \(dz/dt\) when \(x=8\) and \(y=15\).
Solution 

video by The Organic Chemistry Tutor 

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Find the rate of change of the distance between the origin and a moving point on the graph of \( y=x^2 +1 \) at \( (2,5) \) if the ycoordinate is increasing at a rate of 4cm/s.
Problem Statement 

Find the rate of change of the distance between the origin and a moving point on the graph of \( y=x^2 +1 \) at \( (2,5) \) if the ycoordinate is increasing at a rate of 4cm/s.
Solution 

video by The Organic Chemistry Tutor 

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A point is moving on the graph of \(3x^3+4y^3=xy\). When the point is at \(P=(1/7,1/7)\), its \(y\)coordinate is increasing at a speed of 9 units per second. What is the speed of the \(x\)coordinate at that time and in which direction is the \(x\)coordinate moving?
Problem Statement 

A point is moving on the graph of \(3x^3+4y^3=xy\). When the point is at \(P=(1/7,1/7)\), its \(y\)coordinate is increasing at a speed of 9 units per second. What is the speed of the \(x\)coordinate at that time and in which direction is the \(x\)coordinate moving?
Solution 

video by PatrickJMT 

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For the supplydemand equation \(q = 41105e^{0.01p}\), we know that the quantity supplied is decreasing at a rate of 80 units per week. Calculate the rate at which the price is changing when the selling price is $100 per unit.
Problem Statement 

For the supplydemand equation \(q = 41105e^{0.01p}\), we know that the quantity supplied is decreasing at a rate of 80 units per week. Calculate the rate at which the price is changing when the selling price is $100 per unit.
Solution 

video by Krista King Math 

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If you have two resistors in parallel where one resistor is changing at a rate of \( 0.3 \Omega/sec \) and the other is changing at a rate of \( 0.2 \Omega/sec\). Calculate the rate of change of the total resistance.
Problem Statement 

If you have two resistors in parallel where one resistor is changing at a rate of \( 0.3 \Omega/sec \) and the other is changing at a rate of \( 0.2 \Omega/sec\). Calculate the rate of change of the total resistance.
Hint 

The total resistance, \(R_T\), of two resistors in parallel is given by \(\displaystyle{ \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} }\).
Problem Statement 

If you have two resistors in parallel where one resistor is changing at a rate of \( 0.3 \Omega/sec \) and the other is changing at a rate of \( 0.2 \Omega/sec\). Calculate the rate of change of the total resistance.
Final Answer 

\( 0.132 ~ \Omega/sec \)
Problem Statement 

If you have two resistors in parallel where one resistor is changing at a rate of \( 0.3 \Omega/sec \) and the other is changing at a rate of \( 0.2 \Omega/sec\). Calculate the rate of change of the total resistance.
Hint 

The total resistance, \(R_T\), of two resistors in parallel is given by \(\displaystyle{ \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} }\).
Solution 

video by Michel vanBiezen 

Final Answer 

\( 0.132 ~ \Omega/sec \) 
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We have a cylinderpiston configuration which is filled with gas. We are pushing on the piston, reducing it's volume. We assume that the gas follows the ideal gas law. The initial volume is \( 22253 ~ cm^3\) and the initial pressure is \( 400 ~ kPa\). The pressure is changing at a rate of \( 1 ~ kPa/sec \). What is the rate of change of the volume?
Problem Statement 

We have a cylinderpiston configuration which is filled with gas. We are pushing on the piston, reducing it's volume. We assume that the gas follows the ideal gas law. The initial volume is \( 22253 ~ cm^3\) and the initial pressure is \( 400 ~ kPa\). The pressure is changing at a rate of \( 1 ~ kPa/sec \). What is the rate of change of the volume?
Hint 

Although this problem involves a volume, we do not need an equation for the shape of the volume. We need only the ideal gas law, \(PV=nRT\). We assume the temperature is constant.
Problem Statement 

We have a cylinderpiston configuration which is filled with gas. We are pushing on the piston, reducing it's volume. We assume that the gas follows the ideal gas law. The initial volume is \( 22253 ~ cm^3\) and the initial pressure is \( 400 ~ kPa\). The pressure is changing at a rate of \( 1 ~ kPa/sec \). What is the rate of change of the volume?
Final Answer 

\( 5 ~ cm^3/sec \)
Problem Statement 

We have a cylinderpiston configuration which is filled with gas. We are pushing on the piston, reducing it's volume. We assume that the gas follows the ideal gas law. The initial volume is \( 22253 ~ cm^3\) and the initial pressure is \( 400 ~ kPa\). The pressure is changing at a rate of \( 1 ~ kPa/sec \). What is the rate of change of the volume?
Hint 

Although this problem involves a volume, we do not need an equation for the shape of the volume. We need only the ideal gas law, \(PV=nRT\). We assume the temperature is constant.
Solution 

The negative sign in the answer indicates that the volume is being reduced.
video by Michel vanBiezen 

Final Answer 

\( 5 ~ cm^3/sec \) 
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We have a cylinderpiston configuration filled with air. The initial volume is \( 2800 ~ cm^3 \) and the initial pressure is \( 200 ~ kPa\). We also know that the pressure is changing at a rate of \( 2 ~ kPa/sec\). If we assume an adiabatic expansion, at what rate is the volume changing?
Problem Statement 

We have a cylinderpiston configuration filled with air. The initial volume is \( 2800 ~ cm^3 \) and the initial pressure is \( 200 ~ kPa\). We also know that the pressure is changing at a rate of \( 2 ~ kPa/sec\). If we assume an adiabatic expansion, at what rate is the volume changing?
Hint 

Again, although this problem involves a volume, we do not need an equation for the shape of the volume. Assuming the idea gas law, \(PV=nRT\), holds, for an adiabatic expansion we have \(PV^{1.4} = C\) where \(C\) is a constant.
Problem Statement 

We have a cylinderpiston configuration filled with air. The initial volume is \( 2800 ~ cm^3 \) and the initial pressure is \( 200 ~ kPa\). We also know that the pressure is changing at a rate of \( 2 ~ kPa/sec\). If we assume an adiabatic expansion, at what rate is the volume changing?
Final Answer 

\( 20 ~ cm^3/sec \)
Problem Statement 

We have a cylinderpiston configuration filled with air. The initial volume is \( 2800 ~ cm^3 \) and the initial pressure is \( 200 ~ kPa\). We also know that the pressure is changing at a rate of \( 2 ~ kPa/sec\). If we assume an adiabatic expansion, at what rate is the volume changing?
Hint 

Again, although this problem involves a volume, we do not need an equation for the shape of the volume. Assuming the idea gas law, \(PV=nRT\), holds, for an adiabatic expansion we have \(PV^{1.4} = C\) where \(C\) is a constant.
Solution 

video by Michel vanBiezen 

Final Answer 

\( 20 ~ cm^3/sec \) 
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You CAN Ace Calculus
For related rates problems involving similar triangles, it may help you to review how to set up the ratios. You can find a discussion of this on the similar triangles precalculus page. 
external links you may find helpful 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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