Imagine throwing a stone into a body of water. If the body of water is calm enough, as soon as the stone hits the water, a series of ripples begin to form around the spot the stone went in. The ripples continue to expand outwards as the stone stinks to the bottom. As the ripples expand, the radius of the circular wave grows. As a result, the area enclosed by the ripple also grows.
Framing the problem as a related rate, we could measure the rate at which the enclosed area grows in terms of the rate of change of the radius. Related rates problems are one of the toughest problems for Calculus students to conceptualize. However, this article will further define related rates, how they can be applied in Calculus, and a step-by-step methodology for solving.
What are Related Rates in Calculus?
With the example of dropping a stone into the water in mind, let's define the term related rates more technically.
Related rates problems typically involve finding the rate at which one variable changes by relating the variable to one or more variables whose rates are known.
In our example of the stone in a calm body of water, as the ripples expand, both the radius and the area enclosed by the wave change. It's important to note the rate at which the radius changes is likely different than that of the area. If we are given one of the rates of change, we can solve for the other rate of change. We can do this because the formula for the area of a circle \(A=\pi r^2\) is related to the radius \(r\) of the circle.
The Importance of Related Rates
Solving related rates problems utilizes Calculus skills such as Implicit Differentiation and The Chain Rule. Therefore, make sure to go over our articles before diving into related rates! Learning how to solve these problems will help you strengthen your knowledge of Calculus. Solving related rates of change problems also has extensive applications in finance, physics, travel, and transportation. Framing problems in terms of related rates allows us to write a rate of change in terms of another (typically easier to compute) rate of change.
Related Rates Formulas
In basic Calculus, related rates problems typically fall into one of two categories:
Volume or area
Trigonometry
You'll likely need to brush up on volume/area and geometry formulas you learned years ago. Below are some formulas you might find useful during your work on related rates.
Area
Triangle
- \(area=\dfrac{1}{2}b \cdot h\) where \(b\) is base and \(h\) is height
Rectangle or square
- \(area=h \cot w\) where \(h\) is height and \(w\) is width.
Circle
- \(area=\pi r^2\) where \(r\) is radius.
- \(circumference = 2 \pi \cdot r\)
Volume
Rectangular pyramid
- \(volume=\dfrac{1}{3} L \cdot W \cdot H\)where \(L\) is length, \(W\) is width, and \(H\) is height
Cone
- \(volume= \dfrac{1}{3} \pi \cdot r^2 \cdot h\)where \(r\) is radius and \(h\) is height
Cylinder
- \(volume= \pi \cdot r^2 \cdot h\) where \(r\) is radius and \(h\) is height
Rectangular solid
- \(volume= L \cdot W \cdot H\) where \(L\) is length, \(W\) is width, and \(H\) is height
Sphere
- \(volume=\dfrac{4}{3} \cdot \pi r^3\) where \(r\) is radius
Trigonometry
Use the diagram below for symbol reference.
Fig. 1. SOH-CAH-TOA triangle relationships help in some related rates problems.
SOH-CAH-TOA
\[\sin(\theta)=\dfrac{opposite}{hypotenuse}=\dfrac{b}{c}\]
\[\cos(\theta)=\dfrac{adjacent}{hypotenuse}=\dfrac{a}{c}\]
\[\tan(\theta)=\dfrac{opposite}{adjacent}=\dfrac{b}{a}\]
Pythagorean Theorem
Solving Related Rates Problems, Step-by-Step Calculation
Though each related rates problem is different, below is a general methodology for solving related rates of change problems.
Step 1: Draw a diagram
First and foremost, draw a diagram including what you know and labels of things you need to find.
Step 2: Identify known and unknown information
Read the problem again to better understand the information the problem provides and what the problem is asking. You can use the important information to label your diagram.
Step 3: Use equations to relate information in the problem
One of the most challenging parts of solving related rates problems is finding and modeling the relationship between the information you know and the information you are looking for. To relate the two rates of change to each other, think of an equation that involves both variables.
Step 4: Solve using implicit differentiation
Now that you have an equation, you must perform implicit differentiation on both sides of the equation. You should have an equation in terms of two different rates of change.
Step 5: Substitute in known values
Finally, you can substitute any information the problem provided. You should be able to solve for one of the rates of change.
Once you've solved a related rates problem, you should always try to interpret what the rate means to ensure your answer makes sense in the context of the problem. You should check both the sign and the magnitude of your answer.
In the related rates problems in AP Calculus, you'll likely need to take the derivative with respect to time when using implicit differentiation.
Related Rates Problems - Worked Examples
Let's take a look at typical related rates problems. This first one involves a ladder leaning up against a wall.
Example 1
A \(10ft\) tall ladder leans against a wall. The base of the ladder begins to slide away from the wall at a rate of \(2ft/s) As the base of the ladder slides away from the wall, the top of the ladder slides down the wall vertically. When the base of the ladder is \(9ft\) away from the wall, what is the rate at which the top of the ladder slides down the wall?
Step 1: Draw a diagram
Drawing a diagram of the problem will help us to better comprehend our known and unknown values.
Fig. 2. From the horizontal rate of change, we are tasked with finding the vertical rate of change.
Based on our diagram, we are missing the vertical rate of change. However, we
do have the horizontal rate of change and the length of the ladder.
Step 2: Identify known and unknown quantities
Before we can do any Calculus, we must fully understand the problem. We know that a \(10ft\) ladder slides away from a wall horizontally at a rate of \(2ft/s\). The problem wants to know at what rate the top of the ladder moves when the base of the ladder is \(9ft\) from the wall.
Using our diagram in step 1, we can organize known and unknown variable quantities:
\[\dfrac{dy}{dt}=?\]
\[y(t)=?\]
\[x(t)=9\]
\[\dfrac{dx}{dt}=2\]
\[z=10\]
\(x\) and \(y\) are functions of time in this problem, so they are written \(x(t)\) and \(y(t)\). However, the length of the ladder, \(z\), does not change with time, so it is not written with function notation.
Step 3: Use equations to relate information in the problem
Based on the information we have and the information we need, it should be obvious that the Pythagorean Theorem will be useful in this problem.
Looking at the diagram again, notice that the ladder and 2 walls make a right triangle. This is a perfect scenario to use the Pythagorean Theorem!
Remember, while the ladder moves horizontally and vertically, the hypotenuse of the triangle (length of the ladder) does not change.
\[(x(t))^2+ (y(t))^2=z^2\]
\[(x(t))^2+ (y(t))^2=10^2\]
\[(x(t))^2+ (y(t))^2=100\]
Notice that we are given the derivative of \(x\) with respect to time,
\[\dfrac{dx}{dt}\]
We are also asked to find the rate at which the ladder is moving vertically,
\[\dfrac{dy}{dt}\]
How can we make an equation with these variables? Implicit differentiation!
Step 4: Solve using implicit differentiation
Now that we have an equation, let's use implicit differentiation to get the equation in terms of two rates of change. We will take the derivative with respect to time.
\[\dfrac{d}{dt}[(x(t))^2+(y(t))^2]=\dfrac{d}{dt} 100\]
\[2(x(t))\dfrac{dx}{dt}+2(y(t))\dfrac{dy}{dt}=0\]
Step 5: Substitute in known values
Again, we want to find the rate at which the ladder slides down the wall vertically:
\[\dfrac{dy}{dt}\]
We know that \(x=9\) ft and
\[\dfrac{dx}{dt}=2ft/s\]
Plugging in our known values, we get
\[2 \cdot 9 \cdot 2 + 2(y(t))\dfrac{dy}{dt}=0\]
To solve for \(\dfrac{dy}{dt}\), we still need the value of \(y\) when \(x=9\). We can use the Pythagorean Theorem equation we set up earlier to find \(y\), subbing \(x(t)=9\).
\[(x(t))^2+ (y(t))^2=z^2\]
\[9^2+ (y(t))^2=10^2\]
\[ (y(t))^2=19\]
\[ (y(t))=\sqrt{19}\]
Plugging in \(y(t)\) and solving for \(\dfrac{dy}{dt}\).
\[36+2(\sqrt{19})\dfrac{dy}{dt}=0\]
\[\dfrac{dy}{dt}=-\dfrac{36}{2 \sqrt{19}}\]
\[\dfrac{dy}{dt}=-\dfrac{18}{\sqrt{19}}\]
\[\dfrac{dy}{dt}=-4.129ft/s\]
The negative sign in our answer signifies that the ladder moves in the
negative direction (downwards).Therefore, the ladder slides
down the wall at a rate of \(4.129ft/s\) when the base ladder is \(9ft\) from the wall. Considering the ladder moves at a horizontal rate of \(2ft/s\), the magnitude of our answer also makes sense!
Considering a perfectly spherical balloon being filled with air. The balloon expands at a rate of \(3cm^2/s\). When the balloon's radius is \(4cm\), how fast is the radius increasing?
Step 1: Draw a diagram
Fig. 3. From the rate of change of the volume, we are tasked with finding the rate of change of the radius.
Based on our diagram, we are missing the rate of change of the radius. However, we do have the rate of change of the volume.
Step 2: Identify known and unknown quantities
We know that the volume of a spherical balloon increases at a rate of \(3cm^2/s\). We want to know the rate of change of the radius when the balloon has a radius of \(4cm\). Organizing into variables, we have
\[\dfrac{dV}{dt}=3cm^3/s\]
\[r(t)=4cm\]
\[\dfrac{dr}{dt}=?\]
Step 3: Use equations to relate information in the problem
Based on the information we need, and the shape of the balloon, the volume equation of a sphere will be useful in this problem.
\[V=\dfrac{4}{3} \pi \cdot r^3\]
Step 4: Solve using implicit differentiation
Now that we have an equation, let's use implicit differentiation to get the equation in terms of two rates of change. We will take the derivative with respect to time.
\[\dfrac{d}{dt}[V]=\dfrac{d}{dt}\left(\dfrac{4}{3} \pi \cdot r^3 \right)\]
\[\dfrac{dV}{dt}=4 \cdot \pi \cdot r^2 \dfrac{dr}{dt}\]
Step 5: Substitute in known values
We want to find the rate of change of the radius:
\[\dfrac{dr}{dt}\]
We know that \(r=4cm\) and:
\[\dfrac{dV}{dt}=3cm^3/s\]
Plugging in our known values, we get
\[3=4\pi \cdot 4^2 \dfrac{dr}{dt}\]
\[\dfrac{dr}{dt} \approx 0.01492 cm/s \]
\[\dfrac{dr}{dt} \approx 0.015 cm/s \]
The positive sign in our answer signifies that the radius is growing larger in the positive direction.
Therefore, the radius expands at a rate of about \(0.015cm/s\). Clearly, the radius grows at a very slow rate. However, this makes sense considering the volume also grows at a relatively slow rate.
Related Rates - Key takeaways
- Related rates problems typically involve finding the rate at which one variable changes by relating the variable to one or more variables whose rates are known.
- Solving related rates problems allows us to write a rate of change in terms of another (typically easier to compute) rate of change.
- Though each related rates problem is different, a general methodology for solving is:
- Identify known and unknown quantities
- Draw a diagram
- Use equations to relate information in the problem
- Solve using implicit differentiation (with respect to time)
- Substitute in known values
Explanations 
Exams 
Magazine 
