Calcworkshop

Related Rates How To w/ 7+ Step-by-Step Examples!

// Last Updated: February 22, 2021 - Watch Video //

How do you solve related rates?

Jenn (B.S., M.Ed.) of Calcworkshop® teaching how to solve related rates

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

Great question!

And that’s just what you’re going to learn how to do in today’s calculus lesson.

Let’s go!

What Are Related Rates (Real Life Examples)

Have you ever watched a baseball player who is rounding third and heading for home and wondered if they had enough speed to make it before getting tagged out by the thrower?

Or have you ever watched a basketball player shoot a free-throw and speculate if the ball has enough height and distance?

Or perhaps you’ve listened to a guitar solo and contemplated the number of vibrations per second needed to make the guitar strings hum at the perfect pitch?

If you have, and even if you haven’t, all of these queries have something in common — something is changing with respect to time .

This is the idea behind related rates .

The baseball player’s distance to the home plate is changing with respect to the runner’s speed per second. The success of a free-throw is related to the ball’s projectile motion and the instantaneous rate of change of the height and distance traveled. And when a guitar string is plucked, the rate of the guitar string’s vibration (frequency) produces high or low pitches, which make the music we hear sound pleasing.

What Does It Mean If Two Rates Are Related

Let’s make sense of things using the image to the right.

Imagine a person is outside looking up into the sky and they spot an airplane that is flying at an altitude of 6 miles above the ground. Now, as the plane continues on its flight path, several things are changing with respect to each other.

The angle of elevation (theta), the line of sight (hypotenuse), as well as the horizontal distance are all changing as the plane flies overhead and with respect to time.

And that’s what it means for two or more rates to be related — as one rate changes, so does the other.

Related Rates - Airplane Speed Problem

Calculate the Speed of an Airplane

How To Solve Related Rates Problems

We use the principles of problem-solving when solving related rates. The steps are as follows:

  • Read the problem carefully and write down all the given information.
  • Sketch and label a graph or diagram, if applicable.
  • Find an equation that relates the unknown variable and known variable(s) by looking for geometric shapes, known formulas, ratios such as the Pythagorean theorem, area and volume formulas, or trig identities.
  • Simplify using appropriate substitutions, so that chosen equation has only two variables (known and unknown).
  • Differentiate the equation implicitly with respect to time.
  • Substitute all known values into the derivative and solve for the final answer.

Ex) Cone Filling With Water

Alright, so now let’s put these problem-solving steps into practice by looking at a question that frequently appears in AP calculus, college, and university classes — the cone problem.

If water is being pumped into the tank at a rate of 3 cubic feet per second, find the rate at which the water level is rising when the water is 4ft deep.

Related Rates Cone Problem

Related Rates – How Things Change Over Time

Steps 1 & 2: Read and Sketch

First, we will sketch and label a cone, vertex down, and identify all pertinent information.

\begin{equation} \frac{d V}{d t}=3^{f t^{3}} / \mathrm{sec} \end{equation}

\begin{equation} \frac{d h}{d t}=? \end{equation}

\begin{equation} h=4 f t \end{equation}

related rates cone

Related Rates — Cone Problem

Step 3: Find An Equation That Relates The Unknown Variables

Because we were given the rate of change of the volume as well as the height of the cone, the equation that relates both V and h is the formula for the volume of a cone .

\begin{equation} V=\frac{1}{3} \pi r^{2} h \end{equation}

But here’s where it can get tricky. Our equation has three variables (V, r, and h), but we only have two derivatives, dh, and dV.

Hmmm, that means we have to reduce the number of variables so that the number of variables equals the number of derivatives.

How does that work?

Step 4: Simplify To Get Known & Unknown Variables

We use an incredibly useful ratio found by the similar triangles created from the cone above ( HINT: this ratio will be used quite often when solving related rate problems ). And in so doing, we will also create a proportion, using the conical tank’s original dimensions, and solve for r . This way, we can eliminate the r in volume formula.

\begin{equation} \frac{r}{h}=\frac{3}{5} \end{equation}

\begin{equation} r=\frac{3}{5} h \end{equation}

\begin{equation} V=\frac{1}{3} \pi\left(\frac{3}{5} h\right)^{2} h=\frac{3}{25} \pi h^{3} \end{equation}

\begin{equation} V=\frac{3}{25} \pi h^{3} \end{equation}

Step 5: Implicit Differentiation

We will now use implicit differentiation on both sides with respect to t.

\begin{equation} \frac{d V}{d t}=\frac{9}{25} \pi h^{2}\left(\frac{d h}{d t}\right) \end{equation}

Step 6: Substitute Back In

And lastly, we will substitute our given information and solve the unknown rate, dh/dt.

\begin{equation} \frac{d V}{d t}=\frac{9}{25} \pi h^{2}\left(\frac{d h}{d t}\right) \text { when } \frac{d V}{d t}=3 \text { and } h=4 \end{equation}

\begin{equation} 3=\frac{9}{25} \pi(4)^{2}\left(\frac{d h}{d t}\right) \end{equation}

\begin{equation} \frac{d h}{d t}=\frac{25}{48 \pi} \end{equation}

See, all we have to do is follow the steps and arrive at our answer!

And don’t worry, we will do two more cone problems in the video below, so you’ll become a master at these questions in no time!

Let’s get after it!

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

Related Rates - Ladder Problem

Ladder Sliding Down Wall

  • Overview of Related Rates + Tips to Solve Them
  • 00:02:58 – Increasing Area of a Circle
  • 00:12:30 – Expanding Volume of a Sphere
  • 00:21:15 – Expanding Volume of a Cube
  • 00:26:32 – Calculate the Speed of an Airplane
  • 00:39:13 – Conical Sand Pile
  • 00:51:19 – Conical Water Tank
  • 00:59:59 – Boat & Winch
  • 01:09:13 – Ladder Sliding Down A Wall

Related Rates

  • Related Rates – Overview
  • Example #1 – Increasing Area of a Circle
  • Example #2 – Boat & Winch

Get access to 6 more examples and over 450 HD videos with your subscription

Monthly and Yearly Plans Available

Get My Subscription Now

Still wondering if CalcWorkshop is right for you? Take a Tour and find out how a membership can take the struggle out of learning math.

5 Star Excellence award from Shopper Approved for collecting at least 100 5 star reviews

  • PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
  • EDIT Edit this Article
  • EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
  • Browse Articles
  • Learn Something New
  • Quizzes Hot
  • This Or That Game New
  • Train Your Brain
  • Explore More
  • Support wikiHow
  • About wikiHow
  • Log in / Sign up
  • Education and Communications
  • Mathematics

How to Solve Related Rates in Calculus

Last Updated: December 13, 2023 Fact Checked

This article was reviewed by Joseph Meyer . Joseph Meyer is a High School Math Teacher based in Pittsburgh, Pennsylvania. He is an educator at City Charter High School, where he has been teaching for over 7 years. Joseph is also the founder of Sandbox Math, an online learning community dedicated to helping students succeed in Algebra. His site is set apart by its focus on fostering genuine comprehension through step-by-step understanding (instead of just getting the correct final answer), enabling learners to identify and overcome misunderstandings and confidently take on any test they face. He received his MA in Physics from Case Western Reserve University and his BA in Physics from Baldwin Wallace University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 76,087 times.

Calculus is primarily the mathematical study of how things change. One specific problem type is determining how the rates of two related items change at the same time. The keys to solving a related rates problem are identifying the variables that are changing and then determining a formula that connects those variables to each other. Once that is done, you find the derivative of the formula, and you can calculate the rates that you need.

Interpreting the Problem

Step 1 Read the entire problem carefully.

  • This graphic presents the following problem: “Air is being pumped into a spherical balloon at a rate of 5 cubic centimeters per minute. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.”
  • From reading this problem, you should recognize that the balloon is a sphere, so you will be dealing with the volume of a sphere. You should also recognize that you are given the diameter, so you should begin thinking how that will factor into the solution as well.
  • Drawing a diagram of the problem can often be useful. In the case, you are to assume that the balloon is a perfect sphere, which you can represent in a diagram with a circle. Mark the radius as the distance from the center to the circle.

Step 2 Determine what you are asked to solve.

  • In the problem shown above, you should recognize that the specific question is about the rate of change of the radius of the balloon. Notice, however, that you are given information about the diameter of the balloon, not the radius. This will have to be adapted as you work on the problem. You should see that you are also given information about air going into the balloon, which is changing the volume of the balloon.

Step 3 List the functions and variables.

Setting up the Solution

Step 1 Determine the function that relates the variables.

Solving a Sample Problem Involving Triangles

Step 1 Read and understand the problem.

  • A baseball diamond is 90 feet square. A runner runs from first base to second base at 25 feet per second. How fast is he moving away from home plate when he is 30 feet from first base?
  • You can diagram this problem by drawing a square to represent the baseball diamond. Label one corner of the square as "Home Plate."

Step 2 Determine what you are being asked to solve.

  • One leg of the triangle is the base path from home plate to first base, which is 90 feet.

r

Solving a Sample Problem Involving a Cylinder

Step 1 Read and understand the problem.

  • Water flows at 8 cubic feet per minute into a cylinder with radius 4 feet. How fast is the water level rising?
  • Diagram this situation by sketching a cylinder. Make a horizontal line across the middle of it to represent the water height.

Step 2 Determine what you are being asked to solve.

Community Q&A

Community Answer

Video . By using this service, some information may be shared with YouTube.

  • Double check your work to help identify arithmetic errors. Thanks Helpful 1 Not Helpful 0
  • Remember that if the question gives you a decreasing rate (like the volume of a balloon is decreasing), then the rate of change against time (like dV/dt) will be a negative number. Thanks Helpful 1 Not Helpful 0
  • When you take the derivative of the equation, make sure you do so implicitly with respect to time. Thanks Helpful 1 Not Helpful 0

steps to solve related rates problems

You Might Also Like

Find the Equation of a Tangent Line

  • ↑ http://tutorial.math.lamar.edu/Classes/CalcI/RelatedRates.aspx
  • ↑ https://openstax.org/books/calculus-volume-1/pages/4-1-related-rates
  • ↑ https://faculty.math.illinois.edu/~lfolwa2/GW_101217_Sol.pdf
  • ↑ https://www.matheno.com/blog/related-rates-problem-cylinder-drains-water/

About This Article

Joseph Meyer

  • Send fan mail to authors

Reader Success Stories

Peter Dunkin

Peter Dunkin

Mar 14, 2021

Did this article help you?

Peter Dunkin

Nov 29, 2016

Brett Smiley

Brett Smiley

Mar 28, 2016

Am I a Narcissist or an Empath Quiz

Featured Articles

25+ Pro Tips To Help You Truly Enjoy Life

Trending Articles

Everything You Need to Know to Rock the Corporate Goth Aesthetic

Watch Articles

Cook Fresh Cauliflower

  • Terms of Use
  • Privacy Policy
  • Do Not Sell or Share My Info
  • Not Selling Info

wikiHow Tech Help Pro:

Develop the tech skills you need for work and life

  • 4.1 Related Rates
  • Introduction
  • 1.1 Review of Functions
  • 1.2 Basic Classes of Functions
  • 1.3 Trigonometric Functions
  • 1.4 Inverse Functions
  • 1.5 Exponential and Logarithmic Functions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • 2.1 A Preview of Calculus
  • 2.2 The Limit of a Function
  • 2.3 The Limit Laws
  • 2.4 Continuity
  • 2.5 The Precise Definition of a Limit
  • 3.1 Defining the Derivative
  • 3.2 The Derivative as a Function
  • 3.3 Differentiation Rules
  • 3.4 Derivatives as Rates of Change
  • 3.5 Derivatives of Trigonometric Functions
  • 3.6 The Chain Rule
  • 3.7 Derivatives of Inverse Functions
  • 3.8 Implicit Differentiation
  • 3.9 Derivatives of Exponential and Logarithmic Functions
  • 4.2 Linear Approximations and Differentials
  • 4.3 Maxima and Minima
  • 4.4 The Mean Value Theorem
  • 4.5 Derivatives and the Shape of a Graph
  • 4.6 Limits at Infinity and Asymptotes
  • 4.7 Applied Optimization Problems
  • 4.8 L’Hôpital’s Rule
  • 4.9 Newton’s Method
  • 4.10 Antiderivatives
  • 5.1 Approximating Areas
  • 5.2 The Definite Integral
  • 5.3 The Fundamental Theorem of Calculus
  • 5.4 Integration Formulas and the Net Change Theorem
  • 5.5 Substitution
  • 5.6 Integrals Involving Exponential and Logarithmic Functions
  • 5.7 Integrals Resulting in Inverse Trigonometric Functions
  • 6.1 Areas between Curves
  • 6.2 Determining Volumes by Slicing
  • 6.3 Volumes of Revolution: Cylindrical Shells
  • 6.4 Arc Length of a Curve and Surface Area
  • 6.5 Physical Applications
  • 6.6 Moments and Centers of Mass
  • 6.7 Integrals, Exponential Functions, and Logarithms
  • 6.8 Exponential Growth and Decay
  • 6.9 Calculus of the Hyperbolic Functions
  • A | Table of Integrals
  • B | Table of Derivatives
  • C | Review of Pre-Calculus

Learning Objectives

  • 4.1.1 Express changing quantities in terms of derivatives.
  • 4.1.2 Find relationships among the derivatives in a given problem.
  • 4.1.3 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, V , V , is related to the rate of change in the radius, r . r . In this case, we say that d V d t d V d t and d r d t d r d t 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 4.1

Inflating a balloon.

A spherical balloon is being filled with air at the constant rate of 2 cm 3 / sec 2 cm 3 / sec ( Figure 4.2 ). How fast is the radius increasing when the radius is 3 cm ? 3 cm ?

The volume of a sphere of radius r r centimeters is

Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, t t 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 V ′ ( t ) = 2 cm 3 / sec . V ′ ( t ) = 2 cm 3 / sec . Therefore,

which implies

When the radius r = 3 cm, r = 3 cm,

Checkpoint 4.1

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

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

Problem-Solving Strategy

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.

Example 4.2

An airplane flying at a constant elevation.

An airplane is flying overhead at a constant elevation of 4000 ft . 4000 ft . A man is viewing the plane from a position 3000 ft 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 , 600 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.

As shown, x x denotes the distance between the man and the position on the ground directly below the airplane. The variable s s denotes the distance between the man and the plane. Note that both x x and s 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 . 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 x feet, creating a right triangle.

Step 2. Since x x denotes the horizontal distance between the man and the point on the ground below the plane, d x / d t d x / d t represents the speed of the plane. We are told the speed of the plane is 600 ft/sec. Therefore, d x d t = 600 d x d t = 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 d s / d t d s / d t when x = 3000 ft . x = 3000 ft .

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

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 d s d t d s d t when x = 3000 ft . x = 3000 ft . Since the speed of the plane is 600 ft/sec , 600 ft/sec , we know that d x d t = 600 ft/sec . d x d t = 600 ft/sec . We are not given an explicit value for s ; s ; however, since we are trying to find d s d t d s d t when x = 3000 ft , x = 3000 ft , we can use the Pythagorean theorem to determine the distance s s when x = 3000 x = 3000 and the height is 4000 ft . 4000 ft . Solving the equation

for s , s , we have s = 5000 ft s = 5000 ft at the time of interest. Using these values, we conclude that d s / d t d s / d t 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 x ( t ) x ( t ) and s ( t ) 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 x ( t ) = 3000 into the equation before differentiating, our equation would have been

After differentiating, our equation would become

As a result, we would incorrectly conclude that d s d t = 0 . d s d t = 0 .

Checkpoint 4.2

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 ? 300 ft/sec ?

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

Example 4.3

Chapter opener: a rocket launch.

A rocket is launched so that it rises vertically. A camera is positioned 5000 ft 5000 ft from the launch pad. When the rocket is 1000 ft 1000 ft above the launch pad, its velocity is 600 ft/sec . 600 ft/sec . Find the necessary rate of change of the camera’s angle as a function of time so that it stays focused on the rocket.

Step 1. Draw a picture introducing the variables.

Let h h denote the height of the rocket above the launch pad and θ θ 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 1000 ft off the ground. That is, we need to find d θ d t d θ d t when h = 1000 ft . h = 1000 ft . At that time, we know the velocity of the rocket is d h d t = 600 ft/sec . d h d t = 600 ft/sec .

Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: h h and θ . θ . 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 tan θ tan θ 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 t , t , we obtain

Step 5. We want to find d θ d t d θ d t when h = 1000 ft . h = 1000 ft . At this time, we know that d h d t = 600 ft/sec . d h d t = 600 ft/sec . We need to determine sec 2 θ . sec 2 θ . Recall that sec θ sec θ 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 . 5000 ft . To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is 5000 ft , 5000 ft , the length of the other leg is h = 1000 ft , h = 1000 ft , and the length of the hypotenuse is c c feet as shown in the following figure.

We see that

and we conclude that the hypotenuse is

Therefore, when h = 1000 , h = 1000 , we have

Recall from step 4 that the equation relating d θ d t d θ d t to our known values is

When h = 1000 ft , h = 1000 ft , we know that d h d t = 600 ft/sec d h d t = 600 ft/sec and sec 2 θ = 26 25 . sec 2 θ = 26 25 . Substituting these values into the previous equation, we arrive at the equation

Therefore, d θ d t = 3 26 rad/sec . d θ d t = 3 26 rad/sec .

Checkpoint 4.3

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 4000 ft from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket is 2000 ft 2000 ft off the ground?

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.

Example 4.4

Water draining from a funnel.

Water is draining from the bottom of a cone-shaped funnel at the rate of 0.0 3 ft 3 /sec . 0.0 3 ft 3 /sec . The height of the funnel is 2 ft and the radius at the top of the funnel is 1 ft . 1 ft . At what rate is the height of the water in the funnel changing when the height of the water is 1 2 ft ? 1 2 ft ?

Step 1: Draw a picture introducing the variables.

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

Step 2: We need to determine d h d t d h d t when h = 1 2 ft . h = 1 2 ft . We know that d V d t = −0.03 ft 3 /sec . d V d t = −0.03 ft 3 /sec .

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, r h = 1 2 r h = 1 2 or r = h 2 . r = h 2 . 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 t , t , we obtain

Step 5: We want to find d h d t d h d t when h = 1 2 ft . h = 1 2 ft . Since water is leaving at the rate of 0.0 3 ft 3 /sec , 0.0 3 ft 3 /sec , we know that d V d t = −0.03 ft 3 /sec . d V d t = −0.03 ft 3 /sec . Therefore,

It follows that

Checkpoint 4.4

At what rate is the height of the water changing when the height of the water is 1 4 ft ? 1 4 ft ?

Section 4.1 Exercises

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

Find d y d t d y d t at x = 1 x = 1 and y = x 2 + 3 y = x 2 + 3 if d x d t = 4 . d x d t = 4 .

Find d x d t d x d t at x = −2 x = −2 and y = 2 x 2 + 1 y = 2 x 2 + 1 if d y d t = −1 . d y d t = −1 .

Find d z d t d z d t at ( x , y ) = ( 1 , 3 ) ( x , y ) = ( 1 , 3 ) and z 2 = x 2 + y 2 z 2 = x 2 + y 2 if d x d t = 4 d x d t = 4 and d y d t = 3 . d y d t = 3 .

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

[T] If two electrical resistors are connected in parallel, the total resistance (measured in ohms, denoted by the Greek capital letter omega, Ω ) Ω ) is given by the equation 1 R = 1 R 1 + 1 R 2 . 1 R = 1 R 1 + 1 R 2 . If R 1 R 1 is increasing at a rate of 0.5 Ω / min 0.5 Ω / min and R 2 R 2 decreases at a rate of 1.1 Ω/min , 1.1 Ω/min , at what rate does the total resistance change when R 1 = 20 Ω R 1 = 20 Ω and R 2 = 50 Ω R 2 = 50 Ω ?

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 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 20 ft 20 ft away from the wall, how fast does the ladder move up the wall 5 sec 5 sec after we start pushing?

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

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 12 mph 12 mph north. After you traveled 4 mi, 4 mi, at what rate is the distance between you changing?

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

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

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?

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?

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?

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.

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?

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

The side of a cube increases at a rate of 1 2 1 2 m/sec. Find the rate at which the volume of the cube increases when the side of the cube is 4 m.

The volume of a cube decreases at a rate of 10 m 3 /s. Find the rate at which the side of the cube changes when the side of the cube is 2 m.

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

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

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

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.

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.

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 π / 6 . π / 6 .

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 cm 2 /sec. 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 cm 2 .

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.

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 ft 3 /min?

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 ft 3 /min.

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.

A vertical cylinder is leaking water at a rate of 1 ft 3 /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?

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.

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 5 m 3 /min . 5 m 3 /min . At what rate does the height of the water change when the water is 1 m deep?

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 2 3 2 3 m 3 /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 ft 3 /min. The radius of the pool is 10 ft. The formula for the volume of a partial hemisphere is V = πh 6 ( 3 r 2 + h 2 ) V = πh 6 ( 3 r 2 + h 2 ) where h h is the height of the water and r r is the radius of the water.

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

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

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

Gravel is being unloaded from a truck and falls into a pile shaped like a cone at a rate of 10 ft 3 /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.

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.

You are stationary on the ground and are watching a bird fly horizontally at a rate of 10 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?

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.

A lighthouse, L , is on an island 4 mi away from the closest point, P , 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?

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.

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?

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

[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?

[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.)

[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?

[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?

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License and you must attribute OpenStax.

Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction
  • Authors: Gilbert Strang, Edwin “Jed” Herman
  • Publisher/website: OpenStax
  • Book title: Calculus Volume 1
  • Publication date: Mar 30, 2016
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/calculus-volume-1/pages/1-introduction
  • Section URL: https://openstax.org/books/calculus-volume-1/pages/4-1-related-rates

© Jul 17, 2023 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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

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.

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]?

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

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

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],

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.

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.

Example: 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.

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.

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 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].

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

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

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

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?

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

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.

Example: 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.

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]

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].

  • 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

Footer Logo Lumen Candela

Privacy Policy

Fiveable

Find what you need to study

Practice Quizzes

4.5 Solving Related Rates Problems

1 min read • june 7, 2020

Welcome back to AP Calculus with Fiveable! In this session, we're diving into the world of related rates. Get ready to apply your calculus skills to real-world scenarios and solve problems that involve the rates at which quantities change with respect to time. 🕰️

🔗 Related Rates Basics

Related rates problems involve finding the rate at which a variable changes concerning the rate of change of another related variable. These scenarios may involve geometric figures and equations that connect different variables to time.

To review related rates, check out the previous Fiveable guide: Introduction to Related Rates .

🪜 Steps to Solve Related Rates Problems

When you first look at a related rates problem, you will be presented with so much information that you may feel overwhelmed. It’s okay to take a step back and organize before solving anything out!

The following steps will help you manage a related rates problem efficiently.

  • 📚 Read the Problem Carefully: Identify values that are significant to the problem. You may find it helpful to circle, underline, or rewrite these values off to the side.
  • ✏️ Draw a Diagram: Visualizing the situation by drawing a diagram can help us understand how each of the variables are changing. Make sure to accurately label variables and indicate their rates of change.
  • 🏁 Set up an Equation: Use the information given to set up an equation that relates the variables involved. Usually, these equations are geometric, or given.
  • 💫 Implicit Differentiation: Differentiate the equation implicitly with respect to time (t). This usually involves applying the chain rule to each term containing a variable.
  • 🔌 Substitute Known Values: Substitute any known values and rates of change into the derivative equation.
  • ✅ Solve for the Desired Rate: Solve the final equation for the rate you're asked to find. Be sure to check that your units match what is expected!

Time to put these steps to action…

🧮 Related Rates Practice Problems

Let’s work on a few questions and make sure we have the concept down!

1) Expanding Rectangle

An expanding rectangle has one side of length 6 feet and the other side of length 8 feet. If the length of the shorter side is increasing at a rate of 2 feet per minute, at what rate is the area of the rectangle increasing?

Let’s begin by identifying some key facts and drawing a picture of the scenario.

Known Information

  • l 0 = 8 l_0 = 8 l 0 ​ = 8 ft
  • w 0 = 6 w_0 = 6 w 0 ​ = 6 ft
  • d w d t = + 2 f t m i n \frac {dw}{dt} = +2 \frac{ft}{min} d t d w ​ = + 2 min f t ​

Important Equations

  • A r e a = l e n g t h ⋅ w i d t h = l ⋅ w Area = length \cdot width = l\cdot w A re a = l e n g t h ⋅ w i d t h = l ⋅ w

Screenshot 2024-01-04 at 1.27.58 PM.png

Image created with Virtual Graph Paper

Now we can set up the equation. When looking at the image, we notice that only one of the variables, the width, is changing. Since the length is not changing, we conclude that d l d t = 0 \frac {dl}{dt} = 0 d t d l ​ = 0 .

After directly plugging in any constants into the equation, we can find that the A r e a = 8 ⋅ w Area = 8\cdot w A re a = 8 ⋅ w at any time. Now, let’s differentiate using the chain rule.

Therefore, the area A A A of the rectangle is increasing at a rate of 16 feet per minute. Nice! 👍

2) Sliding Ladder

Consider a 13 13 13 meter long ladder leaning against the wall. If the distance between the wall and the bottom of the ladder is increasing at 3 3 3 m/s, how fast is the distance between the ground and the top of the ladder changing when the bottom of the ladder is 5 5 5 feet away from the wall when it begins sliding?

This question unloaded a lot of information all at once, so take a second to reread the scenario before we get to work. You can do it! 🙌

  • x 0 = 5 x_0 = 5 x 0 ​ = 5 m
  • d x d t = + 3 m s \frac {dx}{dt} = +3 \frac{m}{s} d t d x ​ = + 3 s m ​
  • d y d t = ? m s \frac {dy}{dt} = ? \frac{m}{s} d t d y ​ = ? s m ​
  • Pythagorean Theorem

Screenshot 2024-01-05 at 5.42.52 PM.png

Now we can use the equation relating the side lengths of the triangle made with the ladder, wall, and ground. When looking at the image, we notice that the hypotenuse of the triangle is not changing and we conclude that d z d t = 0 \frac {dz}{dt} = 0 d t d z ​ = 0 .

After directly plugging in any constants into the equation, we can relate the sides with the following equation: x 2 + y 2 = 1 3 2 x^2 + y^2 = 13^2 x 2 + y 2 = 1 3 2 . And now we can finally take the derivative! Don’t forget to use the chain rule since we are taking the derivative with respect to time. Taking the derivative of the main equation gives us:

Now let’s plug in all known variables and see what we have to solve for.

You may have noticed that we have two missing pieces of information: the value of y y y when x = 5 x = 5 x = 5 and d y d t \frac {dy}{dt} d t d y ​ which is our ultimate solution.

We can solve for the value of y y y by using the Pythagorean Theorem! A triangle will a side length of 5 5 5 and a hypotenuse of 13 13 13 is actually part of thre Pythagorean Triple ( 5 , 12 , 13 ) (5,12,13) ( 5 , 12 , 13 ) . But if you didn’t pick that up, we can quickly solve for the missing value of y y y by just plugging into the equation.

We can then isolate y y y to get…

We're almost there!

The final step is to solve for d y d t \frac {dy}{dt} d t d y ​ . Using algebra, we can isolate d y d t \frac {dy}{dt} d t d y ​ and get 2 ( 5 ) ⋅ 3 + 2 ( 12 ) ⋅ d y d t = 0 2(5) \cdot 3 + 2(12)\cdot \frac {dy}{dt} = 0 2 ( 5 ) ⋅ 3 + 2 ( 12 ) ⋅ d t d y ​ = 0 and 15 = − 12 d y d t 15 = -12 \frac {dy}{dt} 15 = − 12 d t d y ​ .

Therefore, we find that the distance between the top of the ladder and the ground is decreasing at a rate of − 5 4 \frac{-5}{4} 4 − 5 ​ meters per second when the bottom of the ladder is five feet away from the wall.

You’re on fire! Amazing job working through this difficult question. 🔥

Great work! 🙌 You now have the tools to tackle related rates problems. These types of questions often appear in the AP Calculus exam as both multiple choice and free response, challenging you to apply calculus concepts to real-world situations. Keep practicing, and you'll be able to solve related rates problems with ease!

Image Courtesy of Giphy

Fiveable

Student Wellness

Stay connected.

© 2024 Fiveable Inc. All rights reserved.

The following problems involve the concept of Related Rates. In short, Related Rates problems combine word problems together with Implicit Differentiation, an application of the Chain Rule. Recall that if $ y=f(x) $, then $ D \{y \} = \displaystyle{ dy \over dx } = f'(x)=y' $. For example, implicitly differentiating the equation $$ y^3+y^2= y+1$$ would be $$ D\{y^3+y^2\} = D\{ y+1 \} \ \ \ \ \longrightarrow $$ $$ 3y^2 \cdot y'+ 2y \cdot y' = y' + 0 $$ If $ x=f(t) $ and $ y=g(t) $, then $ D\{x\} = \displaystyle{ dx \over dt } = f'(t) $ and $ D\{y\} = \displaystyle{ dy \over dt } = g'(t) $ . For example, implicitly differentiating the equation $$ x^3+y^2= x+y+3$$ would be $$ D\{x^3+y^2\} = D\{ x+y+3 \} \ \ \ \ \longrightarrow $$ $$ 3x^2 \cdot \displaystyle{ dx \over dt } + 2y \cdot \displaystyle{ dy \over dt } = \displaystyle{ dx \over dt } + \displaystyle{ dy \over dt } + 0 $$ In all of the following Related Rates Problems, it will be assumed that each variable function $y$ is a function of time $t$. For that reason, I will always use Leibniz notation and not the ambiguous prime notation for derivatives, i.e., i will use $$ \displaystyle{ dy \over dt } \ \ \ \ instead \ of \ \ \ \ y' $$ Here is my strategy for approaching and solving Related Rates Problems: 1.) Read the problem slowly and carefully. 2.) Draw an appropriate sketch. 3.) Introduce and define appropriate variables. Use variables if quantities are changing. Use constants if quantities are not changing. 4.) Read the problem again. 5.) Clearly label the sketch using your variables. 6.) State what information is given in the problem. 7.) State what information is to be determined or found. 8.) Use a given equation or create an appropriate equation relating the given variables. 9.) Differentiate this equation with respect to the time variable $t$. 10.) Plug in the given rates and numbers to the differentiated equation. 11.) Solve for the unknown rate. 12.) Put proper units on your final answer. EXAMPLE 1: Consider a right triangle which is changing shape in the following way. The horizontal leg is increasing at the rate of $ 5 \ in./min. $ and the vertical leg is decreasing at the rate of $ 6 \ in./min $. At what rate is the hypotenuse changing when the horizontal leg is $ 12 \ in. $ and the vertical leg is $ 9 \ in. $ ? Draw a right triangle with legs labeled $x$ and $y$ and hypotenuse labeled $z$, and assume each edge is a function of time $t$. GIVEN: $ \ \ \ \displaystyle{ dx \over dt } = 5 \ in./min. \ $ and $ \ \displaystyle{ dy \over dt }= -6 \ in./min. $ FIND: $ \ \ \ \displaystyle{ dz \over dt } $ when $ x=12 \ in. $ and $ y=9 \ in $. Use the Pythagorean Theorem to get the equation $$ x^2 + y^2 = z^2 $$ Now differentiate this equation with repect to time $t $ getting $$ D \{ x^2 + y^2\} = D \{z^2\} \ \ \ \longrightarrow $$ $$ 2x \displaystyle{ dx \over dt } + 2y \displaystyle{ dy \over dt } = 2z \displaystyle{ dz \over dt } \ \ \ \longrightarrow \ \ $$ (Multiply both sides of the equation by $1/2$.) $$ x \displaystyle{ dx \over dt } + y \displaystyle{ dy \over dt } = z \displaystyle{ dz \over dt } \ \ \ \ \ \ \ \ \ \ \ \ \ \ (DE) $$ Now let $ x=12 $ and $ y=9 $ and solve for $z$ using the Pythagorean Theorem. $$ 12^2+9^2= z^2 \ \ \ \longrightarrow \ \ \ z^2=225 \ \ \ \longrightarrow \ \ \ z=15 $$ Plug in all given rates and values to the equation $(DE)$ getting $$ (12)(5) + (9)(-6) = (15) \displaystyle{ dz \over dt } \ \ \ \longrightarrow $$ $$ 6 = 15 \displaystyle{ dz \over dt } \ \ \ \longrightarrow $$ $$ \displaystyle{ dz \over dt } = {6 \over 15} = { 2 \over 5} \ in/min. $$ In the list of Related Rates Problems which follows, most problems are average and a few are somewhat challenging. PROBLEM 1 : The edge of a square is increasing at the rate of $ \ 3 \ cm/sec $. At what rate is the square's $ \ \ \ \ $ a.) perimeter changing $ \ \ \ \ $ b.) area changing when the edge of the square is $10 \ cm.$ ? Click HERE to see a detailed solution to problem 1.

  • Exponent Laws and Logarithm Laws
  • Trig Formulas and Identities
  • Differentiation Rules
  • Trig Function Derivatives
  • Table of Derivatives
  • Table of Integrals
  • Calculus Home

Jump to solved problems

  • Evaluating Limits
  • Limits at Infinity
  • Limits at Infinity with Square Roots
  • Calculating Derivatives
  • Equation of a Tangent Line
  • Mean Value Theorem & Rolle’s Theorem
  • Garden fence
  • Least expensive open-topped can
  • Printed poster
  • Snowball melts
  • Snowball melts, area decreases at given rate
  • How fast is the ladder’s top sliding
  • Angle changes as a ladder slides
  • Lamp post casts shadow of man walking
  • Water drains from a cone
  • Given an equation, find a rate

Have a question you could use some help with? Please post on our Forum:

Community.Matheno.com

It's free for your use, just to support your learning. We'd love to help!

Related Rates

Calculus related rates problem solving strategy.

We will use the steps outlined below to solve each Related Rates problem on this site, step-by-step, every single time. We hope that this will help you see the strategy we’re using so you can learn it too, and then be able to apply it to all of your problems, especially those on your exams. That’s as opposed to learning just how to solve a particular problem on your homework, say, since you may well never see that specific problem again.

  • Draw a picture of the physical situation.
  • Be sure to label as a variable any value that changes as the situation progresses; don’t substitute a number for it yet.
  • a simple geometric fact (like the relation between a sphere’s volume and its radius, or the relation between the volume of a cylinder and its height); or
  • a trigonometric function (like $\tan{\theta}$ = opposite/adjacent); or
  • similar triangles; or
  • the Pythagorean theorem.
  • Take the derivative with respect to time of both sides of your equation. Remember the chain rule.
  • Solve for the quantity you’re after.

On our Related Rates page , we solve an example of each of the approaches listed in 2.B above: i. Geometric fact. Typical problems: A circle’s radius changes, a snowball melts, a rectangle’s height and/or width changes. ii. Trig function. Typical problems: A searchlight rotates, a rocket takes off, a kite travels horizontally. iii. Similar triangles. Water fills a cone or trough, sand falls onto a conical pile, person walks away from a light pole that casts a shadow. iv. Pythagorean theorem. Typical problems: Cars/ships/joggers move along 90 degree paths, baseball players run along the diamond, boat is pulled toward a dock. To see an example of each type, please visit our Related Rates page .

Share a link to this screen:

  • Share on Facebook
  • Share on Pinterest
  • Share on Reddit
  • Share on Telegram
  • Share on WhatsApp
  • Share on SMS
  • Email this Page

This site is free?!?

We don't charge for anything on this site, we don't run ads, and we will never sell your personal information.

We're passionate educators with a goal:

Provide high-quality, interactive materials to dedicated learners everywhere in the world, regardless of ability to pay, so they (you!) can learn well and excel.

☕ Buy us a coffee We're working to add more, and would appreciate your help to keep going! 😊

top

  • Terms of Use
  • Privacy Policy

Matheno ®

Berkeley, California

LinkedIn

AP ® is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this site.

© 2014–2024 Matheno, Inc.

  • Your answers to multiple choice questions;
  • Your self-chosen confidence rating for each problem, so you know which to return to before an exam (super useful!);
  • Your progress, and specifically which topics you have marked as complete for yourself.

Your selections are for your use only, and we do not share your specific data with anyone else. We do use aggregated data to help us see, for instance, where many students are having difficulty, so we know where to focus our efforts.

You will also be able to post any Calculus questions that you have on our Forum , and we'll do our best to answer them!

We believe that free, high-quality educational materials should be available to everyone working to learn well. There is no cost to you for having an account, other than our gentle request that you contribute what you can, if possible, to help us maintain and grow this site.

Please join our community of learners. We'd love to help you learn as well as you can!

Library homepage

  • school Campus Bookshelves
  • menu_book Bookshelves
  • perm_media Learning Objects
  • login Login
  • how_to_reg Request Instructor Account
  • hub Instructor Commons
  • Download Page (PDF)
  • Download Full Book (PDF)
  • Periodic Table
  • Physics Constants
  • Scientific Calculator
  • Reference & Cite
  • Tools expand_more
  • Readability

selected template will load here

This action is not available.

Mathematics LibreTexts

5.2: Related Rates

  • Last updated
  • Save as PDF
  • Page ID 83939

  • Mike May, S.J. & Anneke Bart
  • Saint Louis University

As we have seen, \(\frac{dy}{dx}\) is the instantaneous rate of change of \(y\) with respect to \(x\text{.}\) In chapter 4 we learned techniques for finding \(\frac{dy}{dx}\) when \(y\) is defined as a function of \(x\text{.}\) In the last section we learned how to use implicit differentiation to find \(\frac{dy}{dx}\) when we were given an equation in \(x\) and \(y\text{.}\) In this section we want find \(\frac{dy}{dx}\) when \(x\) and \(y\) are both described in terms of another variable. As with the section on related rates, we will start with an example where we can solve the problem by eliminating the extra variable before differentiating, and then look at how to solve with related rates.

Example 5.2.1: Change in Revenue with Respect to Expense, Doable Two Ways.

We can buy widgets for $10. The demand price of widgets is $20 minus 0.1 times the quantity to be sold. Find the derivative of revenue with respect to expense.

Solution a : The revenue and cost functions for widgets depend on the quantity (q). The formulas for revenue and cost are:

\[ revenue=q(20-0.1q)=20q-0.1q^2. \nonumber \]

\[ cost=10q. \nonumber \]

We can solve the second equation for quantity and substitute back into the first equation. This now gives us the revenue function in terms of cost (c).

\[ quantity =.1*c. \nonumber \]

\[ revenue=2c-0.001 c^2. \nonumber \]

It is straightforward to take the derivative.

\[ \frac{d\ revenue}{d\ cost}=2-0.002*cost. \nonumber \]

Note that the derivative is positive for cost between $0 and $1000. This implies that the revenue is rising until the cost is $1000. After we hit a cost of $1000, the derivative becomes negative. This indicates that the revenue will actually decrease.

The alternative method is to differentiate the equations for revenue (\(r\)) and cost (\(c\)) with respect to quantity (\(q\)), and find the two derivatives \(\frac{d\ r}{d\ q}\) and \(\frac{d\ c}{d\ q}\text{,}\) then treat them as fractions. The derivative we want is the quotient of these fractions.

Solution b : The revenue and cost functions for widgets are the same as above.

\[ revenue=20q-0.1 q^2. \nonumber \]

We now differentiate

\[ \frac{d\ r}{d\ q} =20-0.2 q. \nonumber \]

\[ \frac{d\ c}{d\ q}=10. \nonumber \]

We divide these derivatives to get the desired derivative.

\[ \frac{change\ in\ revenue}{change\ in\ cost}: \frac{d\ r}{d\ c}=\frac{d\ r}{d\ q}/\frac{d\ c}{d\ q}=(20-0.2 q)/10 \nonumber \]

Substituting q =.1 c gives the same solution we had from the first method.

When using the method of related rates, we act as if the derivatives are fractions that we can multiply or divide to obtain the appropriate fraction. We want to use a bit of caution with that approach, because it does not work with higher order derivatives, or with derivatives of functions of several variables. However, for derivatives of one variable the intuition works. Once again, if we zoom in far enough, the curve will look like a straight line and the derivative is the quotient of rise over run.

For the first example we could use both methods. We either use algebra to eliminate the extra variable or find two rates of change and combine them to find the rate we are interested in. For some problems we will only have one choice, either because the algebra is too hard, or because we have been given partial information and the algebraic method is impossible.

Example 5.2.2: Change in Revenue with Respect to Expense, Q Elimination Hard.

The cost (\(c(q)\)) and revenue (\(r(q)\)) equations for gizmos are both given in terms of quantity (q)

\[ r(q)=30q-0.1 q^2-0.001 q^3. \nonumber \]

\[ c(q)=500+10q-0.01 q^2. \nonumber \]

Find the derivative of revenue with respect to cost (i.e. \(\frac{dr}{dc}\) when \(q=50\text{.}\)

Since the cost is quadratic in quantity, solving for revenue as a function of cost involves more work than we need for this problem. The appropriate derivatives are:

\[ \frac{d\ r}{d\ q} =30-0.2 q-0.003 q^2. \nonumber \]

\[ \frac{d\ c}{d\ q}=10-0.02 q. \nonumber \]

When q =50, we have

\[ \frac{d\ r}{d\ q} =30-0.2*50-0.003*50^2=12.5. \nonumber \]

\[ \frac{d\ c}{d\ q}=10-0.02*50=9. \nonumber \]

\[ \frac{d\ r}{d\ c} =\frac{d\ r}{d\ q}/\frac{d\ c}{d\ q} =\frac{12.5}{9}\approx 1.389 \nonumber \]

This means that when \(quantity=50\text{,}\) there is an increase of $1.39 for every dollar increase in cost of investment.

Example 5.2.3: Change in Revenue with Respect to Expense, Long Variable Names.

We have the following cost and revenue information for whatchamacallits:

\[ revenue=50*quantity-0.01*quantity^2. \nonumber \]

\[ \frac{d\ cost}{d\ quantity}=15. \nonumber \]

Find the derivative of revenue with respect to cost when quantity=100.

In this example we do not have a formula that lets us solve for revenue as a function of cost, so we must use the method of related rates. The other derivatives is:

\[ \frac{d\ revenue}{d\ quantity} =50-0.02*quantity. \nonumber \]

When \(quantity=100\text{,}\) we have \(\frac{d\ revenue}{d\ quantity} =50-0.02*100=48\text{.}\) Thus

\[ \frac{d\ revenue}{d\ cost} =\frac{d\ revenue}{d\ quantity}/\frac{d\ cost}{d\ quantity} =\frac{48}{15}=3.2. \nonumber \]

Related rates are also useful when we are looking at a two-step process and we are interested in the rate of the combined process.

Example 5.2.4: Composition of Functions.

We are producing widgets (w). The manufacturing process turns goop (g) into sludge (s) and sludge into widgets. The yield equations in the appropriate units are:

\[ widgets(sludge)=4*sludge-0.1*sludge^2, \nonumber \]

or in shorthand notation: \(w(s)=4 s-0.1 s^2\)

\[ sludge(goop)=3*goop+.1*goop^2. \nonumber \]

or in shorthand notation: \(s(g)=3 g+.1 g^2\)

Find the derivative of widgets with respect to goop when \(goop=10\text{.}\)

We note that when \(g=10\text{,}\) we have \(s=3*10+.1*10^2=40\text{.}\) In this example we will take the derivatives of our equation. We will then multiply them to get the derivative we want.

\[ \frac{d\ widgets}{d\ sludge}=\frac{dw}{ds}=4-0.02*s. \nonumber \]

\[ \frac{d\ sludge}{d\ goop}=\frac{ds}{dg}=3+.2*g. \nonumber \]

When \(goop=10\text{,}\) \(\frac{d\ w}{d\ s} =(4-0.02*40)=3.2\text{,}\) and \(\frac{d\ s}{d\ g}\text{.}\) We need to multiply the derivatives to cancel the \(d\ s\text{.}\)

\[ \frac{dw}{dg} =\frac{dw}{ds}*\frac{ds}{dg}=(3.2)(5)=16. \nonumber \]

Thus the rate of widget production is increasing by 16 units per increase in on unit of goop at that point.

We often run into situations where several quantities are related by some constraint or equation. In such situations we will want to know the rate at which quantities are changing with time. The technique of related rates gives us a way to move from one rate with respect to time to another. Recall the Cobb-Douglas equation from the last section.

\[ Y=AL^\alpha K^\beta, \nonumber \]

where \(Y\text{,}\) \(L\text{,}\) and \(K\) represent total production, labor, and capital, respectively. If we know the rate of investment in capital equipment, we will be interested in the rate of change of labor with respect to time. An interesting question is to ask for the rate of change of capital with respect to labor, or how increasing or reducing capital investment will raise or lower labor costs.

Example 5.2.5: Cobb-Douglas.

A gizmo manufacturer has a production function given by

\[ Y=50L^.75 K^.25. \nonumber \]

The manufacturer currently uses 16 units of labor and 81 units of capital. The total production is constant but the manufacturer is investing in automation. The derivative of capital with respect to time is 5. How fast is the amount of labor needed changing?

We are gong to assume that both labor and capital are functions of time and the Y is a constant. We start by implicitly differentiating our equation with respect to time.

\[ \frac{d}{dt}(Y=50L^.75 K^.25) \nonumber \]

\[ 0=50*(0.75*L^{-0.25}*\frac{dL}{dt}*K^{.25}+L^{.75}*.25*K^{-0.75}*\frac{dK}{dt}) \nonumber \]

We now substitute in for the values of \(K\text{,}\) \(L\text{,}\) and \(\frac{dK}{dt}\text{,}\) which were given.

\[ 0=50*(0.75*16^{-0.25}*\frac{d\ L}{(d\ t}*81^{.25}+16^{.75}*.25*81^{-0.75}*2) \nonumber \]

\[ 0=3/4*1/2*\frac{dL}{dt}*3+8*1/4*1/27*2 \nonumber \]

\[ \frac{dL}{dt}=-32/243\approx -0.1317 \nonumber \]

If capital is increasing at a rate of 2 per unit of time, then labor is decreasing at a rate of -0.1317 per unit of time.

The related rates technique is an application of the chain rule. We use this technique when we have either three variables. We may want the rate of change of one variable with respect to a second and those variables may be connected through equations using a third variable. We may also want to relate the rate of change of two variables with respect to time. We take advantage of the fact that we can think of a derivative as a fraction of two small values. We either want to multiply or divide theses fractions to obtain the desired derivative.

Solver Title

Practice

Generating PDF...

  • Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Mean, Median & Mode
  • Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems
  • Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
  • Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
  • Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
  • Linear Algebra Matrices Vectors
  • Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
  • Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
  • Physics Mechanics
  • Chemistry Chemical Reactions Chemical Properties
  • Finance Simple Interest Compound Interest Present Value Future Value
  • Economics Point of Diminishing Return
  • Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time
  • Pre Algebra
  • Pre Calculus
  • Linear Algebra
  • Trigonometry
  • Conversions

Click to reveal more operations

Most Used Actions

Number line.

  • x^{2}-x-6=0
  • -x+3\gt 2x+1
  • line\:(1,\:2),\:(3,\:1)
  • prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x)
  • \frac{d}{dx}(\frac{3x+9}{2-x})
  • (\sin^2(\theta))'
  • \lim _{x\to 0}(x\ln (x))
  • \int e^x\cos (x)dx
  • \int_{0}^{\pi}\sin(x)dx
  • \sum_{n=0}^{\infty}\frac{3}{2^n}

step-by-step

related rates

  • Practice Makes Perfect Learning math takes practice, lots of practice. Just like running, it takes practice and dedication. If you want... Read More

Please add a message.

Message received. Thanks for the feedback.

Read our research on: Immigration & Migration | Podcasts | Election 2024

Regions & Countries

How americans view the situation at the u.s.-mexico border, its causes and consequences, 80% say the u.s. government is doing a bad job handling the migrant influx.

steps to solve related rates problems

Pew Research Center conducted this study to understand the public’s views about the large number of migrants seeking to enter the U.S. at the border with Mexico. For this analysis, we surveyed 5,140 adults from Jan. 16-21, 2024. Everyone who took part in this survey is a member of the Center’s American Trends Panel (ATP), an online survey panel that is recruited through national, random sampling of residential addresses. This way nearly all U.S. adults have a chance of selection. The survey is weighted to be representative of the U.S. adult population by gender, race, ethnicity, partisan affiliation, education and other categories. Read more about the ATP’s methodology .

Here are the questions used for the report and its methodology .

The growing number of migrants seeking entry into the United States at its border with Mexico has strained government resources, divided Congress and emerged as a contentious issue in the 2024 presidential campaign .

Chart shows Why do Americans think there is an influx of migrants to the United States?

Americans overwhelmingly fault the government for how it has handled the migrant situation. Beyond that, however, there are deep differences – over why the migrants are coming to the U.S., proposals for addressing the situation, and even whether it should be described as a “crisis.”

Factors behind the migrant influx

Economic factors – either poor conditions in migrants’ home countries or better economic opportunities in the United States – are widely viewed as major reasons for the migrant influx.

About seven-in-ten Americans (71%), including majorities in both parties, cite better economic opportunities in the U.S. as a major reason.

There are wider partisan differences over other factors.

About two-thirds of Americans (65%) say violence in migrants’ home countries is a major reason for why a large number of immigrants have come to the border.

Democrats and Democratic-leaning independents are 30 percentage points more likely than Republicans and Republican leaners to cite this as a major reason (79% vs. 49%).

By contrast, 76% of Republicans say the belief that U.S. immigration policies will make it easy to stay in the country once they arrive is a major factor. About half as many Democrats (39%) say the same.

For more on Americans’ views of these and other reasons, visit Chapter 2.

How serious is the situation at the border?

A sizable majority of Americans (78%) say the large number of migrants seeking to enter this country at the U.S.-Mexico border is eithera crisis (45%) or a major problem (32%), according to the Pew Research Center survey, conducted Jan. 16-21, 2024, among 5,140 adults.

Related: Migrant encounters at the U.S.-Mexico border hit a record high at the end of 2023 .

Chart shows Border situation viewed as a ‘crisis’ by most Republicans; Democrats are more likely to call it a ‘problem’

  • Republicans are much more likely than Democrats to describe the situation as a “crisis”: 70% of Republicans say this, compared with just 22% of Democrats.
  • Democrats mostly view the situation as a major problem (44%) or minor problem (26%) for the U.S. Very few Democrats (7%) say it is not a problem.

In an open-ended question , respondents voice their concerns about the migrant influx. They point to numerous issues, including worries about how the migrants are cared for and general problems with the immigration system.

Yet two concerns come up most frequently:

  • 22% point to the economic burdens associated with the migrant influx, including the strains migrants place on social services and other government resources.
  • 22% also cite security concerns. Many of these responses focus on crime (10%), terrorism (10%) and drugs (3%).

When asked specifically about the impact of the migrant influx on crime in the United States, a majority of Americans (57%) say the large number of migrants seeking to enter the country leads to more crime. Fewer (39%) say this does not have much of an impact on crime in this country.

Republicans (85%) overwhelmingly say the migrant surge leads to increased crime in the U.S. A far smaller share of Democrats (31%) say the same; 63% of Democrats instead say it does not have much of an impact.

Government widely criticized for its handling of migrant influx

For the past several years, the federal government has gotten low ratings for its handling of the situation at the U.S.-Mexico border. (Note: The wording of this question has been modified modestly to reflect circumstances at the time).

Chart shows Only about a quarter of Democrats and even fewer Republicans say the government has done a good job dealing with large number of migrants at the border

However, the current ratings are extraordinarily low.

Just 18% say the U.S. government is doing a good job dealing with the large number of migrants at the border, while 80% say it is doing a bad job, including 45% who say it is doing a very bad job.

  • Republicans’ views are overwhelmingly negative (89% say it’s doing a bad job), as they have been since Joe Biden became president.
  • 73% of Democrats also give the government negative ratings, the highest share recorded during Biden’s presidency.

For more on Americans’ evaluations of the situation, visit Chapter 1 .

Which policies could improve the border situation?

There is no single policy proposal, among the nine included on the survey, that majorities of both Republicans and Democrats say would improve the situation at the U.S.-Mexico border. There are areas of relative agreement, however.

A 60% majority of Americans say that increasing the number of immigration judges and staff in order to make decisions on asylum more quickly would make the situation better. Only 11% say it would make things worse, while 14% think it would not make much difference.

Nearly as many (56%) say creating more opportunities for people to legally immigrate to the U.S. would make the situation better.

Chart shows Most Democrats and nearly half of Republicans say boosting resources for quicker decisions on asylum cases would improve situation at Mexico border

Majorities of Democrats say each of these proposals would make the border situation better.

Republicans are less positive than are Democrats; still, about 40% or more of Republicans say each would improve the situation, while far fewer say they would make things worse.

Opinions on other proposals are more polarized. For example, a 56% majority of Democrats say that adding resources to provide safe and sanitary conditions for migrants arriving in the U.S. would be a positive step forward.

Republicans not only are far less likely than Democrats to view this proposal positively, but far more say it would make the situation worse (43%) than better (17%).

Chart shows Wide partisan gaps in views of expanding border wall, providing ‘safe and sanitary conditions’ for migrants

Building or expanding a wall along the U.S.-Mexico border was among the most divisive policies of Donald Trump’s presidency. In 2019, 82% of Republicans favored expanding the border wall , compared with just 6% of Democrats.

Today, 72% of Republicans say substantially expanding the wall along the U.S. border with Mexico would make the situation better. Just 15% of Democrats concur, with most saying either it would not make much of a difference (47%) or it would make things worse (24%).

For more on Americans’ reactions to policy proposals, visit Chapter 3 .

Sign up for our Politics newsletter

Sent weekly on Wednesday

Report Materials

Table of contents, fast facts on how greeks see migrants as greece-turkey border crisis deepens, americans’ immigration policy priorities: divisions between – and within – the two parties, from the archives: in ’60s, americans gave thumbs-up to immigration law that changed the nation, around the world, more say immigrants are a strength than a burden, latinos have become less likely to say there are too many immigrants in u.s., most popular.

About Pew Research Center Pew Research Center is a nonpartisan fact tank that informs the public about the issues, attitudes and trends shaping the world. It conducts public opinion polling, demographic research, media content analysis and other empirical social science research. Pew Research Center does not take policy positions. It is a subsidiary of The Pew Charitable Trusts .

MarketBeat

7 growth stocks that will prove growth is back in 2024

Posted: February 15, 2024 | Last updated: February 15, 2024

<p><span>If you've been investing for any length of time, you've probably heard a lot of time-honored investment maxims. Things like "time in the market beats timing the market." One that I like to keep in mind is to "skate where the puck is moving."</span> </p> <p><span>There's a reason for that. Investors frequently believe they need special knowledge to be successful at investing. Let's be clear: you do have to put in the work. But the information you need to be a successful investor is not unknowable, even without a background in finance or accounting. </span> </p> <p><span>However, investors can choose from thousands of stocks, ETFs, and mutual funds. And that's just equities. There are also bonds, precious metals, real estate, and – for those so inclined – cryptocurrency to consider. It's impossible to stay on top of every emerging story. </span> </p> <p><span>Sometimes you need a little nudge. </span> </p> <p><span>This special presentation focuses on growth stocks that may be flying under investor's radars. Some of these stocks are already growing – and have room to grow some more. Others haven't participated in the rally but have strong growth potential in 2024 and beyond.</span> </p> <br> <br>

If you've been investing for any length of time, you've probably heard a lot of time-honored investment maxims. Things like "time in the market beats timing the market." One that I like to keep in mind is to "skate where the puck is moving."  

There's a reason for that. Investors frequently believe they need special knowledge to be successful at investing. Let's be clear: you do have to put in the work. But the information you need to be a successful investor is not unknowable, even without a background in finance or accounting.  

However, investors can choose from thousands of stocks, ETFs, and mutual funds. And that's just equities. There are also bonds, precious metals, real estate, and – for those so inclined – cryptocurrency to consider. It's impossible to stay on top of every emerging story.  

Sometimes you need a little nudge.  

This special presentation focuses on growth stocks that may be flying under investor's radars. Some of these stocks are already growing – and have room to grow some more. Others haven't participated in the rally but have strong growth potential in 2024 and beyond.  

<p>The first company on this list is from the fintech sector. But <a href="https://www.marketbeat.com/stocks/NYSE/NU/"><strong>Nu Holdings Ltd. (NYSE: NU)</strong></a> may not be well-known to investors. That's because it's not a U.S. company. However, Nu is the largest fintech bank in North America.  </p> <p>The digital-first bank was founded in 2013 as a way to disrupt the Latin American banking system, which is dominated by a small number of large banks. Among other things, this trapped customers in an ecosystem of high fees for limited services. </p> <p>The company has already signed up five million customers and has a total addressable market that can potentially bring in millions more. Revenue is growing year-over-year, and the bank is solidly profitable. Nu Holdings is also projecting earnings growth of 75% in the next 12 months.  </p> <p>The concern is how much of that growth is priced into a stock up 97% in the last 12 months. The <a href="https://www.marketbeat.com/stocks/NYSE/NU/price-target/">Nu Holdings analyst ratings on MarketBeat</a> show that analysts are beginning to bid NU stock higher, with <a href="https://www.marketbeat.com/stocks/NYSE/UBS/"><strong>UBS Group AG (NYSE: UBS)</strong></a> reiterating its Buy rating with a price target of $11.50.  </p>

#1 - Nu Holdings (NYSE:NU)

The first company on this list is from the fintech sector. But Nu Holdings Ltd. (NYSE: NU) may not be well-known to investors. That's because it's not a U.S. company. However, Nu is the largest fintech bank in North America.  

The digital-first bank was founded in 2013 as a way to disrupt the Latin American banking system, which is dominated by a small number of large banks. Among other things, this trapped customers in an ecosystem of high fees for limited services. 

The company has already signed up five million customers and has a total addressable market that can potentially bring in millions more. Revenue is growing year-over-year, and the bank is solidly profitable. Nu Holdings is also projecting earnings growth of 75% in the next 12 months.  

The concern is how much of that growth is priced into a stock up 97% in the last 12 months. The Nu Holdings analyst ratings on MarketBeat show that analysts are beginning to bid NU stock higher, with UBS Group AG (NYSE: UBS) reiterating its Buy rating with a price target of $11.50.  

<p>Artificial intelligence is driving a super cycle in the chip sector. <a href="https://www.marketbeat.com/stocks/NASDAQ/QUIK/"><strong>QuickLogic Corporation (NASDAQ: QUIK)</strong></a> is a fabless chipmaker. That means the company designs and markets semiconductors and owns its intellectual property. But since it's a fabless company, it doesn't fabricate (i.e., fab) them.  </p> <p>QuickLogic has seen a sharp spike in revenue largely fueled by unprecedented demand for chips to handle AI applications. In its most recent quarter, the company posted positive earnings. And the company is projecting a full year of positive earnings.  </p> <p>QUIK stock up 116% in the last 12 months. So, it's logical to wonder if it can move any higher. And the company is not widely covered by analysts. However, here's something to consider. QuickLogic is mainly known for designing chips for industrial and defense applications. That niche is likely to grow due to demand from aerospace and defense contractors. The company also has a history of beating analysts' expectations.  </p>

#2 - QuickLogic (NASDAQ:QUIK)

Artificial intelligence is driving a super cycle in the chip sector. QuickLogic Corporation (NASDAQ: QUIK) is a fabless chipmaker. That means the company designs and markets semiconductors and owns its intellectual property. But since it's a fabless company, it doesn't fabricate (i.e., fab) them.  

QuickLogic has seen a sharp spike in revenue largely fueled by unprecedented demand for chips to handle AI applications. In its most recent quarter, the company posted positive earnings. And the company is projecting a full year of positive earnings.  

QUIK stock up 116% in the last 12 months. So, it's logical to wonder if it can move any higher. And the company is not widely covered by analysts. However, here's something to consider. QuickLogic is mainly known for designing chips for industrial and defense applications. That niche is likely to grow due to demand from aerospace and defense contractors. The company also has a history of beating analysts' expectations.  

<p>Chinese stocks took a beating in 2023. <a href="https://www.marketbeat.com/stocks/NASDAQ/LI/"><strong>Li Auto Inc. (NASDAQ: LI)</strong></a> was a notable exception. The stock is up 18% in the last 12 months despite being down 25% in the last three months. Li Auto is the leading manufacturer of electric vehicles (EVs) within the People's Republic of China.  </p> <p>However, this appears to be a case of a rose getting buried among the thorns. A lack of demand in the United States has beaten down the EV sector. The same can't be said of China. In the company's third quarter 2023 earnings report, Li Auto reported a <a href="https://www.marketbeat.com/originals/li-auto-smashes-estimates-proving-evs-can-be-profitable/">271% year-over-year increase in revenue</a>. And the bottom line grew at a similar year-over-year pace.  </p> <p>Elon Musk has already sounded the alarm about the <a href="https://www.axios.com/2024/02/14/chinese-ev-electric-vehicles-sold-america">potential dominance of Chinese EV makers</a>. In fact, Li Auto outsold <a href="https://www.marketbeat.com/stocks/NASDAQ/TSLA/"><strong>Tesla Inc. (NASDAQ: TSLA)</strong></a> in October 2023 and now leads China in EV sales.  </p> <p>The <a href="https://www.marketbeat.com/stocks/NASDAQ/LI/price-target/">Li Auto analyst ratings on MarketBeat</a> project a stock price gain of 164% in that same time. That's likely due to the company's expectation that it will increase earnings by more than 83.5% in the next 12 months. </p>

#3 - Li Auto (NASDAQ:LI)

Chinese stocks took a beating in 2023. Li Auto Inc. (NASDAQ: LI) was a notable exception. The stock is up 18% in the last 12 months despite being down 25% in the last three months. Li Auto is the leading manufacturer of electric vehicles (EVs) within the People's Republic of China.  

However, this appears to be a case of a rose getting buried among the thorns. A lack of demand in the United States has beaten down the EV sector. The same can't be said of China. In the company's third quarter 2023 earnings report, Li Auto reported a 271% year-over-year increase in revenue . And the bottom line grew at a similar year-over-year pace.  

Elon Musk has already sounded the alarm about the potential dominance of Chinese EV makers . In fact, Li Auto outsold Tesla Inc. (NASDAQ: TSLA) in October 2023 and now leads China in EV sales.  

The Li Auto analyst ratings on MarketBeat project a stock price gain of 164% in that same time. That's likely due to the company's expectation that it will increase earnings by more than 83.5% in the next 12 months. 

<p>No matter how you feel about <a href="https://www.marketbeat.com/cryptocurrencies/bitcoin/"><strong>Bitcoin (BTC)</strong></a> as an asset class, you must acknowledge that it's been one of the best-performing assets in 2024. <a href="https://www.marketbeat.com/stocks/NASDAQ/RIOT/"><strong>Riot Platforms Inc. (NASDAQ: RIOT)</strong></a> gives you a way to invest in a blockchain future without owning the digital currency.  </p> <p>Here's why. Bitcoin is "mined" via specialized, high-speed computers that compete to solve complex cryptographic problems. Riot operates one of the largest blockchain mining networks in the world. As a result, it has the lowest mining costs, which means it's a very efficient company. That's not the case with many Bitcoin miners. </p> <p>The company's low mining costs will stand out as the next Bitcoin halving occurs in April 2024. This means miners' profits will be cut in half (as Bitcoin nears its maximum supply of 21 million). This will benefit efficient operators like Riot even as RIOT stock is up 161% in the last 12 months.  </p>

#4 - Riot Platforms (NASDAQ:RIOT)

No matter how you feel about Bitcoin (BTC) as an asset class, you must acknowledge that it's been one of the best-performing assets in 2024. Riot Platforms Inc. (NASDAQ: RIOT) gives you a way to invest in a blockchain future without owning the digital currency.  

Here's why. Bitcoin is "mined" via specialized, high-speed computers that compete to solve complex cryptographic problems. Riot operates one of the largest blockchain mining networks in the world. As a result, it has the lowest mining costs, which means it's a very efficient company. That's not the case with many Bitcoin miners. 

The company's low mining costs will stand out as the next Bitcoin halving occurs in April 2024. This means miners' profits will be cut in half (as Bitcoin nears its maximum supply of 21 million). This will benefit efficient operators like Riot even as RIOT stock is up 161% in the last 12 months.  

stock chart graphic

#5 - Enphase Energy (NASDAQ:ENPH)

Solar stocks zigged when they were supposed to zag last year. And if you were an investor in Enphase Energy Inc. (NASDAQ: ENPH) , you know how painful it's been. ENPH stock is down 39.5% in the last 12 months, and that's after a 41% increase in the stock price in the last three months.  

The issue was the company's guidance, which became prescient in its most recent quarter as revenue and earnings fell sharply year-over-year. However, the maker of solar-focused semiconductor-based home energy solutions continues to have a strong long-term story as the United States continues its transition to renewable energy.  

Enphase is forecasting 87% earnings growth in the next 12 months. A reason to believe in that forecast is that the company has the highest margins in the industry. Lower interest rates could also be a potential catalyst. And the E nphase Energy analyst ratings on MarketBeat show a 16% upside for ENPH stock.  

<p>Lithium stocks were expected to be attractive investments in 2023. But as supply outpaced demand on declining EV demand, many of these investments dragged down portfolios. <a href="https://www.marketbeat.com/stocks/NYSE/SQM/"><strong>Sociedad Quimica y Minera de Chile (NYSE: SQM)</strong></a> was no different. SQM stock is down over 55% in the last 12 months.  </p> <p>The Chilean-based company has a diversified portfolio, but about 75% of its revenue comes from lithium. That could hold the stock price down as lithium prices are expected to be down through at least the first half of 2024. </p> <p>However, the opportunity for the company comes in its brine asset, the Salar de Atacama, which has the highest lithium concentration in the world. The company is also taking steps to secure more lithium production in Australia and China. </p> <p>As the lithium supply-demand dynamic flips in its favor, SQM stock is an attractive long-term investment, trading at just 5.4x forward earnings.  </p>

#6 - Sociedad Quimica y Minera de Chile (NYSE:SQM)

Lithium stocks were expected to be attractive investments in 2023. But as supply outpaced demand on declining EV demand, many of these investments dragged down portfolios. Sociedad Quimica y Minera de Chile (NYSE: SQM) was no different. SQM stock is down over 55% in the last 12 months.  

The Chilean-based company has a diversified portfolio, but about 75% of its revenue comes from lithium. That could hold the stock price down as lithium prices are expected to be down through at least the first half of 2024. 

However, the opportunity for the company comes in its brine asset, the Salar de Atacama, which has the highest lithium concentration in the world. The company is also taking steps to secure more lithium production in Australia and China. 

As the lithium supply-demand dynamic flips in its favor, SQM stock is an attractive long-term investment, trading at just 5.4x forward earnings.  

<p><a href="https://www.marketbeat.com/stocks/NYSE/HD/"><strong>Home Depot (NYSE: HD)</strong></a> is one part of a virtual duopoly with <a href="https://www.marketbeat.com/stocks/NYSE/LOW/"><strong>Lowe's Companies Inc. (NYSE: LOW)</strong></a> in the United States. HD stock presents an anomaly for investors. The stock is up 11% in the last 12 months, even though revenue and earnings have fallen year-over-year.  </p> <p>But a closer look at the HD chart shows that much of the growth has come in the last three months. That might suggest that investors are becoming bullish on interest rate cuts that could boost the housing market.  </p> <p>However, with the amount and timing of those rate cuts in question, Home Depot still appears to be a buy because the company has successfully integrated e-commerce and omnichannel services into its business model. That's important because home improvement is a niche market that even <a href="https://www.marketbeat.com/stocks/NASDAQ/AMZN/"><strong>Amazon.com Inc. (NASDAQ: AMZN)</strong></a> has been unable to successfully crack. </p> <p>The <a href="https://www.marketbeat.com/stocks/NYSE/HD/price-target/">Home Depot analyst ratings on MarketBeat</a> show that analysts are moving their price targets for the stock higher, and Home Depot offers a <a href="https://www.marketbeat.com/stocks/NYSE/HD/dividend/">solid dividend</a> that has increased for the last 14 years, has a 2.33% yield and an annual payout of $8.36 per share.   </p>

#7 - Home Depot (NYSE:HD)

Home Depot (NYSE: HD) is one part of a virtual duopoly with Lowe's Companies Inc. (NYSE: LOW) in the United States. HD stock presents an anomaly for investors. The stock is up 11% in the last 12 months, even though revenue and earnings have fallen year-over-year.  

But a closer look at the HD chart shows that much of the growth has come in the last three months. That might suggest that investors are becoming bullish on interest rate cuts that could boost the housing market.  

However, with the amount and timing of those rate cuts in question, Home Depot still appears to be a buy because the company has successfully integrated e-commerce and omnichannel services into its business model. That's important because home improvement is a niche market that even Amazon.com Inc. (NASDAQ: AMZN) has been unable to successfully crack. 

The Home Depot analyst ratings on MarketBeat show that analysts are moving their price targets for the stock higher, and Home Depot offers a solid dividend that has increased for the last 14 years, has a 2.33% yield and an annual payout of $8.36 per share.   

More for You

Donald Trump

New Ad Reimagines Donald Trump's Father Reacting to New York Judgment

Boebert’s X fight

Colorado voters had enough of Lauren Boebert

Tiger Woods confirms suspicions at Genesis Invitational amid withdrawal

Tiger Woods confirms suspicions at Genesis Invitational amid withdrawal

restaurant_dining-2

Another popular restaurant chain files Chapter 11 bankruptcy

US warship crews are learning from battles with anti-ship ballistic missiles, threats no one's ever faced in combat until now, Navy commanders say

US warships are shooting down weapons no one's ever faced in combat before, and a Navy commander says it's a 'great opportunity'

A still of a video posted on Sept. 19, 2023, by the Nationalist Social Club to its public page on Telegram. The neo-Nazi group said it was demonstrating outside the Red Roof Inn hotel in Framingham.

We can’t ignore these fascists, or normalize them

Donald Trump supporters

Russian State TV Mocks Donald Trump Supporters

Trump, Engoron in court

MSNBC host Katy Tur demands: ‘Is this fair’ after judge fines Trump, bans him from business in New York

Why Allen Iverson may not be getting his $32 million trust fund from Reebok

Why Allen Iverson may not be getting his $32 million trust fund from Reebok

Researchers are continuing to investigate the mystery illness affecting dogs

Mystery dog disease sweeping the US - the latest on the illness, states, and symptoms

Julian Assange’s moment of truth has arrived – and the stakes are high

Julian Assange’s moment of truth has arrived – and the stakes are high

Let Them Vote

Let Them Vote

Sandy Gennari, 65, of Waterford said she plans to vote for former President Donald Trump for the fifth time -- counting primaries and general elections -- in Michigan's Feb. 27 Republican primary.

For Trump supporters, a Waterford rally with the former president is like a church meeting

Jason Kelce on the football field

Shaq gives retirement advice to Jason Kelce: 'Don’t be an idiot like me'

Tennessee lawmakers react to group of 'Nazis' marching through downtown Nashville

Tennessee lawmakers react to group of 'Nazis' marching through downtown Nashville

Scientists make breakthrough discovery while experimenting with urine: ‘We can reuse a very significant portion of the cobalt’

Scientists make breakthrough discovery while experimenting with urine: ‘We can reuse a very significant portion of the cobalt’

‘A smear campaign’: A legal expert weighs in on hearing to disqualify Fani Willis from Trump’s GA case

‘A smear campaign’: A legal expert weighs in on hearing to disqualify Fani Willis from Trump’s GA case

‘Some buy the propaganda. Many have changed their minds’: life in Russia after the invasion

‘Some buy the propaganda. Many have changed their minds’: life in Russia after the invasion

The 10 Most Hated Banks in America

The 10 Most Hated Banks in America

Andrew Hitt, the former chairman of the Republican Party of Wisconsin, spoke with Anderson Cooper of

Wisconsin fake elector tells ‘60 Minutes’ he was afraid of Trump supporters

  • Skip to main content
  • Keyboard shortcuts for audio player

K-12 students learned a lot last year, but they're still missing too much school

Cory Turner - Square

Cory Turner

Headshot of Sequoia Carrillo

Sequoia Carrillo

steps to solve related rates problems

From 2022-2023, chronic absenteeism declined in 33 of the 39 states AEI looked at. But it was still a persistent problem: In a handful of places, including Nevada, Washington, D.C., Michigan, New Mexico and Oregon, roughly 1 in 3 students – or more – were chronically absent. LA Johnson/NPR hide caption

From 2022-2023, chronic absenteeism declined in 33 of the 39 states AEI looked at. But it was still a persistent problem: In a handful of places, including Nevada, Washington, D.C., Michigan, New Mexico and Oregon, roughly 1 in 3 students – or more – were chronically absent.

It's going to take aggressive interventions to repair the pandemic's destructive impact on kids' schooling.

That's the takeaway of two big new studies that look at how America's K-12 students are doing. There's some good news in this new research, to be sure – but there's still a lot of work to do on both student achievement and absenteeism. Here's what to know:

1. Students are starting to make up for missed learning

From spring 2022 to spring 2023, students made important learning gains, making up for about one-third of the learning they had missed in math and a quarter of the learning they had missed in reading during the pandemic.

That's according to the newly updated Education Recovery Scorecard , a co-production of Harvard University's Center for Education Policy Research and The Educational Opportunity Project at Stanford University.

6 things we've learned about how the pandemic disrupted learning

6 things we've learned about how the pandemic disrupted learning

The report says, "Students learned 117 percent in math and 108 percent in reading of what they would typically have learned in a pre-pandemic school year."

In an interview with NPR's All Things Considered , Stanford professor Sean Reardon said that's surprisingly good news: "A third or a quarter might not sound like a lot, but you have to realize the losses from 2019 to 2022 were historically large."

When the same team of researchers did a similar review last year, they found that, by spring of 2022, the average third- through eighth-grader had missed half a grade level in math and a third of a grade level in reading. So, the fact that students are now making up ground is a good sign.

These results do come with a few caveats, including that the researchers were only able to review data and draw their conclusions from 30 states this year.

2. Despite that progress, very few states are back to pre-pandemic learning levels

The Harvard and Stanford study of student learning includes one sobering sentence: "Alabama is the only state where average student achievement exceeds pre-pandemic levels in math." And average achievement in reading has surpassed pre-pandemic levels in just three of the states they studied: Illinois, Louisiana and Mississippi. Every other state for which they had data has yet to reach pre-pandemic levels in math and reading.

"Many schools made strong gains last year, but most districts are still working hard just to reach pre-pandemic achievement levels," said Harvard's Thomas Kane, one of the learning study's co-authors.

3. Chronic absenteeism also improved in many places ... slightly

The rate of chronic absenteeism – the percentage of students who miss 10% or more of a school year – declined from 2022 to 2023. That's according to research by Nat Malkus at the conservative-leaning American Enterprise Institute (AEI). He found chronic absenteeism declined in 33 of the 39 states he studied.

Yes, "the differences were relatively small," Malkus writes, but it's improvement nonetheless: "the average chronic absenteeism rate across these states in 2023 was 26 percent, down from 28 percent for the same 39 states in 2022."

Glass half-full: Things aren't getting worse.

4. But, again, chronic absenteeism is still high

Malkus found chronic absenteeism was at 26% in 2023. Before the pandemic, in 2019, those same states reported a rate of 15%. That adds some painful context to the "good news" two-point decline in absenteeism from 2022 to 2023. Sure, it's down, but it's still so much higher than it was and should be.

Think of it this way: In 2023, roughly 1 student out of 4 was still chronically absent across the school year.

In a handful of places, including Nevada, Washington, D.C., Michigan, New Mexico and Oregon, roughly 1 in 3 students – or more – were chronically absent. That's a crisis.

Research shows a strong connection between absenteeism and all kinds of negative consequences for students, including an increased likelihood of dropping out of school.

Chronic absenteeism also hurts the students who don't miss school. That's because, as the learning study's authors point out, when absent students return, they require extra attention and "make it hard for teachers to keep the whole class moving."

5. Poverty matters (as always)

Both the learning and the chronic absenteeism studies capture the headwinds that constantly buffet children in poverty.

"No one wants poor children to foot the bill for the pandemic," said Harvard's Kane, "but that is the path that most states are on."

On learning: Reardon told NPR "the pandemic really exacerbated inequality between students in high-poverty and low-poverty districts and students of different racial and ethnic backgrounds."

In 2023, students' academic recovery was relatively strong across groups, which is good – but it means "the inequality that was widened during the pandemic hasn't gotten smaller, and in some places it's actually gotten larger," Reardon told NPR.

In fact, the report says, "in most states, achievement gaps between rich and poor districts are even wider now than they were before the pandemic." The learning study singles out Massachusetts and Michigan as the states where those gaps in math and reading achievement widened the most between poor and non-poor students.

Similarly, Malkus, at AEI, found that, between 2019 and 2022, rates of chronic absenteeism rose much more in high-poverty districts (up from 20% to 37%) than in low-poverty districts (up from 12% to 23%).

"Chronic absenteeism has increased the most for disadvantaged students," Malkus writes, "those who also experienced the greatest learning losses during the pandemic and can least afford the harms that come with chronic absenteeism."

6. Families must play an important role in learning recovery

Both studies acknowledge that families must play an important role in helping students – and schools – find a healthy, post-pandemic normal. The problem is, surveys show parents and guardians often underestimate the pandemic's toll on their children's learning . "Parents cannot advocate effectively for their children's future if they are misinformed," says the learning study.

To combat this, the learning researchers propose that districts be required to inform parents if their child is below grade-level in math or English. Those parents could then enroll their students in summer learning, tutoring and after-school programs, all of which have benefitted from federal COVID relief dollars. That funding is set to expire this fall, and some of these learning recovery opportunities may dry up, so the clock is ticking.

7. There's a "culture problem" around chronic absenteeism

Reducing chronic absenteeism, Malkus says, will also depend on families.

"This is a culture problem," Malkus tells NPR. "And in schools and in communities, culture eats policy for breakfast every day."

By "culture problem," Malkus is talking about how families perceive the importance of daily attendance relative to other challenges in their lives. He says some parents seem more inclined now to let their students miss school for various reasons, perhaps not realizing the links between absenteeism and negative, downstream consequences.

"Look, the patterns and routines of going to school were disrupted and to some degree eroded during the pandemic," Malkus says. "And I don't think we've had a decisive turn back that we need to have, to turn this kind of behavior around, and it's going to stay with students until that culture changes."

How do you do that? Malkus points to some low-cost options — like texting or email campaigns to increase parental involvement and encourage kids to get back in school – but says these, alone, aren't "up to the scale of what we're facing now."

Higher-cost options for schools to consider could include door-knocking campaigns, sending staff on student home-visits and requiring that families of chronically absent students meet in-person with school staff.

The learning study goes one step further: "Elected officials, employers, and community leaders should launch public awareness campaigns and other initiatives to lower student absenteeism." Because, after all, students can't make up for the learning they missed during the pandemic if they don't consistently attend school now.

What both of these studies make clear is there is no one solution that will solve these problems, and success will require further investment, aggressive intervention and patience.

Malkus says, even the high-cost, high-return options will likely only drive down chronic absenteeism by about four percentage points. A big win, he says, "but four percentage points against 26% isn't going to get us where we need to go."

Edited by: Nicole Cohen Visual design and development by: LA Johnson and Aly Hurt

IMAGES

  1. 4 Steps to Solve Any Related Rates Problem

    steps to solve related rates problems

  2. How to Solve Related Rates in Calculus (with Pictures)

    steps to solve related rates problems

  3. Step by Step Method of Solving Related Rates Problems

    steps to solve related rates problems

  4. PPT

    steps to solve related rates problems

  5. How to Solve Related Rates in Calculus (with Pictures)

    steps to solve related rates problems

  6. How to Solve Related Rates Problems in 5 Steps :: Calculus

    steps to solve related rates problems

VIDEO

  1. How to Solve ANY Related Rates Problem [Calc 1]

  2. ||Related Rates#3.9||Part2

  3. 5 steps on how to solve related rates problems

  4. Related Rates (Calculus)

  5. 4. Related Rates

  6. Slipping Ladder

COMMENTS

  1. Related Rates (How To w/ 7+ Step-by-Step Examples!)

    How To Solve Related Rates Problems We use the principles of problem-solving when solving related rates. The steps are as follows: Read the problem carefully and write down all the given information. Sketch and label a graph or diagram, if applicable.

  2. 4 Steps to Solve Any Related Rates Problem

    Related rates problems will always give you the rate of one quantity that's changing, and ask you to find the rate of something else that's changing as a result. Here are three common problem-scenarios to illustrate: Expanding Circle Example.Given: The radius of a particular circle increases at 1 millimeter each second.

  3. Study Guide

    Figure 1. As the balloon is being filled with air, both the radius and the volume are increasing with respect to time. Answer: The volume of a sphere of radius r r centimeters is V=\frac {4} {3}\pi r^3 \, \text {cm}^3 V = 34πr3 cm3. Since the balloon is being filled with air, both the volume and the radius are functions of time.

  4. How to Solve Related Rates in Calculus (with Pictures)

    The keys to solving a related rates problem are identifying the variables that are changing and then determining a formula that connects those variables to each other. Once that is done, you find the derivative of the formula, and you can calculate the rates that you need. Part 1 Interpreting the Problem Download Article 1

  5. 4.1 Related Rates

    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.

  6. Analyzing problems involving related rates

    Related rates problems are applied problems where we find the rate at which one quantity is changing by relating it to other quantities whose rates are known. Worked example of solving a related rates problem Imagine we are given the following problem: The radius r ( t) of a circle is increasing at a rate of 3 centimeters per second.

  7. 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.

  8. How to Solve Related Rates Problems in 5 Steps :: Calculus

    • Introduction How to Solve Related Rates Problems in 5 Steps :: Calculus Mr. S Math 4.5K subscribers Subscribe Subscribed 1.5K Share 74K views 4 years ago What are Related Rates problems...

  9. 4 Steps to Solve Any Related Rates Problem

    the Pythagorean theorem; or similar triangles. Most frequently (> 80% of the time) you will use the Pythagorean theorem or similar triangles. Take the derivative with respect to time of both sides of your equation. Remember the Chain Rule. Solve for the quantity you're after. [collapse]

  10. 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

  11. Related Rates

    the Pythagorean theorem. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. Solve for the quantity you're after. In the subsections below, we'll solve each problem using the strategy outlined above, step by step each time. We'll look at problems that require each of the approaches listed in 2.B above:

  12. AP Calculus Unit 4.5 Solving Related Rates Problems

    To review related rates, check out the previous Fiveable guide: Introduction to Related Rates. 🪜 Steps to Solve Related Rates Problems. When you first look at a related rates problem, you will be presented with so much information that you may feel overwhelmed. It's okay to take a step back and organize before solving anything out!

  13. Solving Related Rates Problems in Calculus

    Updated: Dec 11, 2023 9:56 AM EST Related rates problems with solutions Photo by Ben Wicks on Unsplash What Are Related Rates? Related rates are calculus problems that involve finding a rate at which a quantity changes by relating to other known values whose rates of change are known.

  14. PDF Related Rates Problems

    The key to solving a related rates problem is the identification of appropriate relationships between the variables in the problem — and putting all of the pieces of information together to produce an answer to the question. General Strategy for Solving Related Rates Problems Step 1: Read the entire problem; identify quantities to be found ...

  15. Calculus I

    Example 1 Air is being pumped into a spherical balloon at a rate of 5 cm 3 /min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm. Show Solution We can get the units of the derivative by recalling that, r′ = dr dt r ′ = d r d t

  16. Solving Related Rates Problems

    1.) Read the problem slowly and carefully. 2.) Draw an appropriate sketch. 3.) Introduce and define appropriate variables. Use variables if quantities are changing. Use constants if quantities are not changing. 4.) Read the problem again. 5.) Clearly label the sketch using your variables. 6.) State what information is given in the problem. 7.)

  17. 3.9: Related Rates

    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.

  18. Related rates problems with inflating and deflating balloons

    To solve a related rates problem, complete the following steps: 1) Construct an equation containing all the relevant variables. 2) Differentiate the entire equation with respect to (time), before plugging in any of the values you know. 3) Plug in all the values you know, leaving only the one you're solving for. 4) Solve for your unknown variable.

  19. Related Rates Problem Solving Strategy

    the Pythagorean theorem. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. Solve for the quantity you're after. On our Related Rates page, we solve an example of each of the approaches listed in 2.B above: i. Geometric fact.

  20. Calculus I

    Section 3.11 : Related Rates. In the following assume that x x and y y are both functions of t t. Given x =−2 x = − 2, y = 1 y = 1 and x′ = −4 x ′ = − 4 determine y′ y ′ for the following equation. 6y2 +x2 = 2 −x3e4−4y 6 y 2 + x 2 = 2 − x 3 e 4 − 4 y Solution. In the following assume that x x, y y and z z are all ...

  21. PDF CHAPTER 29 Related Rates

    The key to solving such a problem is to find an equation relating r and S. In this case we can use the formula for the surface area of a sphere: S = 4 o r2. Don't forget: S and as saying g(t) = 4 o r are really functions of time t, so think of this equation °f (t)¢2. Now di erentiate both sides with respect to t.

  22. 5.2: Related Rates

    Exercises: Related Rates Problems. Exercise 1: Let y = 3x + 5 y = 3 x + 5 and z = 4y + 7. z = 4 y + 7. Find dz dx d z d x when x = 2 x = 2 by solving for z z as a function of x x and taking the derivative, and also by finding dz dy d z d y and dy dx d y d x and using related rates to apply the chain rule. Answer.

  23. related rates

    Solve problems from Pre Algebra to Calculus step-by-step . step-by-step. related rates. en. Related Symbolab blog posts. Practice, practice, practice. Math can be an intimidating subject. Each new topic we learn has symbols and problems we have never seen. The unknowing...

  24. The U.S.-Mexico Border: How Americans View the Situation, Its Causes

    Democrats mostly view the situation as a major problem (44%) or minor problem (26%) for the U.S. Very few Democrats (7%) say it is not a problem. In an open-ended question, respondents voice their concerns about the migrant influx. They point to numerous issues, including worries about how the migrants are cared for and general problems with ...

  25. 7 growth stocks that will prove growth is back in 2024

    Bitcoin is "mined" via specialized, high-speed computers that compete to solve complex cryptographic problems. Riot operates one of the largest blockchain mining networks in the world.

  26. Students are still absent and making up for missed learning post ...

    The rate of chronic absenteeism - the percentage of students who miss 10% or more of a school year - declined from 2022 to 2023. ... The problem is, ... The learning study goes one step ...