Fiveable

Find what you need to study

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.

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

AP®︎/College Calculus AB

Course: ap®︎/college calculus ab   >   unit 4, related rates intro.

  • Related rates (multiple rates)
  • Related rates: Approaching cars
  • Related rates: Falling ladder
  • Related rates (Pythagorean theorem)
  • Related rates: water pouring into a cone
  • Related rates (advanced)
  • Related rates: shadow
  • Related rates: balloon
  • (Choice A)   − 729 ‍   A − 729 ‍  
  • (Choice B)   − 9747 ‍   B − 9747 ‍  
  • (Choice C)   − 3249 ‍   C − 3249 ‍  
  • (Choice D)   − 6859 ‍   D − 6859 ‍  

HIGH SCHOOL

  • ACT Tutoring
  • SAT Tutoring
  • PSAT Tutoring
  • ASPIRE Tutoring
  • SHSAT Tutoring
  • STAAR Tutoring

GRADUATE SCHOOL

  • MCAT Tutoring
  • GRE Tutoring
  • LSAT Tutoring
  • GMAT Tutoring
  • AIMS Tutoring
  • HSPT Tutoring
  • ISAT Tutoring
  • SSAT Tutoring

Search 50+ Tests

Loading Page

math tutoring

  • Elementary Math
  • Pre-Calculus
  • Trigonometry

science tutoring

Foreign languages.

  • Mandarin Chinese

elementary tutoring

  • Computer Science

Search 350+ Subjects

  • Video Overview
  • Tutor Selection Process
  • Online Tutoring
  • Mobile Tutoring
  • Instant Tutoring
  • How We Operate
  • Our Guarantee
  • Impact of Tutoring
  • Reviews & Testimonials
  • Media Coverage
  • About Varsity Tutors

AP Calculus AB : Modeling rates of change, including related rates problems­

Study concepts, example questions & explanations for ap calculus ab, all ap calculus ab resources, example questions, example question #1 : modeling rates of change, including related rates problems­.

ap calculus worksheet related rates

Recall the the general derivative for the inverse tangent function is:

ap calculus worksheet related rates

To determine the rate of change of the surface area of the spherical bubble, we must relate it to something we do know the rate of change of - the volume. 

The volume of a sphere is given by the following:

ap calculus worksheet related rates

The rate of change of the volume is given by the derivative with respect to time:

ap calculus worksheet related rates

The derivative was found using the following rules:

ap calculus worksheet related rates

We must now solve for the rate of change of the radius at the specified radius, so that we can later solve for the rate of change of surface area:

ap calculus worksheet related rates

Next, we must find the surface area and rate of change of the surface area of the sphere the same way as above:

ap calculus worksheet related rates

Plugging in the known rate of change of the surface area at the specified radius, and this radius into the rate of surface area change function, we get

ap calculus worksheet related rates

Example Question #3 : Modeling Rates Of Change, Including Related Rates Problems­

A pizzeria chef is flattening a circular piece of dough. The surface area of the dough (we are only considering the top of the dough) is increasing at a rate of 0.5 inches/sec. How quickly is the diameter of the pizza changing when the radius of the pizza measures 4 inches?

ap calculus worksheet related rates

To find the rate of change of the diameter, we must relate the diameter to something we do know the rate of change of: the surface area.

The surface area of the top side of the pizza dough is given by

ap calculus worksheet related rates

The rate of change, then, is found by taking the derivative of the function with respect to time:

ap calculus worksheet related rates

Solving for the rate of change of the radius at the given radius, we get

ap calculus worksheet related rates

Now, we relate the diameter to the radius of the pizza dough:

ap calculus worksheet related rates

Taking the derivative of both sides with respect to time, we get

ap calculus worksheet related rates

Plugging in the known rate of change of the radius at the given radius, we get

ap calculus worksheet related rates

We could have found this directly by writing our surface area formula in terms of diameter, however the process we used is more applicable to problems in which the related rate of change is of something not as easy to manipulate.

Example Question #4 : Modeling Rates Of Change, Including Related Rates Problems­

ap calculus worksheet related rates

To determine the rate of change of the circumference at a given radius, we must relate the circumference rate of change to the rate of change we know - that of the volume.

Starting with the equation for the volume of the spherical balloon,

ap calculus worksheet related rates

we take the derivative of the function with respect to time, giving us the rate of change of the volume:

ap calculus worksheet related rates

The chain rule was used when taking the derivative of the radius with respect to time, because we know that it is a function of time.

ap calculus worksheet related rates

Now, we use this rate of change and apply it to the rate of change of the circumference, which we get by taking the derivative of the circumference with respect to time:

ap calculus worksheet related rates

Solving for the rate of change of the circumference by plugging in the known rate of change of the radius, we get

ap calculus worksheet related rates

Example Question #5 : Modeling Rates Of Change, Including Related Rates Problems­

Determine the rate of change of the angle opposite the base of a right triangle -whose length is increasing at a rate of 1 inch per minute, and whose height is a constant 2 inches - when the area of the triangle is 2 square inches. 

ap calculus worksheet related rates

To determine the rate of the change of the angle opposite to the base of the given right triangle, we must relate it to the rate of change of the base of the triangle when the triangle is a certain area. 

First we must determine the length of the base of the right triangle at the given area:

ap calculus worksheet related rates

Now, we must find something that relates the angle opposite of the base to the length of the base and height - the tangent of the angle:

ap calculus worksheet related rates

To find the rate of change of the angle, we take the derivative of both sides with respect to time, keeping in mind that the base of the triangle is dependent on time, while the height is constant:

ap calculus worksheet related rates

We know the rate of change of the base, and we can find the angle from the sides of the triangle:

ap calculus worksheet related rates

Plugging this and the other known information in and solving for the rate of change of the angle adjacent to the base, we get

ap calculus worksheet related rates

Example Question #6 : Modeling Rates Of Change, Including Related Rates Problems­

The position of a car is given by the equation

ap calculus worksheet related rates

Example Question #7 : Modeling Rates Of Change, Including Related Rates Problems­

ap calculus worksheet related rates

This is a related rates problem.

ap calculus worksheet related rates

Since the radius is given as 1 unit, we can write this equation as

ap calculus worksheet related rates

Plugging this information in, we get

ap calculus worksheet related rates

Example Question #8 : Modeling Rates Of Change, Including Related Rates Problems­

A man is standing on the top of a 10 ft long ladder that is leaning against the side of a building, when the bottom of the ladder begins to slide out from under it. How fast is the man standing on the top of the ladder falling when the bottom of the ladder is 6 ft from the building and is sliding at 2ft/sec?

ap calculus worksheet related rates

Simplifying the right side gives us

ap calculus worksheet related rates

Using implicit differentiation to find the derivative with respect to time, we get

ap calculus worksheet related rates

Since we are dealing with physical distances, we will only use the positive 8.

Plugging all the information into our derivative equation gives us

ap calculus worksheet related rates

Example Question #9 : Modeling Rates Of Change, Including Related Rates Problems­

The velocity of a car is given by the equation:

ap calculus worksheet related rates

If the car starts out at a distance of 3 miles from its home, how far will it be after 4 hours?

ap calculus worksheet related rates

Report an issue with this question

If you've found an issue with this question, please let us know. With the help of the community we can continue to improve our educational resources.

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC 101 S. Hanley Rd, Suite 300 St. Louis, MO 63105

Or fill out the form below:

Contact Information

Complaint details.

Learning Tools by Varsity Tutors

  • Digital Codes
  • Unit 1.3 Limits from Graphs
  • Unit 1.5 Algebraic Properties
  • Unit 1.6 Algebraic Manipulation
  • Unit 1.8 L'Hospitals & Squeeze
  • Unit 1.9 MULTIPLE REPRESENTATIONS
  • Unit 1.11-1.13 Continuity
  • Unit 1.14 Infinite Limits & VA
  • UNIT 1.15 CONNECTING LIMITS AT INFINITY
  • Unit 1.16 IVT
  • MC Practice
  • Review Optional
  • 1.1 Limits Numerically & Graphically
  • 1.2 Limits Algebraically
  • 1.3 Limits Graphical and Analytical Connections
  • 1.4 Limits of Exponential Functions
  • 1.5 Limits of Trig Functions
  • 1.6 Limit Based Continuity
  • 1.7 Intermediate Value Theorem
  • 1.8 Limits with Infinity
  • Limits Card Sorter
  • 2.1 DEFINING AVERAGE AND INSTANTANEOUS RATES OF CHANGE AT A POINT
  • 2.2 DEFINING THE DERIVATIVE OF A FUNCTION AND USING DERIVATIVE NOTATION
  • 2.3 ESTIMATING DERIVATIVES OF A FUNCTION AT A POINT
  • 2.4 CONNECTING DIFFERENTIABILITY AND CONTINUITY ​- DETERMINING WHEN DERIVATIVES DO AND DO NOT EXIST
  • Quiz 2.1 - 2.4 Details
  • 2.5 APPLYING THE POWER RULE
  • 2.6 DERIVATIVE RULES - CONSTANT, SUM, DIFFERENCE, AND CONSTANT MULTIPLE
  • Quiz 2.5-2.6 Details
  • 2.7 DERIVATIVES OF COS X, SIN X, EX, AND LN X
  • 2.8 THE PRODUCT RULE
  • 2.9 THE QUOTIENT RULE
  • Quiz 2.8-2.9 Details
  • 2.10 FINDING THE DERIVATIVES OF TANGENT, COTANGENT, SECANT, ​AND/OR COSECANT FUNCTIONS
  • Review Unit 2
  • Test unit 2 Details
  • 3.1 Chain Rule
  • FRQ 2012 #4abc
  • FRQ 2017 #6abc
  • 3.2 Implicit Differentiation
  • FRQ 2000 #5
  • FRQ 2008 #6 Form B
  • FRQ 1998 #6
  • FRQ 2015 #6
  • 3.3 Differentiating Inverse Functions
  • 3.4 Differentiating Inverse Trig Functions
  • 3.5 Procedures for Calculating Derivatives
  • 3.6 Calculating Higher-Order Derivatives
  • FRQ 2006 #6
  • Unit 3 Progress Check
  • Review Unit 3
  • 4.1 Interpreting Meaning of Derivative in Context
  • 4.2 Straight Line Motion - Connecting Position, Velocity & Acceleration
  • 4.3 RATES OF CHANGE IN NON-MOTION CONTEXTS
  • Crack the Code Calc-Medic
  • Big 10 Motion +
  • Lovely Ladybug Calc-Medic
  • 4.4 INTRODUCTION TO RELATED RATES
  • 4.5 SOLVING RELATED RATES
  • Speed Dating - MCQ Related Rates
  • FRQ 2002 #5
  • FRQ 2002 #6 Form B
  • FRQ 2005 #5 Form B
  • 4.6 LOCAL LINEARIZATION
  • 4.7 L'Hospital's Rule
  • Progress Check MCQ
  • Progress Check FRQ Part A
  • Progress Check FRQ Part B
  • Review Unit 4 Record
  • 5.1 Mean Value Theorem
  • 5.2 Extreme Value Theorem
  • 5.3, 5.4 Increasing or Decreasing? 1st Derivative Test
  • 5.5 Candidates' Test
  • 5.6 DETERMINING CONCAVITY OF F(X) ON DOMAIN
  • ​5.8 SKETCHING GRAPHS OF F AND F'
  • 5.9 Part 1 Connecting F, F" and F"
  • 5.9 Part 2 Connecting F, F" and F"
  • FRQ 2001 #4
  • FRQ 2006 #2 Form B
  • FRQ 2007 #6
  • 5.10 Intro to Optimization
  • 5.11 Solving Optimization Problems
  • 5.12 Exploring Behaviors of Implicit Differentiation
  • Review Unit 5
  • Midterm Review
  • Unit 6.1 Exploring Accumulation of Change
  • Unit 6.2 Approximating Areas with Riemann Sums
  • Unit 6.3 Riemann Sums, Notation and Definite Integrals
  • Unit 6.4-6.5 Fundamental Th'm of Calculus
  • Unit 6.6 Applying Properties of Definite Integrals
  • Unit 6.7 - 6.8 Fun'l Th'm of Calc & Definite Integrals
  • Unit 6.9 Integrating Using Substitution
  • 6.9 SmacStyle U Substitution
  • FRQ 2003 #4, 1998 #3, 1999 #3
  • FRQ 2005 #2, 2002 #2 form B, 2000 #4
  • FRQ 2003 #3, 2004 #1, 2001 #2
  • FRQ 2001 #3, 2002 #3 Form B
  • Unit 6.10 Integrating Functions Using Long Division & Completing Square
  • Unit 6.14 Selecting Techniques for Antidifferentiation
  • Unit 6 Exam Review
  • Unit 6 Practice Exam
  • Mario Particle Motion
  • Mario Motion Full Video
  • APClassroom FRQ KEY
  • 7.1, 7.2 Diff EQ - Verifying Solutions
  • 7.3, 7.4 Slope Fields
  • 7.6, 7.7 Separation of Variables
  • FRQ 2000 #6 Diff EQ
  • FRQ 2004 #5 FORM B
  • FRQ 2008 #5
  • FRQ 2010 #5 FORM B
  • FRQ 2004 #6
  • 7.8 Diff EQ with Exponential Models
  • Differential Eq Big 10
  • 8.1 Average Value
  • 8.2 Integrals - s(t), v(t), & a(t)
  • 8.3 Accumulation in Applied Context
  • 8.4-8.6 Area between curves
  • 8.7-8.8 Volumes with Cross Sections
  • 8.9-8.10 Volume Disc Method
  • 8.11-8.12 Volume Washer Method
  • FRQ 2013 #5
  • FRQ 2014 #2
  • FRQ 2010 #1 FORM B
  • FRQ 2012 #2
  • Unit 1 Limits
  • Unit 2 Derivative Rules
  • Unit 3 Derivative Rules of Composites
  • Unit 4 Contextual Applications
  • Mixed Bag Review Units 1 - 4
  • Unit 5 Analytic Applications of Derivative
  • Unit 6 Integration & Accumulation of Change
  • Unit 7 Differential EQ
  • Mixed Bag Review Units 5 - 7
  • Trig Inverses
  • Squeeze Theorem
  • Exam Updated Information
  • FRQ #9, #10
  • Released MC
  • Bohemian Rhapsody
  • 1.1 Limits Graphically
  • 1.2 Limits Analytically
  • 1.3 Asymptotes
  • 1.4 Continuity
  • 2.1 Average Rate of Change
  • 2.2 Definition of Derivative
  • 2.3 Differentiability
  • Unit 2 The Derivative
  • Unit 3 Basic Differentiation
  • 4.1 - Exp & Log Derivatives
  • 4.2 - Inverse Derivatives
  • 4.3 - L'Hopital's Rule
  • 5.1 Critical Values
  • 5.2 Optimization
  • 5.3 Maximizing Profit
  • 6.1 Implicit vs Explicit
  • 6.2 Related Rates
  • 7.1 - Rectangular Approximation
  • 7.2 - Trapezoidal Approximation
  • 7.1 - 7.2 Test Prep
  • 8.1 - Definite Integrals
  • 8.2 - First Fundamental Theorem of Calculus
  • 8.3 - AntiDerivatives
  • 9.1 - The 2nd FTC
  • 9.2 - Trig Integrals
  • 9.3 - Average Value
  • 9.4 - Net Change
  • 10.1 Slope Fields
  • 10.2 U-Substitution Indefinite Integrals
  • 10.3 U-Substitution Definite Integrals
  • 10.4 U-Substitution Trig Functions
  • Review U-Substitution
  • 10.5 Differential Equations
  • 10.6 Review
  • 11.1 Area Between Curves
  • 11.2 Solids of Revolution Disc
  • 11.3 Solids of Revolution Washer
  • 11.4 Perpendicular Cross Sections
  • Previous FRQ
  • 2016 Practice Test
  • 2008 Test Prep
  • 1998 Test Prep

4.3 Related Rates

  • Related Rates Video Introduction​  (Day 1)
  • Area of a Circle Video Example  (Day 1)
  • Volume of a Cylinder Video Example  (Day 1)
  • Volume of a Sphere Video Example  (Day 1)
  • AP CALCULUS BC
  • AP CALCULUS AB
  • TRIG/PRE-CALCULUS

AP CALCULUS BC WORKSHEETS (CHAPTER REVIEWS & AP TOPICS)

Chapter 1 limits,  chapter 2 derivatives, ch. 2 related rates, chapter 3 applications of derivatives, chapter 4 integrals,  chapter 5  transcendental functions,  chapter 6 differential equations, chapter 7 area/volume,  chapter 8 part 1  integration techniques, chapter 8 part 2   trig. integrals 8-3/8-4, chapter 9  part 2  taylor series.

  • Practice Problems
  • Assignment Problems
  • Show all Solutions/Steps/ etc.
  • Hide all Solutions/Steps/ etc.
  • Implicit Differentiation
  • Higher Order Derivatives
  • Applications of Derivatives
  • Calculus II
  • Calculus III
  • Differential Equations
  • Algebra & Trig Review
  • Common Math Errors
  • Complex Number Primer
  • How To Study Math
  • Cheat Sheets & Tables
  • MathJax Help and Configuration
  • Notes Downloads
  • Complete Book
  • Practice Problems Downloads
  • Complete Book - Problems Only
  • Complete Book - Solutions
  • Assignment Problems Downloads
  • Other Items
  • Get URL's for Download Items
  • Print Page in Current Form (Default)
  • Show all Solutions/Steps and Print Page
  • Hide all Solutions/Steps and Print Page

Section 3.11 : Related Rates

In this section we are going to look at an application of implicit differentiation. Most of the applications of derivatives are in the next chapter however there are a couple of reasons for placing it in this chapter as opposed to putting it into the next chapter with the other applications. The first reason is that it’s an application of implicit differentiation and so putting it right after that section means that we won’t have forgotten how to do implicit differentiation. The other reason is simply that after doing all these derivatives we need to be reminded that there really are actual applications to derivatives. Sometimes it is easy to forget there really is a reason that we’re spending all this time on derivatives.

For these related rates problems, it’s usually best to just jump right into some problems and see how they work.

The first thing that we’ll need to do here is to identify what information that we’ve been given and what we want to find. Before we do that let’s notice that both the volume of the balloon and the radius of the balloon will vary with time and so are really functions of time, i.e. \(V\left( t \right)\) and \(r\left( t \right)\).

We know that air is being pumped into the balloon at a rate of 5 cm 3 /min. This is the rate at which the volume is increasing. Recall that rates of change are nothing more than derivatives and so we know that,

We want to determine the rate at which the radius is changing. Again, rates are derivatives and so it looks like we want to determine,

Note that we needed to convert the diameter to a radius.

Now that we’ve identified what we have been given and what we want to find we need to relate these two quantities to each other. In this case we can relate the volume and the radius with the formula for the volume of a sphere.

As in the previous section when we looked at implicit differentiation, we will typically not use the \(\left( t \right)\) part of things in the formulas, but since this is the first time through one of these we will do that to remind ourselves that they are really functions of \(t\).

Now we don’t really want a relationship between the volume and the radius. What we really want is a relationship between their derivatives. We can do this by differentiating both sides with respect to \(t\). In other words, we will need to do implicit differentiation on the above formula. Doing this gives,

Note that at this point we went ahead and dropped the \(\left( t \right)\) from each of the terms. Now all that we need to do is plug in what we know and solve for what we want to find.

We can get the units of the derivative by recalling that,

The units of the derivative will be the units of the numerator (cm in the previous example) divided by the units of the denominator (min in the previous example).

Let’s work some more examples.

The first thing to do in this case is to sketch picture that shows us what is going on.

ap calculus worksheet related rates

We’ve defined the distance of the bottom of the ladder from the wall to be \(x\) and the distance of the top of the ladder from the floor to be \(y\). Note as well that these are changing with time and so we really should write \(x\left( t \right)\) and \(y\left( t \right)\). However, as is often the case with related rates/implicit differentiation problems we don’t write the \(\left( t \right)\) part just try to remember this in our heads as we proceed with the problem.

Next, we need to identify what we know and what we want to find. We know that the rate at which the bottom of the ladder is moving towards the wall. This is,

Note as well that the rate is negative since the distance from the wall, \(x\), is decreasing. We always need to be careful with signs with these problems.

We want to find the rate at which the top of the ladder is moving away from the floor. This is \(y'\). Note as well that this quantity should be positive since \(y\) will be increasing.

As with the first example we first need a relationship between \(x\) and \(y\). We can get this using Pythagorean theorem.

All that we need to do at this point is to differentiate both sides with respect to \(t\), remembering that \(x\) and \(y\) are really functions of \(t\) and so we’ll need to do implicit differentiation. Doing this gives an equation that shows the relationship between the derivatives.

Next, let’s see which of the various parts of this equation that we know and what we need to find. We know \(x'\) and are being asked to determine \(y'\) so it’s okay that we don’t know that. However, we still need to determine \(x\) and \(y\).

Determining \(x\) and \(y\) is actually fairly simple. We know that initially \(x = 10\) and the end is being pushed in towards the wall at a rate of \({\textstyle{1 \over 4}}\)ft/sec and that we are interested in what has happened after 12 seconds. We know that,

So, the end of the ladder has been pushed in 3 feet and so after 12 seconds we must have \(x = 7\). Note that we could have computed this in one step as follows,

To find \(y\) (after 12 seconds) all that we need to do is reuse the Pythagorean Theorem with the values of \(x\) that we just found above.

Now all that we need to do is plug into \(\eqref{eq:eq1}\) and solve for \(y'\).

Notice that we got the correct sign for \(y'\). If we’d gotten a negative value we’d have known that we had made a mistake and we could go back and look for it.

Before working another example, we need to make a comment about the set up of the previous problem. When we labeled our sketch, we acknowledged that the hypotenuse is constant and so just called it 15 ft. A common mistake that students will sometimes make here is to also label the hypotenuse as a letter, say \(z\), in this case.

Well, it’s not really a mistake to label with a letter, but it will often lead to problem down the road. Had we labeled the hypotenuse \(z\) then the Pythagorean theorem and its derivative would have been,

Again, there is nothing wrong with doing this but it does require that we acknowledge the values of two more quantities, \(z\) and \(z'\). Because \(z\) is just the hypotenuse that is clearly \(z = 15\). The problem that some students then sometimes run into is determining the value of \(z'\). In this case, we have to remember that because the ladder, and hence the hypotenuse has a fixed length, its length can’t be changing and so \(z' = 0\).

Plugging both of these values into the derivative give us same equation that we got in the example but required a little more effort to get to. It would have been easier to just label the hypotenuse 15 to start off with and not have to worry about remembering that \(z' = 0\).

When labeling a fixed quantity (the length of the ladder in this example) with a letter it is sometimes easy to forget that it is a fixed quantity and so it’s derivative must be zero. If you don’t remember this, the problem becomes impossible to finish as you will have two unknown quantities that you have to deal with. In any problem were a quantity is fixed and will never over the course of the problem change it is always best to just acknowledge that and label it with its value rather than with a letter.

Of course, if we’d had a sliding ladder that was allowed to change length then we would have to label it with a letter. However, for that kind of problem we would also need some more information in the problem statement in order to actually do the problem. The practice problems in this section have several problems in which all three sides of a right triangle are changing. You should check them out and see if you can work them.

A right triangle with the hypotenuse labeled “x”.  The vertical side has no label and the bottom, horizontal side labeled “50”.  The angle between the bottom and hypotenuse is labeled $\theta$  At the far right is the notation that this where the stationary person is at and at the top left corner there is notation that is the current location of the moving person as well as an arrow pointing upwards.

This example is not as tricky as it might at first appear. Let’s call the distance between them at any point in time \(x\) as noted above. We can then relate all the known quantities by one of two trig formulas.

We want to find \(x'\)and we could find \(x\) if we wanted to at the point in question using cosine since we also know the angle at that point in time. However, if we use the second formula we won’t need to know \(x\) as you’ll see. So, let’s differentiate that formula.

As noted, there are no \(x\)’s in this formula. We want to determine \(x'\) and we know that \(\theta = 0.5\) and \(\theta ' = 0.01\) (do you agree with it being positive?). So, just plug in and solve.

So far we we’ve seen three related rates problems. While each one was worked in a very different manner the process was essentially the same in each. In each problem we identified what we were given and what we wanted to find. We next wrote down a relationship between all the various quantities and used implicit differentiation to arrive at a relationship between the various derivatives in the problem. Finally, we plugged the known quantities into the equation to find the value we were after.

So, in a general sense each problem was worked in pretty much the same manner. The only real difference between them was coming up with the relationship between the known and unknown quantities. This is often the hardest part of the problem. In many problems the best way to come up with the relationship is to sketch a diagram that shows the situation. This often seems like a silly step but can make all the difference in whether we can find the relationship or not.

Let’s work another problem that uses some different ideas and shows some of the different kinds of things that can show up in related rates problems.

  • At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft?
  • At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft?

Okay, we should probably start off with a quick sketch (probably not to scale) of what is going on here. We’ll also be doing the sketch as if we were looking at the tank from directly in front of it (and so the 3D of the tank will not be visible) as this will help a little with seeing what is going on. Showing the 3D nature of the tank is liable to just get in the way. So here is the sketch of the tank with some water in it.

ap calculus worksheet related rates

As we can see, the water in the tank actually forms a smaller cone/triangle (depending on which image we are looking at) with the same central angle as the tank itself. The radius of the “water” cone at any time is given by \(r\) and the height of the “water” cone at any time is given by \(h\). The volume of water in the tank at any time \(t\) is given by,

and we’ve been given that \(V' = - 2\).

For this part we need to determine \(h'\) when \(h = 6\) and now we have a problem. The only formula that we’ve got that will relate the volume to the height also includes the radius and so if we were to differentiate this with respect to \(t\) we would get,

So, in this equation we know \(V'\) and \(h\) and want to find \(h'\), but we don’t know \(r\) and \(r'\). As we’ll see finding \(r\) isn’t too bad, but we just don’t have enough information, at this point, that will allow us to find \(r'\) and \(h'\) simultaneously.

To fix this we’ll need to eliminate the \(r\) from the volume formula in some way. This is actually easier than it might at first look. If we go back to our sketch above and look at just the right half of the tank we see that we have two similar triangles and when we say similar we mean similar in the geometric sense. Recall that two triangles are called similar if their angles are identical, which is the case here. When we have two similar triangles then ratios of any two sides will be equal. For our set this means that we have,

If we take this and plug it into our volume formula we have,

This gives us a volume formula that only involved the volume and the height of the water. Note however that this volume formula is only valid for our cone, so don’t be tempted to use it for other cones! If we now differentiate this we have,

At this point all we need to do is plug in what we know and solve for \(h'\).

So, it looks like the height is decreasing at a rate of 0.1386 ft/hr.

In this case we are asking for \(r'\) and there is an easy way to do this part and a difficult (well, more difficult than the easy way anyway….) way to do it. The “difficult” way is to redo the work in part (a) above only this time use,

to get the volume in terms of \(V\) and \(r\) and then proceed as before.

That’s not terribly difficult, but it is more work that we need to so. Recall from the first part that we have,

So, as we can see if we take the relationship that relates \(r\) and \(h\) that we used in the first part and differentiate it we get a relationship between \(r'\) and \(h'\). At this point all we need to do here is use the result from the first part to get,

Much easier that redoing all of the first part. Note however, that we were only able to do this the “easier” way because it was asking for \(r'\)at exactly the same time that we asked for \(h'\) in the first part. If we hadn’t been using the same time then we would have had no choice but to do this the “difficult” way.

In the second part of the previous problem we saw an important idea in dealing with related rates. In order to find the asked for rate all we need is an equation that relates the rate we’re looking for to a rate that we already know. Sometimes there are multiple equations that we can use and sometimes one will be easier than another.

Also, this problem showed us that we will often have an equation that contains more variables that we have information about and so, in these cases, we will need to eliminate one (or more) of the variables. In this problem we eliminated the extra variable using the idea of similar triangles. This will not always be how we do this, but many of these problems do use similar triangles so make sure you can use that idea.

Let’s work some more problems.

Note that an isosceles triangle is just a triangle in which two of the sides are the same length. In our case sides of the tank have the same length.

Let’s add in some dimensions for the water to the sketch from above.

This is a sketch of the tank.  In the front there two triangles.  The larger represents the tank whose height is shown at 2.  The smaller represents the water in the tank and its height is shown as “h”.  In the back of the tank there are two similar triangles.  The width of the larger (i.e. the tank) is shown as 5 and the width of the smaller (i.e. the water in the tank) is shown as “w”.  The depth of the tank is shown as 8.

Now, in this problem we know that \(V' = 6 \mbox{m}^{3}\mbox{/sec}\) and we want to determine \(h'\) when \(h = 1.2\,{\rm{m}}\). Note that because \(V'\) is in terms of meters we need to convert \(h\) into meters as well. So, we need an equation that will relate these two quantities and the volume of the tank will do it.

The volume of this kind of tank is simple to compute. The volume is the area of the end times the depth. For our case the volume of the water in the tank is,

As with the previous example we’ve got an extra quantity here, \(w\), that is also changing with time and so we need to eliminate it from the problem. To do this we’ll again make use of the idea of similar triangles. If we look at the end of the tank we’ll see that we again have two similar triangles. One for the tank itself and one formed by the water in the tank. Again, remember that with similar triangles ratios of sides must be equal. In our case we’ll use,

Plugging this into the volume gives a formula for the volume (and only for this tank) that only involved the height of the water.

We can now differentiate this to get,

Finally, all we need to do is plug in and solve for \(h'\).

So, the height of the water is rising at a rate of 0.25 m/sec.

In order to answer the second part of this question is not all that difficult.

We will need \(w'\) to answer this part and we have the following equation from the similar triangle that relate the width to the height and we can quickly differentiate it to get a relationship between \(w'\) and \(h'\).

From the first part we know the value of \(h'\) and so all wee need to do is plug that into this equation and we’ll have the answer.

Therefore the width is increasing at a rate of 0.625 m/sec.

  • At what rate is the tip of the shadow moving away from the pole when the person is 25 ft from the pole?
  • At what rate is the tip of the shadow moving away from the person when the person is 25 ft from the pole?

Let’s start off with putting all the relevant quantities into the sketch from above.

There are two triangles in this figure.  The smaller triangle fits inside the larger triangle with the base/hypotenuse of the smaller triangle on the base/hypotenuse of the larger triangle.  The height of the smaller triangle represents the person and is marked as “5.5”.  There is also an arrow here pointing right and marked \({{{x}'}_{p}}\) = 2 ft/sec showing the direction of motion away from the lamp.  The base of the smaller triangle represents the shadow the person casts and is labeled \(x_{s}\).  The height of the larger triangle represents the lamp and is marked as “12”.  The base of the larger triangle represents the distance from the lamp to the person, labeled  \(x{p}\), and the length of the shadow, again labeled \(x_{s}\).  The total length of the base of the larger triangle is labeled x.

Here \(x\) is the distance of the tip of the shadow from the pole, \({x_p}\) is the distance of the person from the pole and \({x_s}\) is the length of the shadow. Also note that we converted the persons height over to 5.5 feet since all the other measurements are in feet.

The tip of the shadow is defined by the rays of light just getting past the person and so we can see they form a set of similar triangles. This will be useful down the road.

In this case we want to determine \(x'\) when \({x_p} = 25\) given that \({x'_p} = 2\).

The equation we’ll need here is,

but we’ll need to eliminate \({x_s}\) from the equation in order to get an answer. To do this we can again make use of the fact that the two triangles are similar to get,

From this we can quickly see that,

We can then plug this into the equation above and solve for \(x\) as follows.

Now all that we need to do is differentiate this, plug in and solve for \(x'\).

The tip of the shadow is then moving away from the pole at a rate of 3.6923 ft/sec. Notice as well that we never actually had to use the fact that \({x_p} = 25\) for this problem. That will happen on rare occasions.

This part is actually quite simple if we have the answer from (a) in hand, which we do of course. In this case we know that \({x_s}\) represents the length of the shadow, or the distance of the tip of the shadow from the person so it looks like we want to determine \({x'_s}\) when \({x_p} = 25\).

Again, we can use \(x = {x_p} + {x_s}\), however unlike the first part we now know that \({x'_p} = 2\) and \(x' = 3.6923{\rm{ ft/sec}}\) so in this case all we need to do is differentiate the equation and plug in for all the known quantities.

The tip of the shadow is then moving away from the person at a rate of 1.6923 ft/sec.

Below is a copy of the sketch in the problem statement with all the relevant quantities added in. The top of the shadow will be defined by the light rays going over the head of the person and so we again get yet another set of similar triangles.

There are two triangles in this figure.  The smaller triangle fits inside the larger triangle with the base/hypotenuse of the smaller triangle on the base/hypotenuse of the larger triangle.  The height of the smaller triangle represents the person and is marked as “5.5”.  There is also an arrow here pointing left and marked \({{x}'\) = 2.5 ft/sec showing the direction of motion away from the spotlight.  The base of the smaller triangle represents the distance of the person from the spotlight and is labeled x.  The height of the larger triangle represents the shadow the person casts on the wall and is labeled y.  The base of the larger triangle represents the distance from the wall to the person, labeled  \(20-x\), and the distance of the person from the spotlight, again labeled x$.  The total length of the base of the larger triangle is  marked as 20.

In this case we want to determine \(y'\) when the person is 8 ft from wall or \(x = 12{\rm{ ft}}\). Also, if the person is moving towards the wall at 2.5 ft/sec then the person must be moving away from the spotlight at 2.5 ft/sec and so we also know that \(x' = 2.5\).

In all the previous problems that used similar triangles we used the similar triangles to eliminate one of the variables from the equation we were working with. In this case however, we can get the equation that relates \(x\) and \(y\) directly from the two similar triangles. In this case the equation we’re going to work with is,

Now all that we need to do is differentiate and plug values into solve to get \(y'\).

The height of the shadow is then decreasing at a rate of 2.0833 ft/sec.

Okay, we’ve worked quite a few problems now that involved similar triangles in one form or another so make sure you can do these kinds of problems.

It’s now time to do a problem that while similar to some of the problems we’ve done to this point is also sufficiently different that it can cause problems until you’ve seen how to do it.

There is a lot to digest here with this problem. Let’s start off with a sketch of the situation that shows each person’s location sometime after both people start riding.

In the upper left corner of this sketch is the label for Person A and an arrow pointing upwards labeled ${x}'$ = 5 m/sec indicated the direction of motion for Person A.  In the lower right corner of this sketch is the label for Person B and an arrow pointing downwards labeled ${y}'$ = 3 m/sec indicated the direction of motion for Person B.   The two labels are connected with a line whose distance is given as z.  In the middle of the two Person labels is a horizontal line that represents where the two people both started.  It’s distance is given as 350.  From the left end of this line up towards the Person A label is a line whose distance is given as x.  From the right end of this line down towards the Person B label is a line whose distance is given as y.    The figure also notes that the total horizontal distance separating the two person labels is x+y.

Now we are after \(z'\) and we know that \(x' = 5\) and \(y' = 3\). We want to know \(z'\) after Person A had been riding for 25 minutes and Person B has been riding for \(25 - 7 = 18\) minutes. After converting these times to seconds (because our rates are all in m/sec) this means that at the time we’re interested in each of the bike riders has rode,

Next, the Pythagorean theorem tells us that,

Therefore, 25 minutes after Person A starts riding the two bike riders are

To determine the rate at which the two riders are moving apart all we need to do then is differentiate \(\eqref{eq:eq2}\) and plug in all the quantities that we know to find \(z'\).

So, the two riders are moving apart at a rate of 7.9958 m/sec.

Every problem that we’ve worked to this point has come down to needing a geometric formula and we should probably work a quick problem that is not geometric in nature.

Suppose that \({R_{\,1}}\) is increasing at a rate of 0.4 \(\Omega \)/min and \({R_{\,2}}\) is decreasing at a rate of 0.7\(\Omega \)/min. At what rate is \(R\) changing when \({R_{\,1}} = 80\,\Omega \) and \({R_{\,2}} = 105\,\Omega \)?

Okay, unlike the previous problems there really isn’t a whole lot to do here. First, let’s note that we’re looking for \(R'\) and that we know \({R'_{\,1}} = 0.4\) and \({R'_{\,2}} = - 0.7\). Be careful with the signs here.

Also, since we’ll eventually need it let’s determine \(R\) at the time we’re interested in.

Next, we need to differentiate the equation given in the problem statement.

Finally, all we need to do is plug into this and do some quick computations.

So, it looks like \(R\) is decreasing at a rate of 0.002045\(\Omega \)/min.

We’ve seen quite a few related rates problems in this section that cover a wide variety of possible problems. There are still many more different kinds of related rates problems out there in the world, but the ones that we’ve worked here should give you a pretty good idea on how to at least start most of the problems that you’re liable to run into.

Classroom Contact Information

Ap exam information.

ap calculus worksheet related rates

404 Not found

Practice Solutions

ap calculus worksheet related rates

Corrective Assignments

ap calculus ab related rates

All Formats

Resource types, all resource types.

  • Rating Count
  • Price (Ascending)
  • Price (Descending)
  • Most Recent

Ap calculus ab related rates

Preview of Pixel Art - AP Calculus AB/BC Related Rates I

Pixel Art - AP Calculus AB /BC Related Rates I

ap calculus worksheet related rates

  • Google Apps™

Preview of The Derivative Test with Related Rates in AP Calculus AB

The Derivative Test with Related Rates in AP Calculus AB

ap calculus worksheet related rates

AP Calculus AB - Related Rates problems

ap calculus worksheet related rates

AP Calculus AB - Related Rates Problems + Google Forms Assessment

  • Google Drive™ folder

Preview of Fall Semester Derivatives Review with Related Rates AP Calculus AB

Fall Semester Derivatives Review with Related Rates AP Calculus AB

ap calculus worksheet related rates

AP Calculus AB Related Rates Activities

ap calculus worksheet related rates

  • Word Document File

Preview of Calculus Related Rates Sort & Match Activity (Unit 4)

Calculus Related Rates Sort & Match Activity (Unit 4)

ap calculus worksheet related rates

AP Calculus AB Free Response Questions

ap calculus worksheet related rates

AP Calculus AB Review Foldables

ap calculus worksheet related rates

AP Calc AB Unit 4 - Derivative Applications

ap calculus worksheet related rates

AP Calc AB Content Review

ap calculus worksheet related rates

AP Calculus : Related Rates Complete Lesson

ap calculus worksheet related rates

AP Calculus AB : - Antiderivatives - Definite Integrals - Activities - Worksheets

ap calculus worksheet related rates

AP Calculus ( AB ) Essential Study Card (all on 1 sheet of ledger paper)

ap calculus worksheet related rates

AP Calculus AB /BC Unit 4: Contextual Application of Derivatives Doodle Review

ap calculus worksheet related rates

AP Calculus AB : Unit Exam Derivative Applications

ap calculus worksheet related rates

AP Calculus AB : Limits and Continuity - Derivatives - Applications - Activities

Preview of AP Calculus AB: Unit Exam Derivative Application Answer Key

AP Calculus AB : Unit Exam Derivative Application Answer Key

Preview of Pixel Art - AP Calculus AB/BC (derivatives)

Pixel Art - AP Calculus AB /BC (derivatives)

Preview of AP Calculus AB: Derivative Applications in the Real World Notes Bundle

AP Calculus AB : Derivative Applications in the Real World Notes Bundle

Preview of Calc Ch 2.6c More Related Rates

Calc Ch 2.6c More Related Rates

ap calculus worksheet related rates

Calc Ch 2.6b More Related Rates

Preview of Calc Ch 2.6a Related Rates

Calc Ch 2.6a Related Rates

Preview of Concept of Instantaneous Rate of Change Guided Notes

Concept of Instantaneous Rate of Change Guided Notes

  • We're hiring
  • Help & FAQ
  • Privacy policy
  • Student privacy
  • Terms of service
  • Tell us what you think

IMAGES

  1. AP Calculus, Section 2.7

    ap calculus worksheet related rates

  2. Related Rates Day 1 AP Calculus

    ap calculus worksheet related rates

  3. Ap calculus related rates mc

    ap calculus worksheet related rates

  4. Ap Calculus

    ap calculus worksheet related rates

  5. Related Rates Worksheet With Answers

    ap calculus worksheet related rates

  6. Calculus: Worksheet (Study Guide) for Related Rates by Julane Crabtree

    ap calculus worksheet related rates

VIDEO

  1. Related rates intro

  2. Analyzing related rates problems: equations (trig)

  3. Related Rates in Calculus

  4. Related Rates Ultimate Study Guide (7 examples for your calculus class)

  5. Analyzing related rates problems: expressions

  6. Introduction to Related Rates

COMMENTS

  1. PDF AP CALCULUS AB/BC: Related Rates Worksheet ilearnmath

    AP CALCULUS AB/BC: Related Rates Worksheet ilearnmath.net Problem 3 A kite is flying 150 m high, where the wind causes it to move horizontally at the rate of 5 m per second. In order to maintain the kite at a height of 150 m, the person must allow more string to be let out.

  2. PDF Related Rates Worksheet

    Calculus 1500 Related Rates page 1 1. An airplane is flying towards a radar station at a constant height of 6 km above the ground. If the distance s between the airplane and the radar station is decreasing at a rate of 400 km per hour when s 10 Ian., what is the horizontal speed of the plane? 2. A light is on the ground 20 m from a building.

  3. PDF Related Rates Date Period

    04 - Related Rates. 2) Oil spilling from a ruptured tanker spreads in a circle on the surface of the ocean. The area of the spill increases at a rate of 9. 3) A conical paper cup is 10 cm tall with a radius of 10 cm. The cup is being filled with water so that the water level rises at a rate of 2 cm/sec.

  4. PDF AP Calculus AB Unit 8 Related Rates

    AP Calculus AB Unit 8 - Related Rates dS 1) Find if S = 4 2 r . dt dV 1 2) Find if V = r dt 3 2 h . dz 3) Find if x 2 + y 2 = z 2 . dt = dy dx dz 2 2 2. 4) If x 4, y = 3, = − 2, and = 0, then find for x + y = z dt dt dt 4 dV dr 5) If V = 3 r and = 3 and r = 6, then find 3 dt dt .

  5. PDF AP Calculus AB Related Rates Practice Worksheet

    A spherical balloon is inflating at a rate of 27 in3/sec. How fast is the radius of the balloon increasing when the radius is 3 inches? Cars A and B leave a town at the same time. Car A heads due south at a rate of 80 km/hr, and car B heads due west at a rate of 60 km/hr. How fast is the distance between the cars increasing after 3 hours?

  6. PDF AP Calculus

    Page 1 of 18 Session Notes Questions that ask for the calculation of the rate at which one variable changes, based on the rate at which another variable is known to change, are usuaJly called related rates.

  7. AP Calculus Unit 4.5 Solving Related Rates Problems

    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

  8. Related rates intro (practice)

    Related rates intro Google Classroom You might need: Calculator The side of a cube is decreasing at a rate of 9 millimeters per minute. At a certain instant, the side is 19 millimeters. What is the rate of change of the volume of the cube at that instant (in cubic millimeters per minute)? Choose 1 answer: − 729 A − 729 − 9747 B − 9747 − 3249 C

  9. PDF Calculus Related Rates Worksheet Solutions

    Calculus Related Rates Problems Worksheet 1) An 8-foot ladder is leaning against a wall. The top of the ladder is sliding down the wall at the rate of 2 feet per second. How fast is the bottom of the ladder moving along the ground at the point in time when the bottom of the ladder is 4 feet from the wall? dx dy dc x + y = c dt dt dt 8 ft. dx dx

  10. AP Calculus AB : Modeling rates of change, including related rates

    Example Question #5 : Modeling Rates Of Change, Including Related Rates Problems­. Determine the rate of change of the angle opposite the base of a right triangle -whose length is increasing at a rate of 1 inch per minute, and whose height is a constant 2 inches - when the area of the triangle is 2 square inches. Possible Answers:

  11. PDF 2.6-day1-SAMPLE PROBLEMS-ANSWRES

    AP CALCULUS BC Section 2.6 (day 1) RELATED RATES - SAMPLE PROBLEMS 1. Zoe is standing on a 6-meter ladder that is leaning against a wall when Jacob begins to pull the bottom of the ... AP CALCULUS BC Section 2.6 (day 1) RELATED RATES - SAMPLE PROBLEMS 3. Find the rate of change of the distance between the origin and a point moving on ...

  12. PDF AP Calculus: Related Rate Worksheet Draw and label a diagram 'real

    AP Calculus: Related Rate Worksheet Calculator permitted. For each of the answers, Draw and label a diagram, label all quantities, find the numeric answer, using appropriate units, then give a 'real-world' explanation of the answer. Show all "position" and "rate" equations. 1. (02 AB) A container has the shape of an open right ...

  13. PDF APCD CALCULUS: RELATED RATES Worksheet to Accompany Exploration, Part 1

    5. Set up an integral for the volume of each tank in the Exploration. Your integral should be written in terms of dy . 6. Verify the volumes given in the Exploration by evaluating your integrals in problem 5, either analytically or using a calculator. 7. Complete the following for each tank. [Part (a) is the same for the four tanks.] dV dh

  14. Calculus I

    Solution A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at a rate of 0.5 m 2 /sec at what rate is the radius decreasing when the area of the sheet is 12 m 2? Solution

  15. PDF Student Session Topic: Related Rate Problems

    Student Session Topic: Related Rate Problems Related Rate problems appear occasionally on the AP calculus exams. Typically there will be a straightforward question in the multiple‐choice section; on the free‐response section a related rate question will be part of a longer question or, occasionally, an entire free-response question.

  16. PDF AP Calculus-AB worksheets by topics

    AP Calculus-AB worksheets by topics Fu n c t i o n s , L i mi t s , & Co n t i n u i t y D i f f e re n t i a t i o n 1. I n te re s t i n g G ra p h s - A few equations to graph that have interesting (and hidden) features. pdf 2. Fu n c t i o n s - Properties of functions and the Rule of Four (equations, tables, graphs, and words).

  17. 4.3 Related Rates

    AP CALCULUS 4.3 Related Rates 2020 Blank Notes (I changed the notes 10/15/2020) 2020 Blank Hw (I changed the hw 10/15/2020) Notes 4.3 Key Related Rates Video Introduction (Day 1) Area of a Circle Video Example (Day 1) Volume of a Cylinder Video Example (Day 1) Volume of a Sphere Video Example (Day 1) Hw 4.3 Key

  18. Bc Sub (Ch Review)

    CH.2 HW 2-3 Product and Quotient Rule.docx: File Size: 138 kb: File Type: docx

  19. 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

  20. AP Calculus AB

    Enrolling in AP Calculus comes with the understanding that you will take the AP exam in May. The 2019 test will be given ... HW #18 - Related Rates worksheet; HW #18 - Answer Key; 2.6: 2.5 - 2.7 Review 2.7: 2.5 - 2.7 Exam 2.8: Extreme Value Theorem. Section 3.1 Notes;

  21. 4.5 Solving Related Rates Problems

    4.5 Solving Related Rates Problems - Calculus | AP Calculus AB Review Topic 3: Related Rates & Theorems ... 4.5 Solving Relate Rates Problems Previously Lesson Next Lesson Need a tutor? Click this link the get choose initial session free!

  22. 4.4 Introduction to Related Rates

    4.4 Introduction to Related Rates Next Lesson Calculus AB/BC - 4.4 Introduction to Related Rates Watch on Packet calc_4.4_packet.pdf Download File Want to save money on printing? Support us and buy the Calculus workbook with all the packets in one nice spiral bound book. Solution manuals are also available. Practice Solutions calc_4.4_solutions.pdf

  23. Results for ap calculus ab related rates

    This Sort & Match Activity is designed for AP Calculus AB students. The concepts will help your students work through the rules for setting up problems, implicit differentiation with respect to time, (including Chain Rule), and solving the basic types of related rates problems.There are eight task cards to present a typical related rate situation.