First, we need to establish an equation that relates all the variables
First, we need to establish an equation that relates all the variables
Alissa Johnson and Tommy
Table of Contents
Title Slide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Related Rates: A Brief Introduction . . . . . . . . . . . . . . . . . . . . . . . . . .3
Real World Applicability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Mathematician Spotlight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Basic Analytical Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
AP Multiple Choice Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
Conceptual Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Graphical Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14
Free Response Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Works Cited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .24
Related Rates: A Brief Introduction
The first step in most related rates problems is
finding an equation that involves all of the
variables which will be used in the problem.
This could be an area or volume formula, as it
is in some cases. It could also involve triangle
trigonometry or some other given equation.
Next, implicitly differentiate the equation with
respect to time. Don’t forget basic derivative
rules (chain rule, product rule, quotient rule,
etc.)(Sometimes some algebraic manipulation
is necessary or helpful prior to this step).
Finally, use the given values from the problem
in order to solve for the requested quantity or
𝐴 = 𝜋𝑟𝑦2
𝑎2 + 𝑏2 = 𝑐 2
= 𝑑𝑡 2 𝑑𝑡
Real World Applicability
Related rates have infinite applications. In fact related rates are often
referred to as the application of the derivative, which means that a related
rates problem by definition is a real-world application! How exciting! As
stated before, anything that can be mapped by an equation can be made into
a related rates problem. In other words, anything changing with respect to
time can be related to something else changing with respect to time. These
relations are important across all disciplines, and most often used in physics.
They could be crucial for things such as the launch of a rocket. Indeed, the
rocket equation 𝑅𝑢 = 𝑚𝑎, which relates the rate at which a rocket is losing
mass to its acceleration, is derived using related rates as are so many other
equations describing fundamental concepts in physics. The concept of
related rates makes is easier to understand and use equations for
displacement, velocity, and acceleration. Related rates applications are truly
limitless. Who does not want to know how quickly the area of a halfpenny
expands when placed on a hot shovel? William Ritchie was certainly curious
(more about him later). Together, with the aid of related rates, we can solve
at what rate the water in a pool is rising, the rate at which Erica moves when
she runs in a circle, or the rates at which a rebel starship and debris fly away
from an exploding Death Star!
The man who brought related rates to their glory goes (or
went if you do not believe the fathers of Calculus are immortal)
by the name of Rev. William Ritchie. He actually started out in
the church, hence the title Reverend. After attending some
conferences in Paris, he found out that he had a knack for
mathematics and eventually became a professor of natural
philosophy at London University. He published a paper, Principles
of Differentiable and Integral Calculus, which included many
examples focusing on related rates. Ritchie’s goal was to make
Calculus less abstract. He wanted to make it understandable and
applicable to people with a simple math background. Which was
an excellent idea, because related rates are so applicable to
The volume of a cylinder is defined by the following…
𝑉 = 𝜋𝑟 2 ℎ
Some house elves are making a BIG cake for the end of the year
celebration. When the cake is baking it expands. However,
because of the container it is in, it only expands upward. Given
that the radius starts at 25 feet, at what rate will the volume be
increasing when the height is increasing at a rate of 5 ft/s.
Working through the example
𝑉 = 𝜋𝑟 2 ℎ
Start with the given equation
= 𝜋 2𝑟 ℎ + 𝑟
Take the derivative of the
equation with respect to time
(t). This will be done using the
= 𝜋 2𝑟 0 ℎ + 𝑟
We know that the radius is not
increasing so therefore
as the over any change in time
(dt) the change of the radius
(dr) will be zero.
= 𝜋 25 2 (5)
We are given all the remaining
problems in the problem. So
just plug everything in and
arrive at the answer!
That’s one fast growing cake!
Multiple Choice Example
Sandy is preparing her rocket for a launch to the moon. She is taking it on a test drive
around Bikini Bottom (which is obviously completely plausible). Plankton tries to thwart
her plan by using a giant fan to create a current in the water and drive the rocket off
course. The rocket is launched from position (0,0) on Sandy’s Map with a constant
upward velocity of 20 m/s. Plankton’s fan creates a constant current in the water
directed due east at 500 m/s. When ѳ = 45 Sandy notices the rocket is at position (10,
10). What is the rate that the angle, ѳ, between the rocket and the ground is changing
when ѳ = 45°?
A) 24 °/s
B) 2 °/s
C) 26 °/s
D) -24 °/s
E) -26 °/s
(𝑥 2 )(sec 𝜃)2
(10 20 − 10 500 )
(sec 45)2 (102 )
First, we need to establish an
equation that relates all the
variables we wish to use and find.
Then, take the derivative of that
equation with respect to time.
Don’t forget quotient rule!
Finally, solve for dθ/dt and insert
known values for the variables in
The final result should be -24°/s,
or choice D.
Why Are The Other Answers Incorrect?
Almost right away, we can eliminate the first 3 answers. Conceptually, we can tell that the
rocket is being pushed MUCH faster in the x direction than it is propelling itself in the y
direction. Since the value of x is increasing faster than the value of y (and arctan 𝑦 = 𝜃),
the angle is decreasing! We can thus conclude that the rate that the angle is changing is
negative. Analytically, we can also see that, when we take the derivative, we know that 𝑑𝑡
is far greater than 𝑑𝑡 (and the x and y values are positive) and thus 𝑑𝑡 is a negative value.
= 𝑑𝑡 2 𝑑𝑡
Both D and E solutions are plausible answers! The true answer is only revealed through
*If you chose choice E, you probably used the quotient rule incorrectly*
*If you chose choice B, you likely differentiated tan incorrectly and/or used cos instead of
*If you arrived at choice A your solution process was correct, but check sign conventions!*
*If you arrived at choice C, be wary of quotient rule AND sign conventions!*
Captain Jack Sparrow was admiring his
reflection for a bit too long, and did not notice
his ship scraping across a rocky cliff.
Unfortunately, he notices too late. There is a
hole and his ship has already begun to fill with
water. Conveniently, the bottom of the ship is
rectangular. Jack Sparrow is unaware of the
exact dimensions of his ship; however, he
does know that the length is equal to three
times the width and the height is twice the
width. Knowing this, he wonders what the
rate of change of the total volume of the
water in the ship would be at the instant
when the height of the water is equal to the
width of the ship. (*Note* Only the height of
the water entering the ship is changing.)
𝑙 = 3𝑤
𝑉 = 3𝑤 2 ℎ
First, we need to establish an equation
that relates all the variables. In this case,
volume because we are looking for the
volume of the water in the ship.
Then, use the given information to
consolidate into two variables. Be careful
not to be thrown off and put everything in
terms of width (this will create a problem
later since dw/dt is zero, yet the height is
Next, differentiate the new equation with
respect to time. Don’t forget product rule!
Finally, we can use the rest of the given
information to further simplify the
equation. Using the fact the width of the
water is not changing eliminates the
second term. Also, we want the rate of
change of the volume when the height is
equal to the width. So, use that
information to get a final equation for
Dv/dt in terms of height and the rate of
change of the height.
R2D2 and C3PO are taking the Millennium Falcon out for a routine
mission. They are paying careful attention to the radar screen and
see the radar arm constantly sweeping out area of the polar curve
𝑟 = 4. Additionally, due to their attentiveness they also happen to
notice a fleck of dust moving across the screen. Coincidentally, it
appears to be moving according to the function 𝑟 = 2 − 2 cos 𝜃,
so that at time t seconds 𝜃 = 𝑡 2 . They then wonder, is there a
time,t, that the rate at which the radar arm is sweeping out area is
equal to the horizontal velocity of the dust particle? (If so, when is
the first time this occurs?) (Honestly, who doesn’t?)
r = 4; 0.0 <= t <= 2pi
r = 2-2cos(t); 0.0 <= t <= 2pi
1 𝜃 2
𝜃 = 𝑡2
𝑑𝜃 = 2𝑡𝑑𝑡
𝐴 t =
First, we must find an equation to
represent the area being swept out by
the radar arm. Since it is a polar curve
the formula to be used is
𝑟 2 𝑑𝜃 .
Next, we need to get the area with
respect to time. Use the given
information 𝜃 = 𝑡 2 and differentiate to
get an expression for d 𝜃 in terms of dt.
Then, use the previous steps to plug in
and obtain an equation for A(t).
𝑥 𝜃 = 𝑟𝑐𝑜𝑠𝜃 = 2 − 2𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃
𝜃 = 𝑡2
𝑥 𝑡 = 2cos(𝑡 2 ) − 2(cos 𝑡 2 )2
Next, we need an equation for x with
respect to time. It is easier to start with
x(𝜃). Again, using the given information
we can then get an equation for x(t).
Continued . . .
= 32𝑡 3
= −4𝑡𝑠𝑖𝑛(𝑡 2 ) + 8𝑡𝑠𝑖𝑛(𝑡 2 )cos(𝑡 2 )
32𝑡 3 = −4𝑡𝑠𝑖𝑛(𝑡 2 ) + 8𝑡𝑠𝑖𝑛(𝑡 2 )cos(𝑡 2 )
Now, we are trying to relate the rates so
we must differentiate both expressions
with respect to time. For A(t), don’t
forget the chain rule and the FTC! Also
beware of multiple chain rules when
Next, since we want to know when the
horizontal velocity (dx/dt) is equal to
the rate of change of area (dA/dt) we
set them equal to each other.
Finally, use the intersect (zero) feature
of your graphing calculator to find the
first value of t where the two are equal
if such a t exists.
The two graphs intersect only at t=0.
Free Response Example
Batman and Jango Fett both are at point (0,0) at time 0.
Jango Fett flies straight up perpendicular to the ground
with a velocity modeled by 𝑣 𝑡 = 𝑡 2 + 1. Batman runs at
4 miles per hour along the ground. Jango Fett’s missile can
only launch 1 mile.
a) Find the distance both have traveled since the
starting point at the when they are 1 mile apart
b) Find the rate that the distance between them is
increasing at the maximum distance apart within the
range that Batman can get hit (when they are 1 mile
When solving a related rates problem, the first step will always be
finding an equation that relates all of the given values as well as
what you are looking for. Then simply take the derivative and
For this problem, we have the rates at which both characters are
moving. This means that, if we can find the time when they are 1
mile apart, we can find the distance they have traveled using
a) 𝐽2 + 𝐵2 = 𝐷2
First we need to find an equation
that relates all our values. Good
𝑡 2 + 1 𝑑𝑡
At some time, x, the two men will
be 1 mile apart. The distances
= 12 traveled can be represented by
the integral of their rates.
4𝑥 − 4 0
Solve the integrals
𝑥 = .24225
Use your graphing
function to find the answer.
4𝑑𝑡 + 0 = .969 𝑚𝑖𝑙𝑒𝑠
Solve for Batman’s distance
(plus zero because he was
initially at point 0)
𝑡 2 + 1 𝑑𝑡 + 0 = .250 𝑚𝑖𝑙𝑒𝑠
Solve for Jango’s distance
(adding zero again for the
2𝐽 + 2𝐵
2 .250 .242252 +1 + 2 .969 4
Find the equation that relates
all the variables (and their
rates) that we wish to use
Take the derivative
Solve for the rate the distance
Plug in all known values. The
distances we solved for in part
a will go in for variables J and
B. The rates are equivalent to
Plug it in to the calculator! Is
Jango Fett skilled enough to
make this shot?
Analytical Problems FOR YOU TO TRY!
1) In an intense game of Quidditch, Harry spots the snitch. Realizing he will not be able
to get there on time, Draco casts a spell on the snitch. This spell causes the
circumference of the spherical snitch to swell in size at a rate of 21 centimeters per
second. Harry can only hold a snitch with a volume of 15 cubic centimeters. When the
volume is at the maximum that Harry can hold, at what rate is the volume changing?
2) The Sarlacc Pitt quite obviously takes the shape of a cone (point down). The
equation for the volume of a cone is 𝑉 = 3 𝜋𝑟 2 ℎ. For this particular cone the height is
equal to the radius for all points in time. When Luke is escaping from Jabba, he is killing
his guards left and right and the pit is filling up! If the height (that has been filled) is 4
meters and the radius is growing at 9 meters per second, at what rate is the volume
3) Hawkeye is standing on top of a roof that is 120 feet tall. Loki is flying his speeder
parallel to the ground at a height of 53 feet moving in a straight path away from the
building. Hawkeyes has a hawk’s eye and can tell that the distance between him and
Loki is changing at 31 feet per second. If Loki’s speeder travels at 45 feet per second,
how far does the arrow have to travel when Loki is 100 feet away from the building?
4) The Lego Starwars gang is stuck in the trash compactor! The volume of the
trash compactor is quickly decreasing (it takes the form of a rectangular
prism). The length of the prism is constant at 25 meters and the height is also
constant at 10 meters. To attempt to slow the process of being smashed, they
place a metal bar (that conveniently forms an isosceles triangle with is height
equal to the height of the prism and its base being equal to the width of the
prism). At the time when the area of this triangle is 20 meters squared and
the volume is decreasing at a rate of 10 cubic meters per second, what is the
rate at which the width is changing?
5) Lego Indiana Jones is running away from a big stone ball down a circular
set of stairs with a decreasing radius. The ball can no longer fit down the
stairs when the radius is 4 meters. At this time if the radius is decreasing at
7.5 meters per second, at what rate is the circumference decreasing?
6) A Lego rocker is putting everything he’s got into a song. However, he’s not
the greatest and as a result fans are leaving the arena. The number of fans in
the arena is modeled by the equation 𝐹 = ln 𝑡 5 − 2𝑚, where F is the number
of fans, t is the time in minutes, and m is the number of notes missed. If the
rocker consistently misses 6 notes per minute, what is the rate that fans are
leaving after 10 minutes?
7) The Lego squad is going old school (building at the skill level of a young
Tommy Martin). The construction crew decides to build a simple house,
just a square base and a flat roof. Every once in a while the crew decides
to add another level. Each level is 4 meters tall and the number of layers is
modeled by 𝐿 𝑡 = 2𝑡 3 − 𝑡 (where t represents time in hours). The base’s
dimensions start off as 12 meters by 12 meters, but supplies are running
low so the length of both sides of the base is decreasing at a rate of .25
meters per LAYER NOT HOUR. When time is equal to 7 hours and the
volume is growing at a rate of 7 𝑚3 /ℎ𝑜𝑢𝑟, at what rate is the height
8) The Justice Leagues Watchtower has a circular orbits around the earth. It
gets closer to and further away from earth depending on the
circumstances (international warfare or interplanetary obviously require
different distances from earth). The distance from the center of the earth
(the orbital’s radius) is represented by 𝑟 = 𝑡 2 sin 𝑡 Flash is decelerating
quickly to avoid crashing into a young child. If Flash is slowing down at
500 2 and at time zero his velocity is 1000 , what is the first time the
circumference of the orbital is changing at he same rate that as Flash is
moving (his velocity)?
9. Frodo accidentally takes the Ring too close to the fires of Orodruin, and the Ring starts
to expand due to this heat exposure. He notices that each radius is increasing at a rate
of 0.5mm/s and the height is increasing at a rate of 0.2 mm/s. He wonders at what rate
the surface area is increasing at the instant when r₁ = 15mm, r₂ = 13.5mm and the height
is 2mm. The surface area of a hollow cylinder is 𝑆𝐴 = 2𝜋(𝑟1
− 𝑟2 2 ) + 2𝜋ℎ(𝑟1 + 𝑟2 ).
10. In an intense Quidditch match against Slytherin, Oliver Wood spots
Marcus Flint obtain the Quaffle back near Slytherin’s goal post. When Flint
has made it halfway up the pitch, (75m from Gryffindor’s goal post) he is
traveling horizontally at a speed of 20 m/s. If the angle his path makes
with the top of the tallest goal post (9.1m) is 30⁰. At what rate is the
distance between Flint and the top of the post changing?
11. Spongebob’s turbo spatula can crank out crabby patties at a rate of 10 patties/s.
The mob of hungry anchovies consumes patties according to the equation
𝐴 𝑡 = Csin 𝑡 + 2𝑡 2 patties/s. Where C is the number of customers entering
per second. The number of customers entering the Krusty Krab per second is
10, 𝑡 ≤ 2
modeled by the piecewise function: 𝐶 𝑡 = 11.25𝑡, 2 < 𝑡 ≤ 4 . Write An
integral expression for the total number of patties at time, t, and use it to find the
rate that the total number of patties is increasing at time t=4s?
12. Sandman spots Spidey and the police chasing him and quickly disguises himself
among a growing pile of sand nearby (how convenient). The sand is falling,
coincidentally, into a square pyramid. The volume of the sand pyramid is increasing at a
rate of 400 cm³/s prior to sandman ‘s sand pile appearing directly behind it. If sandman’s
sand pile stands 67cm tall, how long before the pyramid in front of him is taller than his
sand pile? The initial base of the sand pyramid is 12.5cm and its base is growing at a rate
twice that of the height, which is growing at a rate of 0.05cm/s.
13. Yoda is going for a slow stroll and pondering some new Jedi training techniques. A nearby lamp, 365 cm tall,
causes him to cast a shadow as he walks. Yoda is leisurely strolling at a rate of 20cm/s , and he stands
66 cm tall. When he is 50 cm from the lamp
a) at what rate is the tip of his shadow moving?
b) at what rate is the length of his shadow changing?
14. Bored and distracted, Tony Stark decides to take one of his new suits, mark 17, out for a spin. As he is taking
off from the ground he is rising at a rate of 30m/s. A local fan/stalker is observing Stark 40 m from his take off
point and wants to calculate the rate of change of the angle of elevation at a certain instant. What is the rate
of change of the angle of elevation of iron man fro m his stalker when Stark is 300 m above the ground? (in
15. On the run, and hunting for horcruxes, Hermoine is in a rush and apparates herself, Ron and
Harry to New York City because it just “popped into her head.” When they arrive, the trio
encounters Captain America on a night stroll. Startled by the sudden arrival of three people and
The loud pop of apparation, he instinctively throws his shield at the three wizards. Ron and Harry turn
around and start to run away at a rate of 5m/s as the shield is hurtling towards them at a rate of 10m/s.
They are initially 45m from Captain America. Fortunately, Hermoine quick to react and actually uses her want to
halt the shield in its tracks before it hits them
a) If Hermoine hadn’t stopped the shield, how long after it was thrown would it have hit Harry and Ron
b) The shield was also rotating at a rate of 15 rad/s . The radius of the shield is 45 cm. Considering only
this rotation, find an expression for the linear rate of change ( )of a chip on the edge of the shield
with respect to 𝜃and solve for when 𝜃 = 12rads .
16. Oh no! Bilbo lost his conveniently hobbit sized sword, Sting. Luckily, Gandalf happens to have a spare,
but it is too big. He puts a spell on the sword to shrink it down to Bilbo's size. If the surface area of the sword is
Given by the equation SA = 2𝐴 + 2𝑙𝑤 where A is the area of the top surface of the sword and l and w are the
length and width of the blade respectively. The blade width remains a constant 2cm. At what rate is the top
surface area shrinking if the length is shrinking at a rate of 4cm/s and the total surface area is shrinking at
a rate of 19cm/s?
1. Spongebob is on his way to deliver the very first Krusty Krab pizza.
Unfortunately, he gets lost along the way. Fortunately though,
because Squidward is of no help, he happens across a friendly
Gungan General by the name of Jar Jar Binks. Even though he is a bit
confused himself, Jar Jar agrees to try and help Spongebob deliver
the pizza. Somehow, the two are able to make it to the customers
house; however, Jar Jar clumsily drops the pizza right before they reach the house.
Fortunately though, Spongegbob is the fry cook, and he’s pretty good with
people. He explains the situation to the customer and asks to make a fresh new
pizza for him, free of charge. When they put the pizza in the oven it has an
original radius of 10cm and a thickness of 2cm and the pizza maintains the same
proportions throughout baking. If the radius grows at a rate of 0.05cm/s what is
the radius of the pie at the instant its volume is changing by 15cm³/s.
C. 9.156 cm
D. 16 cm
2) Lego Erica is running in a circle away from Two-Face (she
figures it will confuse him). When the radius is 17 meters, she
will reach her car and be able to get away. If the area of the
circle she creates is changing at a rate of 23 meters squared
per second, at what rate is the circumference of the circle
changing when she reaches her car?
A) 18𝜋 meters/ second
B) -1.353 meters/ second
C) 23 meters/ second
D) 1.353 meters/ second
E) .6765 meters/ second
3) Indiana Jones is trying to escape the clutches of one of his
stereotypical enemies. He comes to a ledge (perpendicular to
the ground) with a wooden plank leaning against it forming a
right triangle with the ground. He leaps on the top of the plank
and it begins to slide towards the ground. He deems it safe to
jump off this 30 meter long plank only when it is 2 meters off
the ground. If the distance between the plank and the wall is
increasing at .2 meters per second, what rate is the plank moving
down the ledge when he jumps?
-2.993 meters/ second
12.007 meters/ second
-12.007 meters/ second
-3.14 meters/ second
4) The jawa’s are roaming the deserts of Tatooine looking for
droids. Today is a bountiful harvest and they are filling up
their vehicle quick. The vehicle has a rectangular floor with a
width of 17 meters. If the droids are brought in so that the
floor fills at 32 meters squared per hour and the width and
length grow at a rate of 1.5 meters per hour, what is the
length when the vehicle is full in width?
A) 38.333 meters
B) .837 meters
C) 3.213 meters
D) 4.333 meters
E) - .75 meters
5) Ron is trying to perform the spell Wingardium Leviosa but is not doing well.
Hermione is trying to explain that, in order to do the spell, one must flick
their wrist so that the change of angle in their wand (relative to a parallel line
to the ground) must be precisely = −12
. Ron’s wand is 10 inches
and stays that way throughout the whole spell. If the distance between the
tip of the wand and Ron’s hand, distance d, is 1 inch at the end of the spell, at
what rate is this distance changing at the end of the spell?
A) 119 inches/second
B) -1.2 inches/second
C) -11,940 inches/second
D) -231 inches/ second
E) -199 inches/second
While Golem and Bilbo are fighting over the Ring, they are unaware of the fact that
Golem’s cave is slowly filling with water. The bottom of the cave takes on the rough
shape of a wide cone. Volume of a cone = 𝜋𝑟 2 ℎ
At what rate is the volume of the water in the cave increasing when the height is
equal to 120cm if the height of the water is changing at a rate of 0.2cm/s. The
dimensions of the cave are r = 3000cm, h=600cm.
At this point the two suddenly notice the water and consequently the ring is launched
as a projectile with an initial horizontal velocity of 5m/s and an initial vertical velocity
of 3m/s upward. The acceleration due to gravity is 9.8 m/s². Find the position vector
of the particle in terms of time (t).
Find the maximum height of the ring and the time where this occurs. Justify your
Find the total distance that the ring traveled from its release to its maximum height.