1 chapter 2 differentiation: related rates – day 1

1

Chapter 2Chapter 2

Differentiation:Differentiation:Related Rates – Day 1

When we talk about a quantity change with respect to time (whether something is whether something is increasing, decreasing, growing, increasing, decreasing, growing, shrinking, etc.shrinking, etc.), we are basically talking about the derivative of that quantity with respect to time.

For example, let’s say little Johnny is growing at a rate of 2 inches per year. We can write this rate mathematically:

Related RatesRelated Rates

yearindt

dh/.2

Basically, this tells me that the change in height with respect to time is 2 in. per year.

Determine the mathematical expressions for each of the following:

1. The radius of a circle is increasing at 4 ft/min

2. The volume of a cone is decreasing at 2 in3/second

3. The account is shrinking by $50,000/day


min/4 ftdt

dr

sindt

dV/2 3

daydollarsdt

da/000,50

Related RatesRelated Rates Often times, we want to know how one one

variablevariable is relatedrelated to another variableanother variable with respect to timetime. In other words, we want to know how ratesrates are relatedrelated; thus, we study related ratesrelated rates!!!

In order to determine the relationship among rates, we need to differentiate with respect to time; in other words, we need to be able to determine:

dt

d

For example, let’s say that a rectangle is 10in x 6in whose sides are changing. The formulas for both the perimeter and area in terms of l and w are:

1. The rate of change for the perimeter is:

2. The rate of change for the area is:


wlP 22 lwA

wlPdt

d22

dt

dw

dt

dl

dt

dP22

lwAdt

d

dt

dlw

dt

dwl

dt

dA

For example, let’s say that a rectangle is 10in x 6in whose sides are changing.

1. The formulas for the rates of change if the length and width are increasing at a rate of 2 in/s are:

2. Given this information, we can determine the rate of change of perimeter:


sindt

dl/2 sin

dt

dw/2

sin /8 )/2(2/22 sinsindt

dP

dt

dw

dt

dl

dt

dP22 Since

For example, let’s say that a rectangle is 10in x 6in whose sides are changing.

1. The formulas for the rates of change if the length and width are increasing at a rate of 2 in/s are:

2. Given this information, we can also determine the rate of change of area:


sindt

dl/2 sin

dt

dw/2

sin /32 2)/2)(6()/2)(10( sininsinindt

dA

dt

dlw

dt

dwl

dt

dA Since

Related RatesRelated Rates Related rates can be difficult for some to

master, so I’ve come up with 5 steps to help you solve these problems:

1. Make a sketch of the situation, introducing all known and unknown variables and information

2. Identify the given rates of change and determine the rate of change to be found (all all rates of change should be written as rates of change should be written as derivativesderivatives)

3. Write the equation that relatesrelates the variables. You may have to write several equations.

4. Differentiate with respect to time, t5. Substitute known information to solve for the

desired rate of change.

Related RatesRelated Rates NoteNote: It is a common MISTAKEMISTAKE to try to

substitute known information before differentiating implicitly. Do not try to substitute the known information until you have completed the implicit differentiation!!!

An oil tanker hits an iceberg and starts to spill oil creating a circular region. The oil is rapidly spilling out of the tanker and the radius of the circle increases at rate of 2 ft/s. How fast is the area of the oil increasing when the radius is 60 ft?

Step 1Step 1: Make a sketch: Make a sketch

Example 1:Example 1:

At this exact point in time, the radius is 60 ft (rr = 60 = 60 ftft).


Step 2Step 2: Identify given rate(s) and rate(s) to : Identify given rate(s) and rate(s) to be foundbe found


sftdt

dr/2 radiusin change

areain change dt

dA


Step 3Step 3: Write the equation(s) that relate the : Write the equation(s) that relate the variablesvariables The area of a circle formula:


2rA


Step 4Step 4: Differentiate with respect to time: Differentiate with respect to time


2rAdt

d

dt

drr

dt

dA 2


Step 5Step 5: Substitute and solve for the desired : Substitute and solve for the desired raterate


)/2)(60(2 sftftdt

dA

s

ft

dt

dA 2

240

dt

drr

dt

dA 2

ft r sftdt

dr60 and /2 that Recall

A ladder 20 ft. long leans against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 2 ft. per second, how fast is the ladder sliding down the wall when the top of the ladder is 12 feet above the ground?



At this exact point in time, the length of the ladder is 20 ft and the height of the ladder is 12 ft (y = 12 fty = 12 ft). So, by the Pythagorean Theorem, the base must be 16 ft away from the wall (x = x = 16 ft16 ft).




sftxdt

dx/2 in change

ydt

dyin change


Step 3Step 3: Write the equation(s) that relate the : Write the equation(s) that relate the variablesvariables The Pythagorean Theorem will related the

variables:


222 20 yx




222 20 yxdt

d

022 dt

dyy

dt

dxx




02)/2(2 dt

dyysftx

0)12(2)/2)(16(2 dt

dyftsftft

022 dt

dyy

dt

dxx

sftdt

dyft /6424 2

sftft

sft

dt

dy/

3

8

24

/64 2

HomeworkHomework Related Rates

Read Sections 2.6 P. 154: 1 – 35 (odd) and 36 WS: Related Rates Homework WS: Related Rates Turvey WS: Related Rates Problem Set

21

Chapter 2Chapter 2


A space shuttle is launched and is rising at a rate of 880 ft/s when it is 4000 ft up. If a camera is mounted on the ground 3000 feet away from where the shuttle is launched, how fast must a camera angle change at that instant in order to keep the shuttle in sight?



At this exact point in time, the height of the rocket is 4000 ft (y = 4000 fty = 4000 ft) and the height of the horizontal distance from the rocket to the camera is 3000 ft (this is constant, this is constant, so we don’t give it so we don’t give it a variable!a variable!)

θθ




sftdt

dy/088 height in change

angle in the change dt

d


Step 3Step 3: Write the equation(s) that relate the : Write the equation(s) that relate the variablesvariables The equation that will relate y with θ is tan θ :


3000tan

y




ft

y

dt

d

3000tan

dt

dy

ftdt

d

3000

1sec2




dt

dy

dt

d

3000

1sec2

3000

4000tan4000,yWhen

radians927.0

Before we can substitute and solve, we need to Before we can substitute and solve, we need to know know dydy//dtdt and and θθ. . We were given We were given dy/dtdy/dt, but , but we still need to solve for we still need to solve for θθ at this point in at this point in time.time.

3

4tan 1




s

ft

dt

d 880

3000

1)927.0(sec2

sft /1057.0

dt

dy

dt

d

3000

1sec2

)927.0(cos880

3000

1 2

s

ft

dt

d

You are trying to fill a tank in the shape of a cone with an altitude of 12 ft and a radius of 6 ft with water. The water is getting pumped into the tank at a rate of 10 gallons per minute. Find the rate at which the water rises when the water is 3 feet deep. Note 1Note 1: 1 gallon : 1 gallon = .1337 cu ft.= .1337 cu ft.; Note 2Note 2: The water fills rapidly at : The water fills rapidly at first, but slows down as the cone gets wider.first, but slows down as the cone gets wider.



At this exact point in time, the height of the water is 3 ft (hh = 3 ft = 3 ft); however, the height of the entire cone is 12 ft. and the radius of the entire cone is 6 ft.

h

You are trying to fill a tank in the shape of a cone with an altitude of 12 ft and a radius of 6 ft with water. The water is getting pumped into the tank at a rate of 10 gallons per minute. Find the rate at which the water rises when the water is 3 feet deep.



heightin changedt

dh

radius in the change dt

dr

min/10 volumein the change galdt

dV

NOTE: There are too NOTE: There are too many unknown many unknown variables!!! So, we variables!!! So, we will eventually need will eventually need to eliminate one!!! to eliminate one!!!


Step 3Step 3: Write the equation(s) that relate the : Write the equation(s) that relate the variablesvariables The equation that will relate r and hh with VV is:


hrV 2

3

At this point, we need to eliminate one of the variables; the rate of volume is given and we are looking for the rate that water rises (height), so we need to try to eliminate the radiusradius!

h

In order to eliminate the radius from the equation (i.e. write volume using only height), we must go back to geometry and consider similar triangles:


Set up a ratio of similar triangles and solve for rr.

h

12

6

h

r

2

hr h

hh

hV

4323so

22 3

12h


Step 3Step 3: Write the equation(s) that relate the : Write the equation(s) that relate the variablesvariables The equation that will relate hh with VV is:


3

12hV

Note: When we complete step 3, the equation Note: When we complete step 3, the equation

should should ONLYONLY include the variables for which include the variables for which

the rates are the rates are GIVENGIVEN and the variable whose and the variable whose rate we are finding!!!rate we are finding!!!




3

12hV

dt

d

dt

dhh

dt

dV 2

4




gal

ftgal 31337.0

min10

min

337.1 3ft

fthgaldt

dV3 and min/10 Recall

Before we can solve, we need to convert Before we can solve, we need to convert from gallons to feet:from gallons to feet:

gal

ft. 313370 is ratio conversion The




dt

dhft

min

189.0

dt

dhft

ft 23

)3(4min

337.1

dt

dhft

ft

min

337.1

9

4 3

2

dt

dhftft

4

9

min

337.1 23 dt

dhh

dt

dV 2

4

37

Chapter 2Chapter 2


An 11 foot by 20 foot swimming pool is being filled at a constant rate of 5 ft3 per minute. The pool has a depth of 4 feet in the shallow end and 9 feet in the deep end. (Assume the pool’s depth increases constantly between the shallow end and the deep end.) How fast is the height of the water increasing when the water has filled 2 feet of the deep end?



At this exact point in time, the height of the water is 2 ft (hh = 2 ft = 2 ft)




min

5in volume change

3ft

dt

dV

height in the change dt

dh

base in the change dt

db

NOTE: There are NOTE: There are too many too many unknown unknown variables. So, variables. So, we will we will eventually need eventually need to eliminate one.to eliminate one.

Step 3Step 3: Write the equation(s) that relate the : Write the equation(s) that relate the variablesvariables This is a tough one! The shape of the pool is a 3-dimensional

trapezoid or a trapezoidal prism. It will be very difficult to work with the equation:

So what do we do? We need to look at the shape in another way. Well, if we think about how water rises in the

shape, it starts at the bottom of the deep end.


whbb

V2

21

Step 3Step 3: Write the equation(s) that relate the : Write the equation(s) that relate the variablesvariables While the water is rising from 0 to 5 feet (since

9ft – 4ft = 5 ft), our 3-dimensional shape is a triangular prism – a much easier shape to solve:


bhwV2

1

This part is superfluous

Now finding the Now finding the volume of a volume of a triangular prism is triangular prism is much easier:much easier:Width is constant, so:Width is constant, so: bhV

2

11

There are still too many variables. We know dV/dtdV/dt and we are looking for dh/dtdh/dt, so we should try to eliminate bb (in other words, we need to write the equation for volume in terms of height alone).


Set up a ratio of similar triangles and solve for bb.

5

20

h

bhb 4

hhbhV )4(2

11

2

11so 222h




222hVdt

d

dt

dhh

dt

dV44




dt

dhft

min88

5dt

dhftft

ft )2(44

min

5 3

dt

dhh

dt

dV44

dt

dhft

min057.0

1 chapter 2 differentiation: related rates – day 1

Documents

given rates of change

desired rate of change

quantity change

related rates day

oil tanker

related ratesnote

related ratesfor example

related ratesbasically