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Two or more quantities can have a constant ratio. For example, the circumference of any circle is proportional to its diameter, and its ratio is always equal to a constant,

π .

In this lesson, equations in two or more variables will be created to represent relationships between quantities and their graphs will be analyzed.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

Number of Phone Calls Per Day Between Two Cities

The average number of phone calls per day between two cities $N$ varies directly with the populations of the cities $P_{1}$ and $P_{2},$ and inversely with the square of distance $d$ between the two cities.

External credits: TUBS

a Which equation models the given situation?

b In

2010,

the population of San Francisco was about

806000,

and the population of Portland was about

585000 .

The distance between the two cities is about

650

miles. In the given year, the average number of calls between the cities was about

42000 .

Use the model determined in Part A to Find the constant

k .

Round the answer to three decimal places.

c In

2010,

the average number of daily phone calls between San Francisco and Los Angeles, with a population of about

3800000,

was about

806000 .

Use the model to find the approximate distance between the two cities.

Discussion

Joint Variation

A joint variation, also known as joint proportionality, occurs when one variable varies directly with two or more variables. In other words, if a variable varies directly with the product of other variables, it is called joint variation.

$z = k x y$

Here, the variable $z$ varies jointly with $x$ and $y,$ and $k$ is the constant of variation. Here are some examples of joint variation.

Examples of Joint Variation
Example	Rule	Comment
The area of a rectangle	$A = ℓ w$	Here, $ℓ$ is the rectangle's length, $w$ its width, and the constant of variation $k$ is $1 .$
The volume of a pyramid	$V = \frac{1}{3} ℓ w h$	Here, $ℓ$ and $w$ are the length and the width of the base, respectively, while $h$ is the pyramid's height. The constant of variation $k$ is $\frac{1}{3} .$

Lastly, it is important to note that joint variation is closely related to other types of variation.

Example

Television Series

Vincenzo and Emily are having a lively chat about television series they love. Emily managed to watch $164$ episodes of The Flash in just $50$ days! Each episode typically lasts $40$ minutes.

Emily and Vincenzo agree that the number of days it takes to watch an entire show

d

is jointly proportional to the number of episodes

e

and the length of the episodes

ℓ .

d varies jointly with e and ℓ .

Vincenzo wants to watch Sherlock, which has

13

episodes, and each episode is about

90

minutes long. How long will it take to watch the series if Vincenzo watch the series at the same rate as Emily did? Round the answer to the nearest integer.

Hint

Use the fact that if $z$ varies jointly with $x$ and $y,$ the equation of variation is $z = k x y$ , where $k$ is the constant of variation.

Solution

Begin with recalling that if

z

varies jointly with

x

and

y,

the equation of variation is

z = k x y .

z = k \cdot x \cdot y

Here,

k

is the constant of variation and

k \neq = 0 .

In this case, the number of days it takes to watch a show

d

is jointly proportional to the number of episodes

e

and the length of the episodes

ℓ .

This means the equation of the variation can be written as follows.

d = k \cdot e \cdot ℓ

Now,

k

will be determined by using the given information about the series Emily watched. It is given that it took

50

days to watch

164

episodes that are each about

40

minutes long. Substitute

d = 50,

e = 164,

and

ℓ = 40

into the equation and solve for

k .

d = k e ℓ

SubstituteValues

Substitute values

50 = k \cdot 164 \cdot 40

▼

Solve for

k

Multiply

50 = k \cdot 6560

DivEqn

$LHS / 6560 = RHS / 6560$

\frac{5 0}{6 5 6 0} = k

ReduceFrac

$\frac{a}{b} = \frac{a / 1 0}{b / 1 0}$

\frac{5}{6 5 6} = k

RearrangeEqn

Rearrange equation

k = \frac{5}{6 5 6}

Using

k,

the time it will take to watch Sherlock can be determined. Recall that it is given that there are

13

episodes and each is

90

minutes long.

d = \frac{5}{6 5 6} e ℓ

SubstituteValues

Substitute values

d = \frac{5}{6 5 6} \cdot 13 \cdot 90

▼

Evaluate right-hand side

Multiply

d = \frac{5}{6 5 6} \cdot 1170

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

d = \frac{5 8 5 0}{6 5 6}

CalcQuot

Calculate quotient

d = 8.917682 \dots

RoundInt

Round to nearest integer

d \approx 9

It will take Vincenzo about

9

days to watch Sherlock. Time to stock up on snacks!

Explore

Recognzing Inverse Variation

Use the applet to investigate the relationship between the width and the length of rectangles.

For rectangles with a fixed area, say

64

square units, complete the table.

Width	Length
$2$
$4$
$8$
$16$
$32$

Draw a scatter plot of the data. Does it make sense to say that the width varies inversely with the length? Explain.

Discussion

Inverse Variation

An inverse variation, or inverse proportionality, occurs when two non-zero variables have a relationship such that their product is constant. This relationship is often written with one of the variables isolated on the left-hand side.

$x y = k or y = \frac{k}{x}$

The constant

k

is the constant of variation. When

k = 0,

the relationship is not an inverse variation. In the following example, the constant of variation is

k = 2 .

A graph of a function y=2/x with a point on the graph that can be moved

The constant of variation may be any real number except

0 .

Here are some examples of inverse variations.

Examples of Inverse Variation
Example	Rule	Comment
The gas pressure in a sealed container if the container's volume is changed, given constant temperature and constant amount of gas.	$P = \frac{n R T}{V}$	The variables are the pressure $P$ and the volume $V .$ The amount of gas $n,$ temperature $T,$ and universal gas constant $R$ are fixed values. Therefore, the constant of variation is $n R T .$
The time it takes to travel a given distance at various speeds.	$t = \frac{d}{s}$	The constant of variation is the distance $d$ and the variables are the time $t$ and the speed $s .$

Inverse variation is closely related to other types of variation.

Pop Quiz

Identifying the Type of Variation

Determine whether the relationship between the variables in the table shows direct or inverse variation, or neither.

Example

Number of Songs on Emily's Phone

Emily, tired of watching shows, wants to update the playlist on her phone before starting a family road trip from Portland to San Francisco. The number of songs that can be stored on her phone varies inversely with the average size of a song.

Emily's phone can store $4100$ songs when the average size of a song is $4$ megabytes (MB).

a How many songs with an average size of

5

megabytes does Emily's phone store?

b List the number of songs that will fit on the phone when the average size of a song is

3 MB,

4 MB,

5 MB,

and

6 MB,

respectively. Round the answer to the nearest integer whenever necessary.

Hint

a Begin by finding the constant of variation.

b Use the inverse variation equation to create a table.

Solution

a It is given that the number of songs

y

that can be stored on Emily's phone varies inversely with the average size

x

of a song. Recall the form of an inverse variation equation that relates

x

and

y .

y = \frac{k}{x}

In the equation,

k

is the constant of variation. It is known that when the average size of a song is

x = 4

megabytes, the phone can store up to

y = 4100

songs. Therefore, substitute

y = 4100

and

x = 4

into the inverse variation equation and solve for

k .

y = \frac{k}{a}

SubstituteII

$x = 4$ , $y = 4100$

4100 = \frac{k}{4}

MultEqn

$LHS \cdot 4 = RHS \cdot 4$

16400 = k

RearrangeEqn

Rearrange equation

k = 16400

The constant of variation is

16400 .

Using this information, the number of songs with an average size of

5 MB

can be found.

y = \frac{1 6 4 0 0}{a}

Substitute

$x = 5$

y = \frac{1 6 4 0 0}{5}

CalcQuot

Calculate quotient

y = 3280

b In the previous part, the inverse variation equation was found.

y = \frac{1 6 4 0 0}{x}

Now a table that shows the number of songs when the average size of a song is

3 MB,

4 MB,

5 MB,

and

6 MB

can be made.

Size, $x$	$\frac{1 6 4 0 0}{x}$	Number of Songs, $y$
$3$	$\frac{1 6 4 0 0}{2}$	$5466$
$4$	$\frac{1 6 4 0 0}{4}$	$4100$
$5$	$\frac{1 6 4 0 0}{5}$	$3280$
$6$	$\frac{1 6 4 0 0}{6}$	$\approx 2733$

In the table, as the size gets larger, the number of songs that the phone can store gets smaller. Therefore, the number of songs decreases as the average size increases.

Example

Emily's Trip to San Francisco

Now that the updated playlist and everything else is ready, Emily's journey from Portland to San Francisco can begin.

External credits: TUBS

The time it takes to reach San Francisco varies inversely with their average rate of speed.

a If Emily lives just outside of the main City of Portland

640

miles from San Francisco by car, write an equation that relates the travel time to the average speed. In addition to the equation, draw its corresponding graph.

b Determine the minimum average speed that Emily would need to drive to reach San Francisco within

12

hours. Round the answer to the nearest integer.

Answer

a Example Equation:

t = \frac{6 4 0}{r}

Example Graph:

b About

53

mph

Hint

a Use the standard equation of the inverse variation,

y = \frac{k}{x},

where

k

is the constant of variation.

b Substitute

12

for the time into the equation from Part A and solve for the rate of speed.

Solution

a It is given that the time it takes Emily to reach San Francisco varies inversely with her average rate of speed. Let

t

be the time and

r

be the rate of speed. Then, the following equation can be written.

t = \frac{k}{r}

Here,

k

is a constant of variation. It is also given that Emily lives

450

miles away from San Francisco. This is the value of

k .

Recall that a distance is a product of the time and rate of speed. Thus, the time can be expressed as a quotient of the distance by the rate of speed.

d = r t \Rightarrow t = \frac{d}{r}

Comparing this formula with the equation, it is seen that

k

indeed represents the distance. Therefore,

k

can be substituted with

640 .

t = \frac{6 4 0}{r}

Now make a table of values to graph the equation.

$r$	$\frac{6 4 0}{r}$	$t$
$10$	$\frac{6 4 0}{1 0}$	$64$
$20$	$\frac{6 4 0}{2 0}$	$32$
$30$	$\frac{6 4 0}{3 0}$	$\approx$ $21$
$40$	$\frac{6 4 0}{4 0}$	$16$
$50$	$\frac{6 4 0}{5 0}$	$12.8$
$60$	$\frac{6 4 0}{6 0}$	$\approx$ $10.7$

Ordered pairs $(r, t)$ are the coordinates of the points on the graph. Plot the points and connect them with a smooth curve.

b To determine the minimum average speed that will allow Emily to arrive to San Francisco within

12

hours, substitute

12

for

t

into the equation from Part A and solve it for

r .

t = \frac{6 4 0}{r}

Substitute

$t = 12$

12 = \frac{6 4 0}{r}

MultEqn

$LHS \cdot r = RHS \cdot r$

12 r = 640

DivEqn

$LHS / 12 = RHS / 12$

r = 53.333333 \dots

RoundInt

Round to nearest integer

r \approx 53

Emily should travel at a minimum speed of

53

miles per hour.

Discussion

Combined Variation

A combined variation, or combined proportionality, occurs when one variable depends on two or more variables, either directly, inversely, or a combination of both. This means that any joint variation is also a combined variation.

$z = \frac{k x}{y}$

The variable $z$ varies directly with $x$ and inversely with $y,$ and $k$ is the constant of variation. Therefore, this is a combined variation. Here are some examples.

Examples of Combined Variation
Example	Rule	Comment
Newton's Law of Gravitational Force	$F = \frac{G m _{1} m _{2}}{d ^{2}}$	The gravitational force $F$ varies directly as the masses of the objects $m_{1}$ and $m_{2},$ and inversely as the square of the distance $d^{2}$ between the objects. The gravitational constant $G$ is the constant of variation.
The Ideal Gas Law	$P = \frac{n R T}{V}$	The pressure $P$ varies directly as the number of moles $n$ and the temperature $T,$ and inversely as the volume $V .$ The universal gas constant $R$ is the constant of variation.

Combined variation is closely related to other types of variation.

Example

Number of T-Shirts Sold

Emily is wandering around a gift shop to buy gifts for some of her friends. Emily overhears a conversation between the shopkeeper and an employee. The shopkeeper says that the number of t-shirts sold is directly proportional to their advertising budget and inversely proportional to the price of each t-shirt.

When $$ 1200$ are spent on advertising and the price of each t-shirt is $$ 4.80,$ the number of t-shirts sold is $6500 .$ How many t-shirts are sold when the advertising budget is $$ 1800$ and the price of each t-shirt is $$ 6 ?$

Hint

Use the equation of the combined variation, $z = \frac{k x}{y},$ where $k$ is the constant of variation.

Solution

When one quantity varies with respect to two or more quantities, this variation can be regarded as a combined variation.

Combined Variation	Equation Form
$a$ varies jointly with $b$ and $c .$	$a = k b c$
$a$ varies jointly with $b$ and $c,$ and inversely with $d .$	$a = \frac{k b c}{d}$
$a$ varies directly with $b$ and inversely with the product $d c .$	$a = \frac{k b}{d c}$

Based on this table, an equation that models the given variation can be written. It is given that the number of t-shirts sold

N

varies directly with the advertising budget

b

and inversely with the price of a t-shirt

p .

N = \frac{k b}{p}

Here

k

is the constant of variation and cannot be

0 .

With an advertising budget of

$ 1200

and the t-shirt price of

$ 4.80,

6500

t-shirts are sold. Using this information, the value of

k

can be found. To do so, substitute

N = 6500,

b = 1200,

and

p = 4.80

in the above equation.

N = \frac{k b}{p}

SubstituteValues

Substitute values

6500 = \frac{k ( 1 2 0 0 )}{4 . 8 0}

▼

Solve for

k

MultEqn

$LHS \cdot 4.80 = RHS \cdot 4.80$

31200 = k (1200)

DivEqn

$LHS / 1200 = RHS / 1200$

\frac{3 1 2 0 0}{1 2 0 0} = k

CalcQuot

Calculate quotient

26 = k

RearrangeEqn

Rearrange equation

k = 26

Now that the value of

k

is known, the equation that models the variation can be written.

N = \frac{2 6 b}{p}

The shopkeeper increases the budget to

$ 1800

and the price of a t-shirt to

$ 6 .

To find the value of the

N

under these circumstances, substitute

b = 1800

and

p = 6

into the equation.

N = \frac{2 6 b}{p}

SubstituteII

$b = 1800$ , $p = 6$

N = \frac{2 6 ( 1 8 0 0 )}{6}

▼

Evaluate right-hand side

Multiply

N = \frac{4 6 8 0 0}{6}

CalcQuot

Calculate quotient

N = 7800

Alternative Solution

The number of t-shirts sold

N

varies directly with the advertising budget

b

and inversely with the price of a t-shirt

p .

Additionally,

N = 6500

when

b = 1200

and

p = 4.80 .

The goal is to find

N

when

b = 1800

and

p = 6 .

To find the value of

N,

write two equations that involve the constant of variation

k .

N_{1} = \frac{k b _{1}}{p _{1}} ⇕ k = \frac{N _{1} p _{1}}{b _{1}} N_{2} = \frac{k b _{2}}{p _{2}} ⇕ k = \frac{N _{2} p _{2}}{b _{2}}

Since

k

is equal to both

\frac{x _{1} z _{1}}{y _{1}}

and

\frac{x _{2} z _{2}}{y _{2}},

by the Transitive Property of Equality these two expressions must be equal. Using this information, a proportion can be written.

\frac{N _{1} p _{1}}{b _{1}} = \frac{N _{2} p _{2}}{b _{2}}

Next, substitute

N_{1} = 6500,

b_{1} = 1200,

p_{1} = 4.8,

b_{2} = 1800,

and

p_{2} = 6

and solve for

N_{2} .

\frac{N _{1} p _{1}}{b _{1}} = \frac{N _{2} p _{2}}{b _{2}}

SubstituteValues

Substitute values

\frac{6 5 0 0 ( 4 . 8 )}{1 2 0 0} = \frac{N _{2} ( 6 )}{1 8 0 0}

▼

Solve for

N_{2}

CrossMult

Cross multiply

6500 (4.8) (1800) = N_{2} (6) (1200)

Multiply

56160000 = N_{2} (7200)

DivEqn

$LHS / 7200 = RHS / 7200$

\frac{5 6 1 6 0 0 0 0}{7 2 0 0} = N_{2}

CalcQuot

Calculate quotient

7800 = N_{2}

RearrangeEqn

Rearrange equation

N_{2} = 7800

Pop Quiz

Finding the Value of $z$

In the applet, various types of variations are shown randomly. Find the value of $z$ by using the given values. If necessary, round the answer to the two decimal places.

Closure

Number of Phone Calls Per Day Between Two Cities

In this lesson, variation types are explained with real-life examples. Considering those examples, the challenge presented at the beginning of the lesson can be solved with confidence. Recall that the average number of phone calls per day between two cities varies directly with the populations of the cities and inversely with the square of the distance between the two cities.

External credits: TUBS

a Which equation models the given situation?

b In

2010,

the population of San Francisco was about

806000

and the population of Portland was about

585000 .

The distance between the two cities is roughly

650

miles. In the given year, the average number of calls between the cities was about

42000 .

Use the model determined in Part A to Find the constant

k .

Round the answer to three decimal places.

c In

2010,

the average number of daily phone calls between San Francisco and Los Angeles, with a population of about

3800000,

was about

806000 .

Use the model to find the approximate distance between the two cities.

Hint

a Use the equation of the combined variation,

z = \frac{k x}{y},

where

k

is the constant of variation.

b Substitute the given values to find the value of

k .

c Substitute the given values into the variation equation.

Solution

a Recall the labels of the quantities.

N : P_{1} : P_{2} : d : Number of phone calls per day Population of one of the cities Population of the other city Distance between the cities

It is known that

N

varies directly with

P_{1}

and

P_{2},

and inversely with the square of

d .

Therefore,

N

is equal to the product of

k,

P_{1},

and

P_{2}

divided by the square of

d .

N = \frac{k P _{1} P _{2}}{d ^{2}}

Here

k

is the constant of variation and cannot be

0 .

This equation models the given variation.

b The value of

k

can be found by using the given information and the equation written in Part A.

	San Francisco	Portland
Population	$806000$	$585000$
Distance	$650$
Number of Calls	$42000$

These values represent

N = 42000,

P_{1} = 806000,

P_{2} = 585000,

and

d = 650 .

Now, substitute them into the equation.

N = \frac{k P _{1} P _{2}}{d ^{2}}

SubstituteValues

Substitute values

42000 = \frac{k ( 8 0 6 0 0 0 ) ( 5 8 5 0 0 0 )}{6 5 0 ^{2}}

▼

Solve for

k

Multiply

42000 = \frac{k ( 4 7 1 5 1 0 0 0 0 0 0 0 )}{6 5 0 ^{2}}

CalcPow

Calculate power

42000 = \frac{k ( 4 7 1 5 1 0 0 0 0 0 0 0 )}{4 2 2 5 0 0}

MovePartNumLeft

$\frac{a \cdot b}{c} = a \cdot \frac{b}{c}$

42000 = k \cdot \frac{4 7 1 5 1 0 0 0 0 0 0 0}{4 2 2 5 0 0}

CalcQuot

Calculate quotient

42000 = k \cdot 1116000

DivEqn

$LHS / 1116000 = RHS / 1116000$

\frac{4 2 0 0 0}{1 1 1 6 0 0 0} = k

RearrangeEqn

Rearrange equation

k = \frac{4 2 0 0 0}{1 1 1 6 0 0 0}

CalcQuot

Calculate quotient

k = 0.037634 \dots

RoundDec

Round to $3$ decimal place(s)

k \approx 0.038

The constant of variation is about

0.038 .

c Using the value of

k

found in Part B, the equation that models the variation can be expressed.

N = \frac{0 . 0 3 8 P _{1} P _{2}}{d ^{2}}

For clarity, make a table to organize the given information.

	San Francisco	Los Angeles
Population	$806000$	$3800000$
Distance	$d$
Number of Calls	$806000$

To find the value of the

d,

substitute

N = 806000,

P_{1} = 806000,

and

P_{2} = 3800000

into the equation and solve for

d .

N = \frac{0 . 0 3 8 P _{1} P _{2}}{d ^{2}}

SubstituteValues

Substitute values

806000 = \frac{0 . 0 3 8 ( 8 0 6 0 0 0 ) ( 3 8 0 0 0 0 0 )}{d ^{2}}

▼

Solve for

d

Multiply

806000 = \frac{1 1 6 3 8 6 4 0 0 0 0 0}{d ^{2}}

MultEqn

$LHS \cdot d^{2} = RHS \cdot d^{2}$

d^{2} (806000) = 116386400000

DivEqn

$LHS / 806000 = RHS / 806000$

d^{2} = \frac{1 1 6 3 8 6 4 0 0 0 0 0}{8 0 6 0 0 0}

CalcQuot

Calculate quotient

d^{2} = 144400

SqrtEqn

$LHS = RHS$

d^{2} = 144400

$a^{2} = \pm a$

d = \pm 144400

CalcRoot

Calculate root

d = \pm 380

Since the distance cannot be negative, choose the positive value and not that the distance between the two cities is

380

miles.

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Catch-Up and Review

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Hint

Solution

Alternative Solution

Hint

Solution