7. Scaling and Descriptive Modeling
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Chapter 1
7.
Scaling and Descriptive Modeling
Scaling and descriptive modeling are essential tools in Algebra 1 for understanding relationships between quantities. The lesson covers various real-life situations, such as calculating population density, determining the cost of paint needed for a room, and understanding unit rates in different contexts like phone plans and shopping deals. These mathematical tools help in making informed decisions, whether it is choosing the most cost-effective paint or selecting a phone plan based on call duration and cost. The importance of using appropriate units for different quantities is also emphasized, making the concepts more relatable and easier to apply in everyday life.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Scaling and Descriptive Modeling
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In real-world situations, quantities are accompanied by appropriate units so that the situations can be easily understood and interpreted. Before analyzing situations involving too large or too small quantities, their models are created by keeping the ratio between corresponding lengths constant. In this lesson, this ratio will be named and used to analyze real-life situations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Scaling the Solar System
In the model of the Solar System below, the initial labeled distance corresponds to 93 000 000 miles. Zoom in or out to see how the labeled distance is affected.
Think about the following questions:
- Is it possible to make a model of the Solar System using the actual sizes of planets and distances between planets?
- Is it possible to draw a model of the Solar System on a sheet of paper? If so, how many miles should be represented by 1 inch?
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Different Presentations in Modeling
When modeling a real-life situation, it is sometimes impossible to represent an object or scenario using the original dimensions. In these situations, it is more convenient to work with manageable units while still being able to maintain the same properties as the original. This can be done by making a scale drawing.
Concept
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Scale Drawing
A scale drawing is a two-dimensional drawing that is similar to an actual object or place. In a scale drawing, the ratio of any length on the drawing to the actual length is always the same. drawing→/actual→ l_1/L_1 = l_2/L_2 ←drawing/←actual Possible examples of scale drawing are floor plans, blueprints, and maps.
In the case of the original real-life situation involving a three-dimensional object, making a 3D model is more useful than a drawing. The idea behind a 3D model is the same as a scale drawing, but the model has three dimensions instead of two.
Concept
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Scale Model
A scale model is a three-dimensional model that is similar to a three-dimensional object. The ratio of a linear measurement of a model to the corresponding linear measurement of the actual object is always the same. model→/actual→ l_1/L_1 = l_2/L_2 ←model/←actual Here is an example scale model of a building.
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Scale
The scale of a model or drawing is the ratio between any length on the model or drawing and its corresponding length on the actual object or place.
lLength on the drawing : lCorresponding length on the actual object
Suppose a drawing has a scale of 1 in:100 ft. This means that 1 inch on the drawing represents 100 feet on the actual object. Apart from the colon notation, a scale can be expressed using an equals sign or as a fraction, as it is a ratio.
| Denoting a Scale | |
|---|---|
| Ratio | 1 in : 100 ft |
| Equals Sign | 1 in = 100 ft |
| Fraction | 1 in/100 ft |
When a scale is written without specifying the units, it is understood that both numbers have the same unit of measure. For example, a scale of 1:2 means that the actual object is twice the size of the model. A scale of 1:0.5 means that the actual object is half the size of the model — whether it be in meters, inches, yards, and so on.
External credits: Derek Quinn
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Using the Scale to Find Actual Dimensions
The Sweden Solar System is the world's largest model of the planetary system. In this model, the Globe Arena in Stockholm represents the Sun, while the planets align to the north. Click on the dots below, representing the planets, to show the dimensions, location, and distance to the Globe Arena.
The scale used to make this model is 1:20 million.
a What is the actual diameter of the Earth? Give the answer in kilometers.
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b What is the actual distance from Pluto to the Sun?
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Solution
a Since the scale does not specify the units, it means that both numbers have the same units. Then, to find the actual diameter of the Earth, multiply both numbers by the diameter of the Earth in the model, 65 centimeters.
1 300 000 000 cm
MultII
Multiply 1 300 000 000 cm by 1 m/100 cm*1 km/1000 m
1 300 000 000 cm * 1 m/100 cm*1 km/1000 m
▼
Simplify
CrossCommonFac
Cross out common factors
1 300 000 000 cm * 1 m/100 cm*1 km/1000 m
CancelCommonFac
Cancel out common factors
1 300 000 000 * 1/100* 1 km/1000
MultFrac
Multiply fractions
1 300 000 000/100 000 km
CalcQuot
Calculate quotient
13 000 km
b Similarly, to find the actual distance from Pluto to the Sun, multiply both numbers in the scale by 300 km.
300km* 1= 20 000 000^(Original scale) * 300km ⇕ 300 km=6 000 000 000 km In conclusion, the distance from Pluto to the Sun is 6 billion kilometers.
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Finding the Scale of a 3D Model
Tom made a 3D model of the Empire State Building for a school project. The model has a height of 55.375 centimeters.
The actual height of the Empire State Building is 443 meters, including its antenna.
a According to the scale Tom used, how many meters correspond to 1 centimeter on the model?
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b What is the actual width of the Empire State Building in meters?
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Hint
a The scale compares a dimension on the 3D model to the corresponding dimension on the original object.
b Multiply each side of the scale by the width of the 3D model.
Solution
a By definition, the scale of a 3D model compares a dimension on the model to the corresponding dimension on the actual object. In this case, compare the height of the model and the actual height of the Empire State Building in order to find the scale.
ccc Model Height& &Actual Height 55.375 cm &:& 443 m According to this scale, 55.375 centimeters correspond to 443 meters. The scale can be simplified by dividing both numbers by 55.375. 55.375/55.375 cm : 443/55.375 m ⇕ 1 cm : 8 m In the scale used by Tom, 1 centimeter corresponds to 8 meters.
b Although the scale was found using the height, the same scale is used for the width as well. Therefore, the actual width can be calculated by multiplying each side of the scale by the width of the 3D model, which is 7.125 centimeters. This time the scale in the form of an equation will be used.
7.125* 1 cm= 8 m * 7.125 ⇕ 7.125 cm= 57 m Therefore, the Empire State Building is 57 meters wide.
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Calculating Population Density
The United States of America has an area of 9 147 420 square kilometers. In mid-2020, the total population was 331 923 317 people.
External credits: @rocketpixel
Answer
In the United States of America, the population density is 94 people per square mile. The population density measures the number of people per unit of area.
Hint
People per square mile can be written as peoplemi^2. This implies that the population density is expressed as a ratio. Remember that 1 square kilometer is equal to 0.386102 square mile.
Solution
According to the problem, the population density needs to be given using the unit of people per square mile. This means that the population density is found by dividing the number of people by the area of a certain region.
Population:& 331 923 317 people Area:& 9 147 420 km^2
Before calculating the quotient, note that the area given is in square kilometers. Therefore, a unit conversion from square kilometers to square miles should be performed. To do this, use the fact that 1 square kilometer is equal to about 0.386102 square miles.
The next step is to divide the population by the area, which gives the population density. Here, the result is rounded to the nearest integer.
Pop. density &= 331 923 317 people/3 531 837.15684 mi^2 [0.25cm]
Pop. density &≈ 94 people/mi^2
Now that the population density is known, write the answer in a sentence form.
The population density measures the number of people per unit of area.
9 147 420 km^2
MultII
Multiply 9 147 420 km^2 by 0.386102 mi^2/1 km^2
9 147 420 km^2 * 0.386102 mi^2/1 km^2
▼
Simplify
CrossCommonFac
Cross out common factors
9 147 420 km^2 * 0.386102 mi^2/1 km^2
MultFrac
Multiply fractions
3 531 837.15684 mi^2
The population density of the United States of America is approximately 94 people per square mile.
This statement assumes that the total population is spread out evenly across the United States.
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Practicing Converting a Scale
Mark wants to paint all of the walls of his bedroom except for the wall that contains the door. Each rectangular wall is 2.8 meters high. The paint he will use is sold in 5-liter cans — the price per liter is $ 25.
After painting for a few minutes, Mark noticed that 5 square meters can be covered with one liter of paint.
a What is the total area, in square meters, that Mark has to paint?
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b What is the minimum number of liters of paint Mark needs?
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c How many cans should Mark buy?
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d How many dollars could Mark save if the paint was sold in 1-liter cans?
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Hint
a Use the formula for the area of a rectangle to calculate the area of each wall.
b How many liters of paint are needed to cover one square meter of wall?
c Compare the number of liters needed to cover all three walls to the amount of paint in one can.
d Calculate the difference between the amount of money that Mark paid and the cost of 7 liters of paint.
Solution
a Start by writing the dimensions of the walls to be painted. Note that two walls have the same dimensions. This means that they also have the same area.
| Dimensions | |
|---|---|
| Wall 1 | l_1 &= 4.75 m h &= 2.8 m |
| Wall 2 | l_2 &= 4.75 m h &= 2.8 m |
| Wall 3 | l_3 &= 3 m h &= 2.8 m |
A_1 = l_1 * h
SubstituteII
l_1= 4.75 m, h= 2.8 m
A_1 = ( 4.75 m)( 2.8 m)
Multiply
Multiply
A_1 = 13.3 m^2
b Saying that 5 square meters can be covered with a liter of paint is equivalent to saying that 15 of a liter of paint can cover 1 square meter.
5 m^2/liter ⇔ 1 liter/5 m^2 Multiplying 15 liters/m^2 by the total area gives the number of liters that Mark needs to paint the three walls. 35 m^2 * 1 liter/5 m^2 = 7 liters
c Mark needs 7 liters of paint, but each can of paint contains 5 liters. This means that Mark will actually have to buy 2 cans to have enough paint to cover all three walls.
d The cost of the 2 cans can be calculated by multiplying the number of liters in both cans by the price per liter. Each can contains 5 liters, so in total there are 10 liters of paint.
Price of2cans &= 10 liters* 25 dollars/liter [0.8em] &= 250 dollars If the paint was sold in 1-liter cans, Mark would need to buy 7 cans. In this case, he would pay 7* 25=$175. The amount of money that Mark could have saved can be found by calculating the difference between the price of 10 liters of paint and 7 liters of paint. 250-175=$ 75
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Practicing Finding or Using the Scale
On the applet, the model and the actual object are shown. Using the given information, find the scale or the size of either the model or the actual object.
- When entering the scale, write it in the form of a fraction with the numerator of 1.
- When entering the size of a model or an actual objects, round to the closest integer.
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Calculating the Scale to Choose Better Phone Plan
Tadeo wants to buy a cheap phone plan for calling his friends and family. He asked Ali and Ramsha how much they pay per call.
- Ali said he paid $10 for his last call.
- Ramsha said she paid $8 for her last call.
Initially, Tadeo decided to take Ramsha's plan since she paid less. Later, he realized that this information is not helpful since he does not know the duration of each of the calls made by Ali and Ramsha.
Price per call is not an appropriate unit.
Therefore, Tadeo decided to ask Ali and Ramsha how long each phone call lasted.
- Ali said his call lasted 40 minutes.
- Ramsha said her call lasted 30 minutes.
Since the duration of the calls is different, Tadeo became confused and made the following diagram to think about the situation. Ali: & $10 → 40 minutes Ramsha: & $8 → 30 minutes After this, Tadeo realized that dividing each call's cost by its duration will give him the price per minute, which is an appropriate unit to compare the plans.
| Person | Scale of the Plan |
|---|---|
| Ali | $10/40 min=$0.25 per minute |
| Ramsha | $8/30 min≈ $0.27 per minute |
Scaling and Descriptive Modeling
Exercise 3.1
Star Trek is about the starship Enterprise and its merry crew. In this fictional universe, mankind has figured out how to travel at warp speed (w) where Warp 1 corresponds to the speed of light. Greater levels of speed can be calculated by using the following formula.
w^3=v/c
Here, v is the speed of the ship and c is the speed of light equal 1 080 000 000 km/h. How many whole laps around the Earth could the Enterprise complete in one second if they traveled at Warp 9?
The Earth has a radius of 6371 kilometers. Assume that the Enterprise travels 150 kilometers above the surface of the Earth.
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To determine the number of laps the Enterprise can make around the Earth in 1 second, we first need to determine how far the ship travels with the speed of Warp 9.
Speed in km/s
To find the speed, let's substitute the warp and speed of light in the given formula.
Now we know the speed in kilometers per hour. However, we need the speed in kilometers per second because we want to know how far the ship travels in 1 second. Since 1 hour equals 3600 seconds, we get the following conversion factor. 1 h/3600 s By multiplying the speed by this conversion factor, we get the speed in kilometers per second.
At Warp 9, the ship travels a distance of 218 700 000 kilometers in one second.
Laps Around the World
To determine how many laps around the Earth the found distance corresponds to, we need to figure out the distance of one lap around the earth. We know that the ship is traveling at an orbit of 150 kilometers above the surface of the earth.
By adding the radius of the Earth and the distance of 150 kilometers, we can calculate the radius of the circular orbit. 6371+150=6521 km With this information, we can calculate the circumference of the circle.
One lap around the Earth at the given orbit equals 13 042π kilometers. This is the distance of one lap. By dividing distance the ship travels in one second by the distance of one lap, we can find the number of laps around the Earth that Enterprise travels in one second. 218 700 000/13 042π=5337.706801... Since we want to know the whole number of laps, we need to round this number down. Therefore, in 1 second, at Warp 9, the ship travels 5337 whole laps around the Earth.
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Scaling and Descriptive Modeling
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