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Lesson Plan 1: Be Direct - Oil Spills on Land
The questions below dealing with direct variation have been selected from various state and national assessments. Although the lesson above may not fully equip students with the ability to answer all such test questions successfully, students who participate in active lessons like this one will eventually develop the conceptual understanding needed to succeed on these and other state assessment questions.
Taken from the California High School Exit Examination (Spring 2002):
The diameter of a tree trunk varies directly with the age of the tree. A 45-year-old tree has a trunk diameter of 18 inches. What is the age of a tree that has a trunk diameter of 20 inches?
A. 47 years
B. 50 years (correct answer)
C. 63 years
D. 90 years
Taken from the California High School Exit Examination (Spring 2001):
Tina is filling a 45-gallon tub at a rate of 1.5 gallons of water per minute. At this rate, how long will it take to fill the tub?
A. 30.0 minutes (correct answer)
B. 43.5 minutes
C. 46.5 minutes
D. 67.5 minutes
Taken from the Florida Comprehensive Assessment Test (Spring 2001):
A person who weighs 63 kilograms burns 56 calories in one hour of sleep and burns twice as many calories in one hour of standing. How many calories does a person who weighs 63 kilograms burn in one half-hour of standing?
Solution: In one hour of standing the person will burn 2 × 56 = 112 calories. In one-half hour, the person will burn 112 × 1/2 = 56 calories. The relationship between duration of standing and calories burned is a direct proportion.
Solution: In one hour of standing the person will burn
2 × 56 = 112 calories.
In one-half hour, the person will burn
112 × 1/2 = 56 calories.
The relationship between duration of standing and calories burned is a direct proportion.
Taken from the Maryland High School Assessment Test (Fall 2002):
The table below shows a relationship between x and y.
Which of these equations represents this relationship?
A. y = x²
B. y = 2x (correct answer)
C. y = 1/2x
D. y = x - 2
Taken from the Massachusetts Comprehensive Assessment, Grade 10 (Fall 2002):
When a diver goes underwater, the weight of the water exerts pressure on the diver. The table below shows how the water pressure on the diver increases as the diver's depth increases.
Water Pressure on a Diver
| Diver's Depth (in feet) |
Water Pressure (in pounds per square inch) |
| 10 |
4.4 |
| 20 |
8.8 |
| 30 |
13.2 |
| 40 |
17.6 |
| 50 |
22.0 |
|
a. Based on the table above, what will be the water pressure on a diver at a depth of 60 feet? Show your work or explain how you obtained your answer.
Solution: For every 10-feet increase in depth, the pressure increases by 4.4 psi. Consequently, the pressure at 60 feet will be 22.0 + 4.4 = 26.4 psi.
b. Based on the table above, what will be the water pressure on a diver at a depth of 100 feet? Show your work or explain how you obtained your answer.
Solution: The relationship between depth and pressure is a direct variation, so the pressure at 100 feet will be double the pressure at 50 feet, which is 44.0 psi.
c. Write an equation that describes the relationship between the depth, D, and the pressure, P, based on the pattern shown in the table.
Solution: P = 0.44D
d. Use your equation from part c to determine the depth of the diver, assuming the water pressure on the diver is 46.2 pounds per square inch. Show your work or explain how you obtained your answer.
Solution: Substituting P = 46.2 into the equation from c gives 46.2 = 0.44D. Solving for D yields  , or D = 105. Consequently, the diver experiences 46.2 psi of pressure at a depth of 105 feet.
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