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Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

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Page 1: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

Section 6.5.2 – Ratio, Proportion, VariationUsing the Vocabulary

Page 2: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

Neglecting air resistance, the distance an object falls (s) varies directly as the square of the duration (t) of the fall. An object falls a distance of 144 feet in 3 seconds. How far will it fall in 5 seconds?

2s kt144 9k

k 16

2s 16t

2s 16 5

s 400

Page 3: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

The stopping distance (d) of an automobile is directly proportional to the square of its speed (s). A car required 75 feet to stop when its speed was 30 miles per hour. Find the stopping distance if the brakes are applied when the car is traveling at 50 miles per hour.

2

dk

s

2

75k

30

1k

12

2

d 1

s 12

2

d 1

50 12

625d

3

Page 4: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

A company has found that the demand (d) for its product varies inversely as the price of the product (p). When the price is $3.75, the demand is 500 units. Approximate the demand when the price is $4.25.

kd

p

k500

3.75

1875 k

1875d

4.25

d 441.176

Page 5: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

The distance a spring is stretched (or compressed) (D) variesdirectly as the force on the spring (F). A force of 220 newtonsstretches a spring 0.12 meters. What force is required to stretch the spring 0.16 meters?

D Fk

00.12 k 22

2

0.12

20k

D Fk

0.16 Fk

293.333 F

Page 6: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

The stopping distance (d) of an automobile is directly proportionalto the square of its speed (s). A car required 75 feet to stopwhen its speed was 30 mph. Estimate the stopping distanceif the brakes are applied when the car is traveling at 50 mph.

2

d

sk

2

75

30k

2

d

sk

2

d

50k

d 208.333

Page 7: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

Property tax is based on the assessed value of the property. Ahouse that has an assessed value (v) of $150,000 has a property tax (t) of $5520. Find the property tax on a house that has an assessed value of $200,000

v

tk

5

1

5

50000

20k

625k

23

v

tk

200000

tk

t 7360

Page 8: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

The maximum load (L) that can be safely supported by a horizontal beam varies jointly as the width of the beam (w) and the square of its depth (d), and inversely as the length of the beam (x).

a) Determine the change in the maximum safe load if the width and length of the beam are doubled.

2kwdL

x

22wk dL

2x

There is no change to the maximum safe load.

Page 9: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

The maximum load (L) that can be safely supported by a horizontal beam varies jointly as the width of the beam (w) and the square of its depth (d), and inversely as the length of the beam (x).

b) Determine the change in the maximum safe load if the width and depth of the beam are doubled

2kwdL

x 2

2wkL

2d

x

The maximum safe load becomes eight times theoriginal maximum safe load.

Page 10: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

The maximum load (L) that can be safely supported by a horizontal beam varies jointly as the width of the beam (w) and the square of its depth (d), and inversely as the length of the beam (x).

c) Determine the change in the maximum safe load if all three dimensions are doubled.

2kwdL

x

22w 2k

Ld

2x

The maximum safe load becomes four times theoriginal maximum safe load.

Page 11: Section 6.5.2 – Ratio, Proportion, Variation Using the Vocabulary

The maximum load (L) that can be safely supported by a horizontal beam varies jointly as the width of the beam (w) and the square of its depth (d), and inversely as the length of the beam (x).

d) Determine the change in the maximum safe load if the depth of the beam is halved.

2kwdL

x

2

x

k1

w d2

L

The maximum safe load becomes one-fourth theoriginal maximum safe load.