Скачать презентацию Chapter 7 Direct Variation Advanced Math Section

• Количество слайдов: 39

Chapter 7: Direct Variation Advanced Math

Section 7. 1: Direct Variation, Slope, and Tangent l When two variable quantities have a constant ratio, their relationship is called direct variation. l The constant ratio is called the variation constant.

Page 359 in your textbook… l For the roof, the height of a post depends on its horizontal distance from the eaves. l Although the height increases as the distance decreases, the ratio of these two quantities is constant.

l Ratio of Quantities = variation constant – – l Height/Distance = 0. 75 Height = 0. 75(Distance) You can either use this equation to find the height of a post needed to support the roof at another point.

Sample Question #1 l Suppose the carpenter of the roof shown on page 359 decides that a post should be 3 feet from the eaves. What should the height of the post be? H= 0. 75(D) H = 0. 75(3) H = 2. 25 The post should be 2. 25 feet long.

Try this one on your own… Distance from eaves to first post is 10/4 or 2. 5 feet. H = 0. 75(d) H = 0. 75(2. 5) H = 1. 875 The shortest post should be 1. 875 feet long. l Suppose the carpenter of the roof shown on page 359 wants only three support post between the eaves and the post at the peak. Assuming that the posts are evenly spaced, how long should the shortest support post be?

Slope… l The slope of a line is the ratio of rise to run for any two points on the line. l l Rise = vertical change between two points Run = horizontal change between two points l Slope = rise/run

Sample Question #2 l Find the slope of the roof shown on page 359. Use the eaves and the peak as the two points. Step One: Draw a Sketch. Step Two: Find the slope. The roof rises 7. 5 feet across a distance of 10 feet. Slope = rise/run Slope = 7. 5/10

Try this one on your own… l Slope = rise/run Slope = 9/18 Slope = 1/2 Suppose the post from the horizontal beam to the peak of the roof is 9 feet high and the length of the beam from the eaves to the post is 18 feet.

Tangent… l The tangent ratio of an acute angle compares the two legs of a right triangle. l Tangent = leg opposite / leg adjacent

Sample Question #3 l Look again at the roof on page 359. What is the tangent ratio of the angle between the roof and the horizontal beam? Tan

Section 7. 2: A Direct Variation Model l Read pages 367 -370. l Get with a partner and begin your homework.

Section 7. 3: Circumference and Arc Length overhead

Section 7. 4: Direct Variation with y = kx l You can model direct variation with an equation y = kx. l Y = kx is called the general form for direct variation. l Y = dependent variable l K = variation constant l X = control variable

Sample #1 l l l Y = kx 336 = k (40) K = 8. 4 l l l Y = 8. 4 x Y = 8. 4(130) Y = 1092 In 130 hrs, Cindy would make \$1092. The amount Cindy earns varies directly with the number of hours she works. She earns \$336 in a 40 hour work week. How much will Cindy earn in 130 hours?

Try this one on your own… l Efren set an empty fish tank on a bathroom scale. When he added 20 quarts of water, the weight increase was 41 ¾ pounds. He plans to have 28 quarts of water in the tank when he sets the tank up for his fish. If the weight of the water varies directly with the number of quarts used, how much will the water in the tank weigh? About 58 pounds

Direct Variation Graphs l The graph of a direct variation equation y = kx is a line that passes through the origin. l The variation constant, k, is the slope of the line.

Sample # 2 l Is direct variation a good model for the data shown in each graph? Explain why or why not.

Try these on your own… l Is direct variation a good model for the data shown in each graph? Explain why or why not.

Sample # 3 l The direct variation graph shown is a line with slope 3. Write an equation of the line.

Try this one on your own… l The direct variation graph shown is a line with slope 3/2. Write an equation for the graph.

Sample #4 l Graph y = -2 x. l Try this one on your own… – Graph y = -2/3 x.

Section 7. 5: Using Dimensional Analysis l When you cancel units of measurement as if they are numbers, you are using a problem solving strategy called dimensional analysis. l A conversion factor is a ratio of two equal quantities that are measured in different units. – Example: 12 inches in 1 foot – 12: 1

Sample #1 l The distance Milo travels varies directly with the number of hours he drives. Milo drives 145 miles in 3 hours. – – Identify the control variable and the dependent variable. Express the variation constant as a rate. l Control/Independent Variable = time traveled (hours) l Dependent Variable = distance traveled (miles) l 145 miles / 3 hours = 48. 3 miles per hour

Try this one on your own… l The thickness of a stack of copier paper varies directly with the number of sheets in the stack. A new pack of copier paper is 2 inches thick and contains 500 sheets. l Control/Independent = thickness of stack of paper l Dependent = number of sheets Identify the control variable and dependent variable. Express the variation constant as a rate. l Variation Constant = 2 inches/500 sheets = 0. 004 inches per sheet – –

Sample #2 l How many centimeters are in 3 inches? – l There are 2. 54 centimeters in 1 inch. Write a direct variation equation that helps you convert inches to centimeters. – X = number of inches Y = number of centimeters K = 2. 54 – Equation: y = 2. 54 x – 3 inches x 2. 54 cm/1 in – 7. 62 centimeters l What is the variation constant? – 2. 54 centimeters per inch

Try this one on your own… l How many pounds are in 70. 5 kg? l 155. 4525 pounds l Write a direct variation equation for converting kilograms to pounds. – l Y = 2. 205 X What is the variation constant? – 2. 205 pounds per kilograms l Pedro Zepeda weighs 70. 5 kg. l There are 2. 205 pounds per 1 kilogram.

Sample #3 l l l In 1990 a French train set a speed record of 515 km/h on the Atlantique line. Find its speed in miles per hour? 1 kilometer = 0. 621 miles 515 km / 1 hr = ? / ? 515 km / 1 hr = ? mi / ? km 515 km / 1 hr = 0. 621 mi / 1 km 320 miles per hour

Try this one on your own… l A large crude oil pipeline from Canada to the United States has a flow of 8. 3 million gallons per day. How many liters per day does the pipeline handle? l There are 3. 78531 liters per 1 gallon. 31. 42 million liters of crude oil per day

Sample #4 rate x conversion factors = new rate l Convert 30 mi/hr to ft/s l l 1 mile = 5, 280 feet 1 hour = 60 minutes 1 minute = 60 seconds l Rate x Conversion Factors = New Rate l l

Section 7. 6: Areas of Circles and Sectors l Area Formula – – What is it? Are they units squared or cubed?

Sample #1 l Find the area of a circle whose radius is 3 units. Round the answer to the nearest square unit. l 28. 3 units squared l Try this one on your own… – – Find the area of a circle whose radius is 7. 5 units. Round the answer to the nearest square units. 176. 7 units squared

Sample #2 l In a center-pivot irrigation system, a moving arm sprinkles water over a circular region. How long must the arm be to water an area of 586, 000 meters squared? l 431. 9 = radius

Try this one on your own… l The amount of water a pipe can carry depends on the area of the opening at the end. An engineer wants a large sewer pipe to have an opening of 13 feet squared. What should the radius of the pipe be? Round the answer to the nearest tenth of a foot. l 2. 0 = radius

l The region of a circle formed by a central angle and its arc is called a sector. l area of sector = measure of the central angle area of circle = 360 degrees

Sample #3 l A fixed camera that is part of the TV security system at a shopping mall parking lot has a range of about 250 feet. The angle of vision of the camera is 100 degrees. Over how large an area can the camera see when it is in operation? l Steps to Solve – – Draw a Picture Set up the Equation

Sample #4 l About how many times more pizza do you get when you buy the large size instead of the small size? – – Large is 16 inches in diameter. Small is 12 inches in diameter.

Try this one on your own… l The camera is Sample #3 replaced an older camera that had a viewing angle of 110 degrees but a range of only 175 feet. How many times greater is the viewing area of the newer camera than that of the older camera? l 1. 86 times