5_1 Rate of Change and Slope
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5.1 Rate of Change and Slope Rate of Change: The relationship between two changing quantities Rate of Change = Change in the dependent variable (y-axis) Change in the independent variable (x-axis) Slope: the ratio of the vertical change (rise) to the horizontal change (run). Slope = Vertical Change (y) Horizontal Change (x) = rise run Real World: Rate of Change can be presented in many forms such as: ππππ π¨π ππ‘ππ§π π = πͺπππππ ππ π πππππππ(π) ππππππ ππ ππππ(π) We can use the concept of change to solve the cable problem by using two sets of given data, for example: A band practices their march for the parade over time as follows: Choosing the data from: Time and Distance 1min 260 ft. 2min 520 ft. We have the following: ππππ π¨π ππ‘ππ§π π = = πππ ππ −πππ ππ π πππ −π πππ πππ ππ π πππ Choosing the data from: Time and Distance 1min 260 ft. 3min 780 ft. We have the following: ππππ π¨π ππ‘ππ§π π = = πππ ππ −πππ ππ π πππ −π πππ πππ ππ π πππ = πππππ π πππ Choosing the data from: Time and Distance 1min 260 ft. 4min 1040 ft. We have the following: ππππ π¨π ππ‘ππ§π π = = ππππ ππ −πππ ππ ππππ −π πππ πππ ππ π πππ = πππππ π πππ NOTE: When we get the same slope, no matter what date points we get, we have a CONSTANT rate of change: YOU TRY IT: Determine whether the following rate of change is constant in the miles per gallon of a car. Gallons Miles 1 28 3 84 5 140 7 196 Choosing the data from: Gallons and Miles 1g 28 m 3g 84 m We have the following: ππππ π¨π ππ‘ππ§π π = = πππ −πππ ππ −ππ ππ π ππ Choosing the data from: Gallons and Miles 1g 28 m. 5g 140 m. We have the following: ππππ π¨π ππ‘ππ§π π = = ππππ −ππ π ππ −π π ππ π ππ THUS: the rate of change is CONSTANT. Once Again: Real World Remember: Rate of Change can be presented in many forms: ππππ π¨π ππ‘ππ§π π = πͺπππππ ππ π πππππππ ππππππ ππ ππππ We can use the concept of change to solve the cable problem by using two sets of given data: (x , y) A : Horizontal(x) = 20 Vertical(y) = 30 ο (20, 30) B : Horizontal(x) = 40 Vertical(y) = 35 ο (40, 35) Using the data for A and B and the definition of rate of change we have: (x , y) A : Horizontal = 20 Vertical = 30 ο (20, 30) B : Horizontal = 40 Vertical = 35 ο (40, 35) Rate of Change = π½πππππππ πͺπππππ(π) π―πππππππππ πͺπππππ(π) ππ −ππ Rate of Change = ππ−ππ π Rate of Change = ππ π Rate of Change from A to B = π Using the data for B and C and the definition of rate of change we have: (x , y) B : Horizontal = 40 Vertical = 35 ο (40, 35) C : Horizontal = 60 Vertical = 60 ο (60, 60) Rate of Change = π½πππππππ πͺπππππ(π) π―πππππππππ πͺπππππ(π) ππ −ππ Rate of Change = ππ−ππ ππ Rate of Change = ππ π Rate of Change from B to C = π Using the data for C and D and the definition of rate of change we have: (x , y) C : Horizontal = 60 Vertical = 60 ο (60, 60) D : Horizontal = 100 Vertical = 70 ο (100, 70) Rate of Change = π½πππππππ πͺπππππ(π) π―πππππππππ πͺπππππ(π) ππ −ππ Rate of Change = πππ−ππ ππ Rate of Change = ππ π Rate of Change from B to C = π Comparing the slopes of the three: π Rate of Change from A to B = π π Rate of Change from B to C = π π Rate of Change from C to D = π As we can see right now the pole from B to C is the π one with the biggest change of rate(steepest) = π However, we must find all the combination that we can do. Try from A to C, from A to D and from B to C. Finally: ππ−ππ A to B = ππ−ππ = ππ−ππ C to D = πππ−ππ ππ−ππ A to D = πππ−ππ π π ππ−ππ B to C = ππ−ππ = π π = π π = π π ππ−ππ A to C = ππ−ππ = π π ππ−ππ B to D = πππ−ππ = π π Finally we can conclude that the poles with the steepest path are poles B to C with slope of 5/4. Class Work: Pages: 295-297 Problems: 1, 4, 8, 9, Remember: When we get the same slope, no matter what date points we get, we have a CONSTANT rate of change: When we get the same slope, no matter what date points we get, we have a CONSTANT rate of change: We further use the concept of CONSTANT slope when we are looking at the graph of a line: We further use the concept of rise/run to find the slope: ππππ πππ = π π π π SLOPE= = π run rise Make a right triangle to get from one point to another, that is your slope. CONSTANT rate of change: due to the fact that a line is has no curves, we use the following formula to find the SLOPE: y2-y1 x2-x1 Slope = π¬ππ π −πΊππππ π π¬ππ π −πΊππππ π Slope = πΉπππ πΉππ B(x2, y2) A(x1, y1) = ππ −ππ ππ −ππ A = (1, -1) B = (2, 1) Slope = π−−π π−π = π π YOU TRY: Find the slope of the line: YOU TRY (solution): Slope = −π π Slope = πΉπππ πΉππ Slope = ππ −ππ ππ −ππ Slope = π−π π−π Slope = −π π = (0,4) -4 = −π π (2,0) 2 −π π YOU TRY IT: Provide the slope of the line that passes through the points A(1,3) and B(5,5): YOU TRY IT: (Solution) Using the given data A(1,3) and B(5,5) and the definition of rate of change we have: A( 1 , 3 ) B(5 , 5) (x1, y1) (x2, y2) Slope = ππ −ππ ππ −ππ π−π Slope = π−π π Slope = π π Rate of Change from A to B is = π YOU TRY: Find the slope of the line: YOU TRY IT: (Solution) Choosing two points say: A(-5,3) and B(1,5) and the definition of rate of change (slope) we have: A( -2 , 3 ) B(1 , 3) (x1, y1) (x2, y2) Slope = ππ −ππ ππ −ππ π−π Slope = π−−π π Slope = π Rate of Change (slope) from A to B is = π YOU TRY: Find the slope of the line: YOU TRY IT: (Solution) Choosing two points say: A(-1,2) and B(-1,-1) and the definition of rate of change (slope) we have: A( -1 , 2 ) B(-1 , -1) (x1, y1) (x2, y2) Slope = ππ −ππ ππ −ππ −π−π Slope = −π−−π −π Slope = π We can never divide by Zero thus our slope = UNDEFINED. THEREFORE: Horizontal ( slope of ZERO ) lines have a While vertical ( ) lines have an UNDEFINED slope. VIDEOS: Graphs https://www.khanacademy.org/math/algebra/line ar-equations-and-inequalitie/slope-andintercepts/v/slope-and-rate-of-change Class Work: Pages: 295-297 Problems: As many as needed to master the concept
