Slide1: VARIATIONS
Slide2: Variations Direct Variations Inverse Variations Joint Variation Combined Variation Table of Values Mathematical Equations Graphs Applications Mathematical Equations Applications
Slide3: Direct Variation The statements: “ y varies directly as x ” “ y is directly proportional to x ” and “ y is proportional to x ” may be translated mathematically as y=kx, where k is the constant of variation.
Slide4: Watch This! If the distance d varies directly as the time t , then the relationship can be translated into a mathematical statement as d = kt , where k is the constant of variation.
Slide5: Likewise, if the distance d varies directly as the rate r , then the mathematical equation describing the relation is d=kr.
Slide6: In the variation d=10t , the constant of variation is k =10. Using a convenient scale the graph of the relation d=10t is a line. Distance ( d ) Time ( t )
Slide7: Examples: Solution: Express the statement “ y varies directly as x” as y= kx. 1. If y varies directly as x and y=24 when x=6, find the variation constant and the equation of variation.
Slide8: b. Solve for k by substituting the given values in the equation. y = kx 24 = 6k k = 24 6 k = 4 Therefore, the constant of variation is 4. c. Form the equation of the variation by substituting 4 in the statement, y= kx . Hence, y = 4x .
Slide9: The table below shows that the distance d varies directly as the time t. Find the constant of variation and the equation which describe the relation. Time (hr) 1 2 3 4 5 Distance (km) 10 20 30 40 50
Slide10: Solution: Since the distance d varies directly as the time t , then d = kt. Using one of the pairs of values, (2, 20) from the table, substitute the value of d and t in d=kt and solve for k . d = kt 20 = 2 k k = 20 2 k = 10
Slide11: Therefore, the constant variation is 10. Form the mathematical equation of the variation by substituting 10 in the statement d=kt. Hence, d=10t.
Slide12: We can see that the constant can be solved if one pair of the values of x and y is known. From the resulting equation, other pairs having the same relationship can be obtained. Let us study the next example. 3. If x varies directly as y and x=35 when y=7, what is the value of y when x=25?
Slide13: Solution 1. Since x varies directly as y then the equation of the variation is in the form x=ky. Substitute the given values of x and y to solve for k in the equation. 35 = k (7) k = 35 7 k =5
Slide14: Hence , the equation of variation is x=5y. Solving for y when x =25 25 = 5 y y = 25 5 y = 5 Therefore, y =5.
Slide15: Solution 2. Since x is a constant, then we can write y k=x/y. From here, we can establish a proportion such that x₁ = x₂ y₁ y₂ where x₁=35, y₁=7 and x₂=25.
Slide16: Substituting the values, we get 35 = 25 7 y₂ 5 = 25 y₂ y₂ = 25 5 y₂ = 5 Therefore, y=5 when x=25.
Slide17: Now, let us test what you have learned from the discussions.
Slide18: Activity 6: It’s Your Turn! A. Write an equation for the following statements: The fare F of a passenger varies directly as the distance d of his destination. The cost C of fish varies directly as its weight w in kilograms. An employee’s salary S varies directly as the number of days d he has worked. The area A of a square varies directly as the square of its side s. The distance D travelled by a car varies directly as its speed s .
Slide19: Prepared by: Tolentino, Rochelle S. “The education of a man is never completed until he dies.” ― Robert E. Lee THANK YOU!!