variations
Page 1
Title: VARIATIONS
Page 2
Title: WHAT IS VARIATION?
Definition: Variation is defined by any change in some quantity due to a change in another.
Page 3
Title: TYPES OF VARIATIONS
Types:
Direct Variation
Inverse Variation
Joint Variation
Combined Variation
Page 4
Title: REVIEW
Solve the following:
4 5 = (4) (10) 10 = ( ) (12) 125 = 2 8 = 7
Page 5
Title: LEARNING TARGETS
Objectives:
I can write equations of direct linear variation.
I can identify direct linear variation equations.
I can solve problems involving direct linear variation.
Page 6
Title: DIRECT VARIATION
Definition: In a direct variation, y varies directly as x, or y is directly proportional to x if there is a constant k such that:
y ∝ x
y = kxWhere k is the constant of proportionality.
Page 7
Title: FEATURES OF DIRECT VARIATION
Behavior:
As x increases, y also increases.
As x decreases, y also decreases.
Equation: y = kx
Key Phrases: Directly proportional, varies directly as.
Page 8
Title: EXAMPLE 1
Problem: y varies directly as x. If y = 10 when x = 2.
A. What is the constant of variation (k)?
B. What is the equation of variation?
C. What is x when y = 20?
Solution: A. k = 10/2 = 5
Page 9
Title: EXAMPLE 1 Continued
Problem: y varies directly as x. If y = 10 when x = 4.
A. What is the constant of variation?
B. What is the equation of the variation?
C. What is x when y = 20?
Solution: The equation of variation is y = 5x.
Page 10
Title: SOLVING FOR x
Problem: Using the equation of variation: y = 5x. If y = 20:
20 = 5x
Solve for x:
x = 4
Therefore, the value of x is 4 when y is 20.
Page 11
Title: GRAPH OF DIRECT VARIATION
Graph: This illustrates the variation in example one (y = 5x).
Observation: What can you notice about the graph?
Page 12
Title: EXAMPLE 2
Problem: If v varies directly as d, find k in the proportion:
4v = 2
Solution:
Remember to cross multiply: 4 = 2(4)
Hence, k = 2.
The constant of variation (k) is equal to 2.
Page 13
Title: EXAMPLE 3
Problem: If g varies directly as l and g = 10 when l = 5, find k and express this variation.
Solution:
g = kl -> 10 = k(5)
k = 2
The equation of the variation is g = 2l.
Page 14
Title: EXAMPLE 4
Problem: Perimeter p of a square varies directly with the length of the side s.
Find side length if perimeter is 50 cm.
Formula: p = ks; given p = 50, k = 4,
Solution: Side length s = 12.5 cm.
Page 15
Title: EXAMPLE 5
Problem: ¼ kg of pork costs ₱50. How much for 3 ½ kg?
Solution: Cost of 3 ½ kg of pork is ₱700.
Page 16
Title: TYPES OF VARIATION SUMMARY
Direct Variation: y varies directly as x (y = kx)
Inverse Variation: y varies inversely as x (xy = k)
Joint Variation: y varies directly to two quantities.
Combined Variation: y varies directly to some and inversely to others.
Page 17
Title: PRACTICE PROBLEMS
Solve the following:
If y varies directly as x and y=8 when x=4, find y when x=3.
If y varies directly as x and y=25 when x=15, find y when x=8.
If y is directly proportional to x, and y=22 when x=3, find y when x=5.
If y is directly proportional to x, and y=30 when x=7, find x when y=3.
If a varies directly as the square of b and a=4 when b=3, find a when b=9.
If r varies directly as the cube of s, find r when s=5.
If r varies directly as the fourth power of s, find r when s=3.
If r varies directly as the fifth power of s, find r when s=3.
Page 18
Title: MORE PRACTICE PROBLEMS
Solve:
If (M+1) varies directly as N² and M=17 when N=3, find M when N=4.
If (P-1) is directly proportional to (Q+1)² and P=19 when Q=2, find Q when P=51.
N varies directly as L, N=400 when L=22, find L when N=550.
Cost C is directly proportional to length l, if material costs ₱35 for 2m, find cost for 5m.
If d is doubled and h is halved in V = T(d²h), what is the effect on V?
Page 19
Title: VARIATIONS
Inverse Variation.
Page 20
Title: INVERSE VARIATION LEARNING TARGETS
Objectives:
Recognize relationships involving inverse variation.
Translate statements to mathematical equations.
Solve problems involving inverse variations.
Page 21
Title: INVERSE VARIATION DEFINITION
Definition: y varies inversely as x if there is a nonzero constant k such that:
y = k/x
y ∝ 1/x.
Page 22
Title: EXAMPLE 1 - INVERSE VARIATION
Problem: If y varies inversely as x and y = 6 when x = 3, find y when x is 9.
Solution:k = xy = 6 * 3 = 18y = 18/x, y = 18/9Thus, y = 2.
Page 23
Title: EXAMPLE 1 CONTINUED
Re-Iterate:k = 18y = k/x = 18/xWhen x = 9:y = 18/9Therefore, y = 2.
Page 24
Title: GRAPH OF INVERSE VARIATION
Graph: Illustration of inverse variation for y = 18/x.
Page 25
Title: EXAMPLE 2 - INVERSE VARIATION
Problem: If y varies inversely as x when y = 8 and x = 2.
A. Find the constant of variation.
B. Find the inverse variation equation.
C. Find y when x = 12.
D. Find x when y = 2.
Page 26
Title: EXAMPLE 2 SOLUTION
Solution Steps:
Find k:
Substitute values into y = k/x.
Find k, k = 16.
Inverse variation equation: y = 16/x.
Page 27
TITLE: FINDING y AND x
Finding y: When x = 12, substitute into the equation:
y = 16/12 = 4/3.
Finding x: When y = 2, solve for x:
2 = 16/x, leading to x = 8.
Page 28
TITLE: EXAMPLE 3 - BOYLE'S LAW
Law: Describes that pressure (P) is inversely proportional to volume (V) of gas.
Given: P = 1000 when V = 1.83.
Find: Pressure when volume is increased to V = 3.
Page 29
TITLE: DISTANCE AND WEIGHT
Problem: Distance from center of seesaw varies inversely as weight. Jomel weighs 80kg and sits 4m from the fulcrum. Find Jun's distance if he weighs 40kg.