# VARIATION by sphinxkey2010

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```									VARIATION
BY
EDSEL M. LLAVE
VARIATION
• DIRECT VARIATION
– Definition of Direct Variation
– Direct Variation as the nth Power
• INVERSE VARIATION
– Definition of Inverse Variation
– Inverse Variation as the nth Power
• JOINT VARIATION
– Definition of Joint Variation
• COMBINED VARIATION
DIRECT VARIATION
• The variable y varies directly as the
variable x, or y is directly proportional to
x, if and only if

where k is a constant called constant of
proportionality or the variation constant.
DIRECT VARIATION
• The statement "y varies directly as x," means
that when x increases, y increases by the same
factor.
• In other words, y and x always have the same
ratio:
Example 1: If y varies directly as x, and
x = 12 when y = 9, what is the equation that
describes this direct variation?
• Solution:
Given: x = 12   y=9
Example 2: If y varies directly as x, and the
constant of variation is      , what is y
when x = 9?
• Solution:
Given:    x=9
Example 3: If y varies directly as x, and y = 15
when x = 10, then what is y when x = 6?
• Solution:
Given:
Condition 1:      y = 15    x = 10
Condition 2:      y=?       x=6
1. Find the variation constant. (condition 1)
Example 3: If y varies directly as x, and y = 15
when x = 10, then what is y when x = 6?
• Solution:
Given:
Condition 1:     y = 15      x = 10
Condition 2:     y=?         x=6
2. Find the value of y. (Condition 2)
Example 3: If y varies directly as x, and y = 15
when x = 10, then what is y when x = 6?
• Solution 2: (Proportion)
Given:
Condition 1:      y1 = 15    x2 = 10
Condition 2:      y=?        x=6
Example 3. The distance sound travels varies
directly as the time it travels. If sound travels 1340
meters in 4 seconds, find the distance sound will
travel in 5 seconds.
• Solution:
CONDITION    Distance (d)    Time (t)
Given:            1       d1 = 1340m      t1 = 4sec
2           d=?          t = 5sec
By proportion:
Example 3. The distance sound travels varies
directly as the time it travels. If sound travels 1340
meters in 4 seconds, find the distance sound will
travel in 5 seconds.
• Solution:
CONDITION       Distance (d)    Time (t)
Given:           1           d1 = 1340m      t1 = 4sec
2               d=?          t = 5sec
Find the value of k.
Find the value of d.
Example 4. Hooke’s Law states that the distance a spring
stretches varies directly as the weight on the spring. A
weight of 80 pounds stretches a spring 6 inches. How
far will a weight of 100 pounds stretch the spring?
• Solution:         CONDITION    Distance (d)    Weight (w)
Given:            1         d1 = 6in       w1 = 80lbs
2          d=?           w = 100lbs
Find the value of k.

Find the value of d.
Example 4. Hooke’s Law states that the distance a spring
stretches varies directly as the weight on the spring. A
weight of 80 pounds stretches a spring 6 inches. How
far will a weight of 100 pounds stretch the spring?
• Solution:
CONDITION     Distance (d)    Weight (w)
Given:
1          d1 = 6in       w1 = 80lbs
2           d=?           w = 100lbs
By proportion:
DIRECT VARIATION AS THE nth POWER

• If y varies directly as the nth power of x,
then

where k is constant.
Example 5. The distance s that an object falls from rest
(neglecting air resistance) varies directly as the square of the
time t that it has been falling. If an object falls 64 feet in 2
seconds, how far will it fall in 10 seconds?
• Solution:            CONDITION       Distance (s)    Time (t)

Given:                1          s1 = 64ft       t1 = 2sec
Formula:                   2            s=?           t = 10sec
Example 5. The distance s that an object falls from rest
(neglecting air resistance) varies directly as the square of the
time t that it has been falling. If an object falls 64 feet in 2
seconds, how far will it fall in 10 seconds?
• Solution:             CONDITION      Distance (s)    Time (t)

Given:                1           s1 = 64ft      t1 = 2sec
2             s=?          t = 10sec
By proportion:
Problem 6. The range of a projectile is directly proportional to the
square of its velocity. If a motorcyclist can make a jump of 140
feet by coming off a ramp at 60 mph, find the distance the
motorcyclist could expect to jump if the speed coming off the
ramp were increased to 65 mph.
• Solution:
CONDITION       Distance (s)    Velocity (v)
Given:
1           s1 = 140ft     v1 = 60 mph
2              s=?         v = 65 mph
Formula:
Problem 6. The range of a projectile is directly proportional to the
square of its velocity. If a motorcyclist can make a jump of 140
feet by coming off a ramp at 60 mph, find the distance the
motorcyclist could expect to jump if the speed coming off the
ramp were increased to 65 mph.
• Solution:
CONDITION       Distance (s)    Velocity (v)
Given:
1           s1 = 140ft     v1 = 60 mph
2              s=?         v = 65 mph
By proportion:
INVERSE VARIATION
• The variable y varies inversely as the
variable x, or y is inversely proportional
to x, if and only if

where k is the variation constant
INVERSE VARIATION
• The statement "y varies inversely as x” means
that when x increases, y decreases by the
same factor.
• In other words, the expression xy is constant:

where k is constant of variation.
Example 1: If y varies inversely as x, and y = 6
when         , write an equation describing this
inverse variation.
• Solution :
• Find the value of k.
Write an equation.

or
Example 2: If y varies inversely as x, and the constant of
variation is     , what is y when x = 10?

• Solution:
Example 3: If y varies inversely as x, and y = 10
when x = 6, then what is y when x = 15?
• Solution:
CONDITION         y            x
Given:       1           y1 = 10      x1 = 6
2            y=?         x = 15
Example 4. Boyle’s Law states that the Volume V of a sample of gas (at
constant temperature) varies inversely as the pressure P. The volume of
a gas in a J-shaped tube is 75 milliliters when the pressure is 1.5
atmospheres. Find the volume of the gas when the pressure is increased
to 2.5 atmospheres.
• Solution:          CONDITION Volume(V) Pressure(P)
Given:             1       V1 = 75ml P1 = 1.5atm
2         V=?     P = 2.5atm
Find the value of k.
Example 4. Boyle’s Law states that the Volume V of a sample of gas (at
constant temperature) varies inversely as the pressure P. The volume of
a gas in a J-shaped tube is 75 milliliters when the pressure is 1.5
atmospheres. Find the volume of the gas when the pressure is increased
to 2.5 atmospheres.
• Solution: CONDITION Volume(V) Pressure(P)
Given:    1       V1 = 75ml P1 = 1.5atm
2              V=?           P = 2.5atm
INVERSE VARIATION AS nth POWER
• If y varies inversely as the nth power of x,
then

where k is a constant
Example 5. The illumination a source of light provides is inversely
proportional to the square of the distance from the source. If the
illumination at a distance of 10 feet from the source is 50 footcandles,
what is the illumination at a distance of 15 feet from the source?

• Solution:         CONDITION Illumination(ℓ) Distance(d)
Given:               1              ℓ1 = 50fc           d1 = 10ft
2                ℓ=?               d = 15ft
Find the value of k.
Example 5. The illumination a source of light provides is inversely
proportional to the square of the distance from the source. If the
illumination at a distance of 10 feet from the source is 50 footcandles,
what is the illumination at a distance of 15 feet from the source?

• Solution:          CONDITION Illumination(ℓ) Distance(d)
Given:                1              ℓ1 = 50fc           d1 = 10ft
2                ℓ=?               d = 15ft
JOINT VARIATION
• The variable z varies jointly as the
variables x and y if and only if

where k is constant.

• If one variable, z, varies as the product of
other variables, x and y, it is called joint
variation.
Example 1. The cost of insulating the ceiling of a house varies jointly
with the thickness of the insulation and the area of the ceiling. It costs
Php8,750 to insulate a 2100-square-foot ceiling with insulation 4
inches thick. Find the cost of insulating a 2400-square-foot ceiling with
insulation that is 6 inches thick.
• Solution:
CONDITION       Cost (C)      Area (A)      Thickness (T)
Given:               1        C1 = Php8,750   A1 = 2100ft2     T1 = 4in
2            C=?         A = 2400ft2       T = 6in

Formula:
Example 1. The cost of insulating the ceiling of a house varies jointly
with the thickness of the insulation and the area of the ceiling. It costs
Php8,750 to insulate a 2100-square-foot ceiling with insulation 4
inches thick. Find the cost of insulating a 2400-square-foot ceiling with
insulation that is 6 inches thick.
• Solution:
CONDITION       Cost (C)       Area (A)      Thickness (T)
Given:                 1        C1 = Php8,750   A1 = 2100ft2     T1 = 4in
2            C=?         A = 2400ft2       T = 6in

Formula:
Example 2. The load L that a horizontal beam can safely support varies jointly
as the width w and the square of the depth d. If a beam with width 2 inches and
depth 6 inches safely supports up to 200 pounds, how many pounds can a
beam of the same length that has width 4 inches and depth 4 inches be
expected to support?
• Solution:            CONDITION        Load (L)      Width (w)      Depth (d)
Given:                 1         L1 = 200lbs      w1 = 2in      d1 = 6in
2            L=?           w = 4in         d = 4in
Formula:
Example 2. The load L that a horizontal beam can safely support varies jointly
as the width w and the square of the depth d. If a beam with width 2 inches and
depth 6 inches safely supports up to 200 pounds, how many pounds can a
beam of the same length that has width 4 inches and depth 4 inches be
expected to support?
• Solution:           CONDITION        Load (L)      Width (w)      Depth (d)
1         L1 = 200lbs      w1 = 2in       d1 = 6in
Given:
2             L=?           w = 4in        d = 4in
COMBINED VARIATIONS
• Combined variations involve more than
one type of variation.

• It involves the direct and inverse
variations.
Problem 1. The weight that a horizontal beam with a rectangular cross
section can safely support varies jointly as the width and square of the
depth of the cross section and inversely as the length of the beam. If a
4-inch by 4-inch beam 10 feet long safely supports a load of 256
pounds, what load L can be safely supported by a beam made of the
same material and with a width w of 4 inches, a depth d of 6 inches,
and a length l of 16 feet?
• Solution:          CONDITION    Load (L)     Width (w)   Depth(d)   Length (ℓ)
Given:               1      L1 = 256lbs   w1 = 4in    d1 = 4in    l1 = 10ft
2         L=?         w = 4in     d = 6in    l = 16ft
Formula:
Problem 1. The weight that a horizontal beam with a rectangular cross
section can safely support varies jointly as the width and square of the
depth of the cross section and inversely as the length of the beam. If a
4-inch by 4-inch beam 10 feet long safely supports a load of 256
pounds, what load L can be safely supported by a beam made of the
same material and with a width w of 4 inches, a depth d of 6 inches,
and a length l of 16 feet?
• Solution:         CONDITION     Load (L)     Width (w)   Depth(d)   Length (ℓ)
1       L1 = 256lbs   w1 = 4in    d1 = 4in    l1 = 10ft
Given:
2          L=?         w = 4in     d = 6in    l = 16ft
Formula:
Example 2. The Ideal Gas Law states that the volume V of a gas varies
jointly as the number of moles of gas n and the absolute temperature T
and inversely as the pressure P. What happens to V when n is tripled
and P is reduced by a factor of one-half?

• Solution:
CONDITION   Volume     No. of   Temperature Pressure
Given:                       (V)     Moles (n)     (T)        (P)
1        V1 = V    n1 = n      T1 = T    P1 = P
2        V2 = ?    n2 = 3n     T2 = T   P2 = 0.5P
Formula:
Example 2. The Ideal Gas Law states that the volume V of a gas varies
jointly as the number of moles of gas n and the absolute temperature T
and inversely as the pressure P. What happens to V when n is tripled
and P is reduced by a factor of one-half?

• Solution:          CONDITION   Volume     No. of   Temperature Pressure
(V)     Moles (n)     (T)        (P)
Given:              1        V1 = V    n1 = n      T1 = T    P1 = P
2        V2 = ?    n2 = 3n     T2 = T   P2 = 0.5P

Formula:
Thank you

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