2 7 variations

Variations
Frank Ma © 2011

Variations
We say the variable y varies (directly)to an expression f if
y = k·f
where k is a constant.

Variations
y = k·f
where k is a constant. The formula y = k* f is called the general (direct) variation equation.

Variations
y = k·f
Example A. Translate the following phrases into equations.
y varies directly to x.

Variations
y = k·f
y = kx for some k.

Variations
y = k·f
y = kx for some k.
b. y varies to xz

Variations
y = k·f
y = kx for some k.
b. y varies to xz
y = kxz for some k.

Variations
y = k·f
y = kx for some k.
b. y varies to xz
y = kxz for some k.
c. y varies to x2z2

Variations
y = k·f
y = kx for some k.
b. y varies to xz
y = kxz for some k.
c. y varies to x2z2
y = kx2z2 for some k.

Variations
y = k·f
y = kx for some k.
b. y varies to xz
y = kxz for some k.
c. y varies to x2z2
d. The cost C varies directly with the square of the length L.

Variations
y = k·f
y = kx for some k.
b. y varies to xz
y = kxz for some k.
c. y varies to x2z2
The square of the length L is L2.

Variations
y = k·f
y = kx for some k.
b. y varies to xz
y = kxz for some k.
c. y varies to x2z2
The square of the length L is L2.
Hence the general equation is y = kL2 for some k.

Variations
We say the variable y varies inverselyto an expression f if
y =
k
f

Variations
y =
where k is a constant. The formula y = is called the general inverse variation equation.
k
f
k
f

Variations
y =
k
f
k
f
Example B. Translate the following phrases into equations.
y varies inversely to x.

Variations
y =
k
f
k
f
k
y =
where k is a constant
x

Variations
y =
k
f
k
f
k
y =
x
b. y varies inversely to x2z

Variations
y =
k
f
k
f
k
y =
x
k
y =
x2z

Variations
y =
k
f
k
f
k
y =
x
k
y =
x2z
c. The intensity I of light varies inversely to the square of distance D

Variations
y =
k
f
k
f
k
y =
x
k
y =
x2z
The square of distance D is D2.

Variations
y =
k
f
k
f
k
y =
x
k
y =
x2z
The square of distance D is D2.
k
Hence I =
D2

Variations
In general, a variation problem gives the type of the variation and the values of the variables.

Variations
In general, a variation problem gives the type of the variation and the values of the variables. From these, we solve for the constant k and find the specific (exact) variation equation.

Variations
Example C.
a. Given that y varies directly to x and y = –4 when x = –6.
Find the constant k and the specific variation equation.

Variations
Example C.
Since y varies directly to x, the general equation is y = kx.

Variations
Example C.
Since y varies directly to x, the general equation is y = kx. Set y = –4 and x = –6 in the general equation.

Variations
Example C.
–6 = k(–4)

Variations
Example C.
–6 = k(–4) so
–6
2
k = =
–4
3

Variations
Example C.
–6 = k(–4) so
–6
2
k = =
–4
3
If we put this into the general equation, we have the specific equation:

Variations
Example C.
–6 = k(–4) so
–6
2
k = =
–4
3
If we put this into the general equation, we have the specific equation:
2
y = x
3

Variations
b. Find the value of x if y = 20.

Variations
2
Substitute y = 20 into the specific equation
y = x
3

Variations
2
y = x
3
2
20 = x
3

Variations
2
y = x
3
2
20 = x
3
3
20 * = x
2

Variations
2
y = x
3
2
20 = x
3
3
20 * = x
2
So x = 30

Variations
2
y = x
3
2
20 = x
3
3
20 * = x
2
So x = 30
In application, the language of variation sums up the basic relation of two or more types of variables.

Variations
2
y = x
3
2
20 = x
3
3
20 * = x
2
So x = 30
Example D. The weight W of an object varies inversely to the square of the distance D from the center of the earth.
A person weighs 160 pounds on the surface which is 4000 miles to the center of the earth. What will the person weigh if he’s 6000 miles above the surface of the earth?

Variations
2
y = x
3
2
20 = x
3
3
20 * = x
2
So x = 30
W = 160
4000 m.

Variations
2
y = x
3
2
20 = x
3
3
20 * = x
2
So x = 30
W = ?
6000 m.
W = 160
4000 m.

Variations
The square of the distance D is D2.

Variations
So the general equation is W = .
k
D2

Variations
k
We are to find k .
D2

Variations
k
We are to find k .
D2
Substitute W = 160, D = 4000 into the specific equation.

Variations
k
We are to find k .
D2
k
160 =
(4000)2

Variations
k
We are to find k .
D2
k
160 =
(4000)2
k
160 *16,000,000 =

Variations
k
We are to find k .
D2
k
160 =
(4000)2
k
160 *16,000,000 =
So k = 2.56 * 109

Variations
k
We are to find k .
D2
k
160 =
(4000)2
k
160 *16,000,000 =
So k = 2.56 * 109
(This constant is specific to earth.

Variations
k
We are to find k .
D2
k
160 =
(4000)2
k
160 *16,000,000 =
So k = 2.56 * 109
(This constant is specific to earth. For other planet,
the general equation is the same but k would be different.)

Variations
k
We are to find k .
D2
k
160 =
(4000)2
k
160 *16,000,000 =
So k = 2.56 * 109
Hence the specific equation is
2.56 * 109
W =
D2

Variations
k
We are to find k .
D2
k
160 =
(4000)2
k
160 *16,000,000 =
So k = 2.56 * 109
2.56 * 109
W =
D2
When the person is 6000 miles above the surface D = 10000,

Variations
k
We are to find k .
D2
k
160 =
(4000)2
k
160 *16,000,000 =
So k = 2.56 * 109
2.56 * 109
W =
D2
we have
2.56 * 109
W =
(10000)2

Variations
k
We are to find k .
D2
k
160 =
(4000)2
k
160 *16,000,000 =
So k = 2.56 * 109
2.56 * 109
W =
D2
we have
2.56 * 109
2.56 * 109
W =
=
(10000)2
108

Variations
k
We are to find k .
D2
k
160 =
(4000)2
k
160 *16,000,000 =
So k = 2.56 * 109
2.56 * 109
W =
D2
we have
10
2.56 * 109
2.56 * 109
W =
=
(10000)2
108
108

Variations
k
We are to find k .
D2
k
160 =
(4000)2
k
160 *16,000,000 =
So k = 2.56 * 109
2.56 * 109
W =
D2
we have
10
2.56 * 109
2.56 * 109
W =
=
= 25.6 lb
(10000)2
108
108

Variations
If the expression f is a product of two or more variables,
we also say y varies jointlyto these variables.

Variations
we also say y varies jointlyto these variables. Hence
“y varies directly to xz” is the same as
“y varies jointly to x and z”.

Variations
we also say y varies jointlyto these variables. Hence
“y varies directly to xz” is the same as
“y varies jointly to x and z”.
“y varies to x2z2 ” is the same as
“y varies jointly to x2 and z2”.

Variations
Exercise A. Write down the general equations for the following variation statements. Assign variables if no variable is assigned in the problems,
1. R varies directly to D.
2. R varies inversely to T.
3. S varies directly to T.
4. S varies inversely to N.
5. T varies directly to P.
6. T varies inversely to V.
7. A varies directly to r2.
8. y varies inversely to d2.
9. C varies directly to D3
10. C varies directly to xy.
11. The rate R of a car varies directly to the distance D
it covers (in a fixed unit of time).
12. The rate R of a car varies inversely to the time T it takes (over a fixed distance).
Each person in a group of N people has to chip in to buy a large pizza that cost $T. Let S be the share of each person.
13. Then S varies directly to T
14. N varies inversely to S.

Variations
15. The temperature varies directly to the pressure.
16. The temperature varies inversely to the volume.
17. The area of a circle varies directly to the square of its radius.
18. The light–intensity varies inversely to the square of the distance.
19. The mass required varies inversely to the square of the speed it was traveling.
20. The cost of cheese balls varies directly to the cube of the diameter of the ball.
21. The fun of doing something varies inversely to how much time seemly has passed doing it.
22. The time seemly has passed doing something varies inversely to how much fun there was doing it.

Variations
23. The number of marbles we are able to buy varies inversely to the price of the marbles. We’re able to buy 30 marbles at $45 each. What is the specific equation of the number of the marbles in terms of the price (per marble) and what is the price if we are only able to buy 10 marbles?
24. The light–intensity underwater varies inversely to the square of the depth of the distance. If at 3 feet the intensity is 5 ft–candle, find the specific equation and what is the intensity at the depth of 10 ft.
25. The price of a pizza varies directly to the square of the diameter of the pizza. Given that a 6” pizza is $5.50, find the specific equation for the price in terms of the diameter. What should be the price of a 12” pizza?
26. The cost of a solid chocolate ball varies directly to the cube of the diameter of the ball. Given that a 3” chocolate ball cost $25, find the specific equation of the cost in terms of the diameter and how much is one with a 1– foot diameter?

2 7 variations

by math123b

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2 7 variations Presentation Transcript

2 7 variations

by math123b

Accessibility

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2 7 variations Presentation Transcript

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