Direct variation
We say that a quantity a varies directly as a quantity b, if, when b changes, a changes in the same ratio.
This means that if b doubles in value, a will also double in value. If b increases by a factor of 3, then a will also increase by a factor of 3. While if the value of b becomes half, so will the value of a.
Let the initial values of a and b be a1, b1, and let their final values be
a2, b2. Then, a varies directly as b means: Proportionally,
a2 : a1 = b2 : b1
10 is five times 2. Therefore, a2 will be five times 7.
Problem 1. a varies directly as b. When b = 12, a = 27. What is the value of a when b = 4?
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The value of b has gone from 12 to 4. Its final value, then, is a third of its initial value. Therefore, the final value of a will be a third of 27, which is 9.
Problem 2. When b = 6, a = 42. What is the value of a when b = 9?
a2 : a1 = b2 : b1.
a2 : 42 = 9 : 6.
Alternately, 42 is seven times 6. Therefore, a2 will be seven times 9, which is 63.
Or, 9 is one and a half times 6. (6 + 3 = 9.) Therefore, a2 will be one and a half times 42. 42 + 21 = 63.
The constant of proportionality
When a varies directly as b, we often say, "a is proportional to b." In that case, the relationship between a and b takes this algebraic form:
a = kb.
k is called the constant of proportionality. In actual scientific problems, it is determined by experiment. For example, in what is called Hooke's Law, the force F that a stretched spring exerts is proportional to the distance x that the spring has stretched.
F = kx.
In other words, the greater the stretch, the greater the force.
Example 3.
a) For a given spring, F has the value 35 when the spring has stretched 8
a) inches. What is the constant of proportionality for that spring?
Solution. F = kx. That is,
35 = k· 8
Therefore,
When a varies directly as b, the constant of proportionality is the quotient of any observed or given values.
Note: The units on the right must equal those of a on the left -- distance, time, force, whatever they might be. a1, then, is the numerator and b1 the denominator: The units of b then cancel.
See the following problem.
Problem 3.
a) The distance d that an automobile travels varies directly as the time t
a) that it travels. After 2 hours, the car has traveled 115 miles. Write
a) the equation that relates d and t.
Problem 4. Prove: Varies directly is a transitive relation. That is, if a varies directly as b, and b varies directly as c, then a varies directly as c.
If a = k1b, and b = k2c, then a = k1k2c.
Problem 5. If the side of a square doubles, how will the perimeter change?
![](Alg_img/430.gif)
The perimeter will also double, because the perimeter varies as the side. P = 4s. The constant of proportionality is 4.
Problem 6.
a) If the diameter of a circle doubles, how will the circumference change?
![](alg_IMG/463.gif)
The circumference will also double, because the circumference varies as the diameter.
b) What number is the constant of proportionality?
π. C = πD.
That constant has been the subject of investigation for over 2500 years.
Problem 7. If the diameter of a circle changes from 6 cm to 9 cm, by how much will the circumference change?
In going from 6 cm to 9 cm, the diameter has increased one and a half times; that is the ratio of 9 to 6. Therefore, the circumference will also increase one and a half times.
Problem 8. The circumference C of a circle varies directly as the perimeter of the circumscribed square.
Varies as the square
A quantity a varies as the square of a quantity b, if, when b changes, a changes by the square of that ratio. Thus, if b changes by a factor of 4, then a will change by a factor of 4² = 16. If b changes to one third of its value, then a will change to one ninth of its value.
![](Alg_IMG/432.gif)
Problem 9. a varies as the square of b. When b = 7, a = 4. What is the value of a when b = 35?
In going from 7 to 35, b has changed by a factor of 5. a therefore will change by a factor of 5² = 25. a = 25· 4 = 100.
Problem 10. a varies as the square of b. When b = 20, a = 32. What is the value of a when b = 15?
In going from 20 to 15, b has become three fourths of its value. 15 is three fourths of 20. a therefore will become nine sixteenths of its value.
Theorem. If a varies directly as b, then a² will vary as b².
This is easily proved if we write the ratios in fractional form.
a varies directly as b means:
![](alg_IMG/436.gif)
Therefore, on squaring both sides:
![](alg_IMG/437.gif)
This implies
This means that a² varies as b²; which is what we wanted to prove.
Problem 11. The area A of a circle varies directly as the area of the circumscribed square. That is, as the area of the square changes, the area
![](alg_IMG/435.gif)
of the circle changes proportionally.
a) Show that this implies that the area A of the circle varies as the
a) square of the radius r.
The side of the circumscribed square is equal to the diameter D of the circle. Therefore the area of the circumscribed square is equal to D². Hence the area A of the circle varies as D².
But D varies directly as r -- D = 2r -- and therefore, according to the
theorem, D² varies as r². Therefore, since A varies as D², and D² varies as r², then transitively, A varies as r². The area of the circle varies as the square of the radius.
b) If the radius of a circle changes from 6 cm to 12 cm, how will the
b) area change?
In going from 6 cm to 12 cm, the radius has doubled, that is, it has changed by a factor of 2. The area therefore will change by a factor of 2² = 4. It will be four times larger.
c) What is the constant of proportionality that relates the area A to r²?
π. A = πr².
Example 4. The surface area of a sphere.
![](alg_IMG/440.gif)
The surface area of a sphere is proportional to the surface area of the circumscribed cube.
Now, each face of the cube is a square whose side is equal to the diameter D of the sphere. And a cube has 6 faces. Therefore, the surface area of the cube is equal to 6D².
In other words, the surface area A of a sphere is proportional to the square of its diameter.
Do you know what the constant of proportionality is?
π. A = πD²
Problem 12. Show that the surface area of a sphere varies as the square of its radius. Write the equation that relates the surface area A to the radius r.
Since A = πD², and D = 2r, then A = π(2r)² = 4πr².
Section 2: Varies inversely. Varies as the inverse square.
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