Problem 18
Question
The relationship of the diameter of a circle, \(x\), and the circumference of the circle, \(y\), is a direct variation. The diameter of a circle is \(20 \mathrm{~cm}\), and the circumference is \(62.8 \mathrm{~cm}\). a. Find the constant of proportionality, \(k\). b. Write an equation that represents this relationship. c. Find the circumference of a circle with a diameter of \(40 \mathrm{~cm}\).
Step-by-Step Solution
Verified Answer
a. \ k = 3.14 \, b. \ y = 3.14x \, c. \ 125.6 \ cm.
1Step 1 - Understand Direct Variation
In a direct variation, two quantities increase or decrease proportionally. This relationship can be expressed as \( y = kx \), where \( k \) is the constant of proportionality.
2Step 2 - Identify Given Values
We are given that the diameter of a circle \( x = 20 \) cm and the circumference \( y = 62.8 \) cm.
3Step 3 - Find the Constant of Proportionality (k)
Substitute the given values into the direct variation formula \( y = kx \). So, \( 62.8 = k \times 20 \). Solve for \( k \): \[ \ k = \frac{62.8}{20} = 3.14 \]\.
4Step 4 - Write the Equation
Using the constant of proportionality \( k = 3.14 \), substitute \( k \) back into the direct variation formula to get the equation representing the relationship: \[ y = 3.14x \].
5Step 5 - Find the Circumference for a New Diameter
We are asked to find the circumference of a circle with a diameter of \( 40 \) cm. Use the equation from Step 4: \[ y = 3.14 \times 40 = 125.6 \] cm.
Key Concepts
constant of proportionalitycircumference formuladiameter and circumference relationship
constant of proportionality
When dealing with direct variation, understanding the constant of proportionality is key. This constant, often represented by the letter 'k', shows how two variables, like diameter and circumference, change together.
In a direct variation, the formula we use is: \( y = kx \). Here, 'y' and 'x' are our two variables and 'k' is what links them proportionally.
To find this 'k', we simply rearrange the formula: \[ k = \frac{y}{x} \].
Given that in our problem the diameter (x) is 20 cm and the circumference (y) is 62.8 cm, we can substitute in these values to get: \[ k = \frac{62.8}{20} = 3.14 \].
This means that for every centimeter increase in diameter, the circumference increases by 3.14 cm.
In a direct variation, the formula we use is: \( y = kx \). Here, 'y' and 'x' are our two variables and 'k' is what links them proportionally.
To find this 'k', we simply rearrange the formula: \[ k = \frac{y}{x} \].
Given that in our problem the diameter (x) is 20 cm and the circumference (y) is 62.8 cm, we can substitute in these values to get: \[ k = \frac{62.8}{20} = 3.14 \].
This means that for every centimeter increase in diameter, the circumference increases by 3.14 cm.
circumference formula
One of the most crucial formulas in geometry, especially when dealing with circles, is the circumference formula.
The standard formula for the circumference of a circle is usually written as: \[ C = \pi d \], where 'd' is the diameter of the circle and \( \pi \) (pi) is approximately 3.14.
However, when we talk about direct variation, it's useful to see it as: \[ y = kx \].
With 'k' being our constant of proportionality (3.14 in this case), this formula tells us that the circumference (y) of a circle is directly proportional to its diameter (x).
Having this equation means you can quickly find the circumference if you know the diameter. For example, with a diameter of 40 cm: \[ y = 3.14 \times 40 = 125.6 \] cm.
The standard formula for the circumference of a circle is usually written as: \[ C = \pi d \], where 'd' is the diameter of the circle and \( \pi \) (pi) is approximately 3.14.
However, when we talk about direct variation, it's useful to see it as: \[ y = kx \].
With 'k' being our constant of proportionality (3.14 in this case), this formula tells us that the circumference (y) of a circle is directly proportional to its diameter (x).
Having this equation means you can quickly find the circumference if you know the diameter. For example, with a diameter of 40 cm: \[ y = 3.14 \times 40 = 125.6 \] cm.
diameter and circumference relationship
The relationship between the diameter and circumference of a circle is a beautiful example of direct variation.
Simply put, as the diameter (x) of a circle increases, the circumference (y) also increases proportionally. This is because they are linked by the constant of proportionality (k).
In all circles, this constant is represented by \( \pi \) (pi), which is approximately 3.14.
So, whenever you have a diameter, you can quickly find the circumference by multiplying the diameter by 3.14.
Take for instance our example values, a circle with a diameter of 20 cm has a circumference of 62.8 cm, and doubling the diameter to 40 cm gives us a circumference of 125.6 cm: both perfectly following the equation \[ y = 3.14x \].
This helps in understanding that the circumference is always about three times larger than the diameter, precisely by a factor of \( \pi \), highlighting the interplay between these two fundamental properties of circles.
Simply put, as the diameter (x) of a circle increases, the circumference (y) also increases proportionally. This is because they are linked by the constant of proportionality (k).
In all circles, this constant is represented by \( \pi \) (pi), which is approximately 3.14.
So, whenever you have a diameter, you can quickly find the circumference by multiplying the diameter by 3.14.
Take for instance our example values, a circle with a diameter of 20 cm has a circumference of 62.8 cm, and doubling the diameter to 40 cm gives us a circumference of 125.6 cm: both perfectly following the equation \[ y = 3.14x \].
This helps in understanding that the circumference is always about three times larger than the diameter, precisely by a factor of \( \pi \), highlighting the interplay between these two fundamental properties of circles.
Other exercises in this chapter
Problem 17
For exercises 7-32, simplify. $$ \frac{z^{2}-7 z-18}{z^{2}+4 z+4} \cdot \frac{z^{2}-4 z-12}{z^{2}-11 z+18} $$
View solution Problem 17
For exercises 1-66, simplify. $$ \frac{y+9}{y^{2}+9 y} $$
View solution Problem 18
For exercises \(9-24\), evaluate or simplify. $$ \frac{\frac{r^{2}+11 r+24}{9 r}}{\frac{r^{2}-64}{27 r^{3}}} $$
View solution Problem 18
For exercises 13-24, rewrite each expression as an equivalent expression with the given denominator. $$ \frac{4}{15 x+6} ; 12(5 x+2) $$
View solution