Solve the Pi (y/2π)² = y²/4π Mystery

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In summary, the "Solve the Pi (y/2π)² = y²/4π Mystery" is a mathematical equation that involves solving for the value of y. The value of π (pi) is a constant in this equation and represents the ratio of a circle's circumference to its diameter. To solve the equation, you can use algebraic manipulation to isolate the variable y. This equation has real-world applications in fields such as engineering, physics, and architecture. Other methods of solving this equation include using the quadratic formula or graphing the equation and finding the points of intersection. However, algebraic manipulation is the most common and straightforward approach.
  • #1
powp
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How does pi(y/2pi)^2 = y^2/4pi?

I am confussed.

Thanks
 
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  • #2
[tex]\pi \left(\frac {y} {2\pi}\right)^2 = \frac {\pi y^2} {(2\pi)^2} = \frac{\pi y^2} {4\pi^2} = \frac {y^2} {4\pi}[/tex]
 
Last edited:
  • #3
In short and abstract form

[tex] \left(\frac{a}{b}\right)^{2}=\frac{a^{2}}{b^{2}} [/tex]

and

[tex] \left(ab\right)^{2}=a^{2}b^{2} [/tex]

,okay?

Daniel.
 
  • #4
what happens to the pi on top?
 
  • #5
It cancels with one of the pi's on the bottom.

[tex]\frac{\pi y^2} {4\pi^2} = \frac{\pi y^2} {4\pi*\pi} = \frac{\pi(y^2)}{\pi(4\pi)}[/tex]

Do you see how you can cancel?
 
  • #6
Yes Thanks
 

Related to Solve the Pi (y/2π)² = y²/4π Mystery

What is the "Solve the Pi (y/2π)² = y²/4π Mystery"?

The "Solve the Pi (y/2π)² = y²/4π Mystery" is a mathematical equation that involves solving for the value of y. It is commonly used in geometry and trigonometry to find the value of a variable in a given equation.

What is the significance of π in this equation?

The value of π (pi) is a constant in this equation and represents the ratio of a circle's circumference to its diameter. It is approximately equal to 3.14 and is commonly used in mathematical calculations involving circles and spheres.

How do you solve this equation?

To solve the equation, you can use algebraic manipulation to isolate the variable y. This involves expanding the squared terms, simplifying the equation, and then taking the square root of both sides. The result will give you the value of y.

What are some real-world applications of this equation?

This equation is commonly used in fields such as engineering, physics, and architecture to calculate the dimensions of circles and spheres. It can also be applied in financial calculations, such as calculating compound interest.

Are there any other ways to solve this equation?

Yes, there are other methods to solve this equation, such as using the quadratic formula or graphing the equation and finding the points of intersection. However, algebraic manipulation is the most common and straightforward approach.

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