Math CentralQuandaries & Queries


Question from Bakshani, a student:

how do you a mark a pipe 5 inch diameter and cut it to form a 45 degree angle

Hi Bakshani.

Two ways come to mind: an easy way and a more "mathematical" way:

Easy way:
Submerge the end of the pipe in water at a 45 degree angle. Use a grease pencil to mark the waterline.

Mathematical way:
If you were to "unroll" the pipe, you would find that the cut would be a sine wave. So if you draw a sine wave of the right amplitude and wavelength on a piece of paper, you can wrap it around the pipe to show where to cut it. The tricky part is to determine the amplitude
and wavelength.

Amplitude - The height from the top to the bottom is double the amplitude. Since the pipe is 5 inches in diameter and this is a 45 degree cut, you can picture the diameter and the height of the cut forming a 90-45-45 triangle, so the height is 5 inches. Therefore the amplitude is 5/2 inches.

Wavelength - One entire circumference of the pipe is the wavelength (the cut has to return to the original spot in exactly one rotation of the pipe). So the wavelength is 5 π.

Now use the amplitude and wavelength to write a sine wave equation in the form y = A sin(Bx). The domain of x is 0 inches to 5 π inches. So the variable B "converts" this domain to a degree between 0 and 360. Thus B is 360 / (5 π). A is just the amplitude. Finally, the simplified equation is y = (5/2) Sin [360x / (5 π) ]. That simplifies to just y = 2.5 Sin ( 72x / π )

Draw this equation on graph paper with the x going from 0 to 5 π inches. You should find that y ranges from -2.5 to 2.5 inches.

It you wrap this around the pipe, you will see where the cut must be.

The water method is clearly simpler and quicker, but I think the math method is more fun!

Stephen La Rocque.

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