Ask Me Help Desk - Applied math, arc length, change of parameter

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)

- Mathematics (https://www.askmehelpdesk.com/forumdisplay.php?f=199)

- - Applied math, arc length, change of parameter (https://www.askmehelpdesk.com/showthread.php?t=57898)

1 Attachment(s)

Applied math, arc length, change of parameter

Here's a arc length problem maybe you'd like to tackle:

"Copper tubing with an outside diameter of 1/2 inches is to be wrapped in a circular helix around a cylinder core that has a diameter of 12 inches. What length of tubing will make one complete turn around the cylinder in a distance of 20 inches measured along the axis of the cylinder?".

Here's a haphazard diagram. Best I could do with Paint.

Here's my approach. I would enjoy seeing others.

Parametrizing. $x={a}cos(t), \;\ y={a}sin(t), \;\ z=ct$

$\frac{dx}{dt}={-a}sin(t), \;\ \frac{dy}{dt}={a}cos(t), \;\ \frac{dz}{dt}=c$

$L=\int_{0}^{t_{0}}\sqrt{a^{2}sin^{2}(t)+a^{2}cos^{ 2}(t)+c^{2}}dt=\int_{0}^{t_{0}}\sqrt{a^{2}+c^{2}}d t=t_{0}\sqrt{a^{2}+c^{2}}$

$a=\frac{25}{4}=6.25$ =the distance from the center of the helix to the center of the tubing.

$c=\frac{20}{2{\pi}}=\frac{10}{\pi}$ , because the tubing makes one revolution about the cylinder.

Therefore, we have:

$2{\pi}\sqrt{(\frac{25}{4})^{2}+(\frac{10}{\pi})^{2 }}=\frac{5\sqrt{25{\pi}^{2}+64}}{2}\approx{44.07} \;\ inches$

Hmm I get a slightly different answer, trying to do it without integrals :D

I tried to calculate the circumference of an ellipse using Ramanujan's approximation.

I take $a = \sqrt{6.25^2+5^2}$
and $b = 6.25$

Calculating circumference using:
$C \approx \pi a \left[ 3 (1+\sqrt{1-\varepsilon^2}) - \sqrt{(3+ \sqrt{1-\varepsilon^2})(1+3 \sqrt{1-\varepsilon^2})} \right] \!\,$

where $\varepsilon = \sqrt{1 - \frac{b^2}{a^2}}$

It comes out to about 41.14 inches, which does seem a little small.

Putting in $a = \sqrt{6.25^2+5.25^2}$ gives 45.48 inches which I am much happier with using an approximation.
Using $a = \sqrt{6.25^2+5.125^2}$ which I think is a slightly better approximation gives 45.21 inches

I don't know how to derive the equation for the helix length, but accepting the arc length as $L = t sqrt{r^2 + c^2}$ as mentioned by galactus, where r is the radius of the helix, 2πc is the vertical distance between the tubing loops, and t is the number of times the tubing is looped around the helix as measured in radii, I calculate the length needed as follows.

In this case c = 10/π and t = 2π.
I would say that r is fully 6.5. The tube length needed will be longer because the surface of the tube that is closest to the helix is much more likely to crumple than the outside surface is likely to stretch. You should also take into account that the tube, being crumpled, is not going to lie exactly flat. The stretch that the tube does take on will be offset by this inability to lie flat.

So the length needed would be $L = 2\pi sqrt{6.5^2 + (\frac{10}{\pi})^2} = 45.47$ .

I think we're talking about a tube made of a perfect material that doesn't suffer from those admin :p

You big pedant

The problem stated "copper" not "perfect material". :P

:x

This is my technique:

Imagine you unroll the helix against the wall. The helix becomes the line L in the following diagram:

Code:

/| / | L/ | / | H / | / | ------- C

C is the circumference of the helix and H is the pitch (vertical distance traveled during one revolution of the helix).
H is obviously 20"
C is $2\pi r$
and r is... debatable it seems :) I'll go with 6.25 so that it's obvious that my answer matches the original posted by galactus.
So now $L=\sqrt{C^2 + H^2}=\sqrt{\left(2\pi r\right)^2 + H^2}\approx 44.07$

Same answer, no calculus :p

r is the distance from the center of the cylinder to the center of the copper tubing.

The cylinder is 12" in diameter, therefore, the radius is 6 inches. The copper tubing is 1/2 inch in diameter, therefore, the radius is 1/4 inches. 6+1/4=6.25 from center of cylinder to center of tubing.

But, like admin said, does the inner surface of the tube compress, or the outer surface stretch, or both?

Technically, a helix does not have any width. I think your real question is, what is the length of the helix that is defined by the center of the tube. Clearly the outside of the tube is longer than the inside of the tube.

Don't read too much into the problem. I just got it from an old calc text. That's the way it was written. After seeing different approaches, I managed to find the solution they have.
They come up with 44.07 also, by using calc similarly to my method. Though, that's just one way. Maybe the problem was phrased poorly.

All times are GMT -7. The time now is 12:34 PM.