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I have found this riddle in my book and so far which may require the use of calculus (integrals) to which I'm familiar but not very savvy with it. Since, I've not yet come with an answer. I wonder if such problem can be also solved using sums in the scope of college precalculus like an approximation (proven) which could be solved by hand even the calculus method which I also feel might also help me.

The problem is as follows:

In a research facility in Taiwan a group of technicians built a new optical disk which stores information in a spiral engraved in its bottom face named "lecture side". Under the microscope it can be seen that the spiral begins in a region starting from $\textrm{2.6 cm}$ from its axis of rotation and it ends at $\textrm{5.7 cm}$ from the center of the disk. It can also be seen that individual turns of the spiral are $0.74\,\mu\textrm{m}$. Using this information calculate the length of the entire track.

So far I've only come with the idea of using the spiral of Archimedes, whose formula is given as follows:

$$r=a+b\phi$$

However, I'm not very familiar with the realm of polar coordinates or how can this equation be used to solve my problem. To better illustrate the situation, however I've drawn this sketch to show how I'm understanding the problem.

Sketch of the problem

I've included a cartesian grid, which well "may not be" in scale. But gives an idea of how I believe it is intended to be said.

I've really wanted to offer more into this such as an attempt into solving, but so far I've ran out of ideas.

However, I've come with the idea that the solution may be linked with finding how many turns are in the "readable sector" which is alluded in the problem.

To calculate this what I did was the following:

$\textrm{number of turns}= \left( 5.7 - 2.6 \right)\times 10^{-2}\textrm{m}\times \frac{\textrm{1 turn}}{0.74\times 10^{-6}\textrm{m}}$

By evaluating this short conversion factors I obtained

$\textrm{number of turns} = 41891.89189\,\textrm{turns or rad}$

And that's it. But from there I don't know how to relate this information with what would be needed to solve this riddle. The answer in the book states that the track's length is $1.09\times 10^{4}\,\textrm{m}$. But again. I don't know what can I do to get there.

So far I've did some preliminary research in the community and I've found this but it isn't really very helpful as the example relates to a vertical helix going upwards around a cylinder and my situation doesn't fit into this. Other proposed methods such as the one seen here propose using calculus and this also is referenced here. Needless to say that the only existing solution doesn't give much details of how the supposed formula works. It also contributes to the problem that I'm not very savvy with it and since this problem was obtained from a precalculus book, I aimed my question to. Does it exist a method to make this computation relying in college algebra?. With this I don't intend to totally discard a calculus approach, but an answer with would really be very pedagogical for me is one which can show me the two methods, so I can compare which can be better according to my current knowledge so I can practice more in what I feel I'm lacking.

Well that's it. I do really hope somebody could help me with the two methods, one using algebra and another using calculus.

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As an approximation for the first loop, as the length would be
between the circumference of a circle of radius 2.6 cm and
the circumference of a circle of radius 2.6 cm + 0.74 $\mu$m,
is the average of those two circumferences.

Doing likewise for the other loops will give
you an approximation for the total length.
Take into account the last loop is 1/4 of a circle.

Another approximation for s the length of the spiral.
Let c = 2$\pi$, r the inner radius, w the width
between each arc. n the number of turns. Then

sum(k=0,n-1) c(r + wk) = crn + cwn(n - 1)/2 < s <
sum(k=1,n) c(r + wk) = crn + cwn(n + 1)/2.
n is an integer. Add as needed a fraction of an additional turn,

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  • $\begingroup$ I was thinking into a more precise computation (hence I not totally discouraging calculus). The method of summing those averages may be a good idea if the number of loops is smaller (and taking for granted the omission of demonstrating the average is the length of the loop), but in this case with only $0.74\,\mu\textrm{m}$ thickness to cover a distance of $3.5\,\textrm{cm}$ would meant a huge pack of loops. Doing it individually will make it impractical. Does it exist a more reasonable method?. $\endgroup$ – Chris Steinbeck Bell Jan 16 '19 at 0:31
  • $\begingroup$ @ChrisSteinbeckBell. See edit. $\endgroup$ – William Elliot Jan 16 '19 at 8:47
  • $\begingroup$ The equations you have posted do not display properly. Mind using Mathjax instead?. I would like to know if there is any limitation in using the sum because in my case the number of turns is not an integer. See $N=\left(\frac{5.7-2.6}{0.74\times 10^{-6}}\right)\times 10^{-2}=\left(\frac{31\times 10^{5}}{74}\right)$. What you proposed it was something which I also thought or intended to use meant as a sum but in this case they are not concentrical disks but rather a spiral. Regarding the notation you use do you meant $s$ as being the length?. $\endgroup$ – Chris Steinbeck Bell Jan 16 '19 at 13:39
  • $\begingroup$ I'd voice my opinion that you may try to evaluate the sums that you propose using the numbers in my problem to see if at least comes closer to the answer. Again I believe that an answer which would help me is one that considers how to find the length of an spiral of archimedes even if resorting to some calculus or maybe the precalculus sum that you are intending to use. To better illustrate what you are trying please enlighten me with some of the computation with numbers. :) $\endgroup$ – Chris Steinbeck Bell Jan 16 '19 at 13:42
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This can be solved using the formula for area: $A=L\times W$, or equivalently: $L=\frac AW$

The entire track occupies an area of $\pi(0.057^2-0.026^2)$ square meters. The width of the track is given to be $0.74\times10^{-6}$ meters. The length is therefore: $$\frac{\pi(0.057^2-0.026^2)}{0.74\times10^{-6}}\approx1.09\times10^4\text{ meters}$$

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  • $\begingroup$ I've found a similar approach here but it was not documented on how it was obtained such formula. In your notation does $\textrm{W}$ stands for width right?. It would be very valuable to add that the difference between the squares comes from finding the area of the "flat doughnut". However as mentioned this is an approximation and although it checks, I wonder if there was a more "elaborate" way to find the lenght, hence accuracy. The integral of the function the spiral for example. $\endgroup$ – Chris Steinbeck Bell Jan 16 '19 at 14:52
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    $\begingroup$ @Chris It is not possible to get a more accurate measure of the length without more accurate measure of the radii. The given distances of $2.6$ cm and $5.7$ cm could each be +/- $0.05$ cm. This gives a variance in total length of +/- $350$ meters. This variance also means your 'number of turns' may be off by more than $1000$ $\endgroup$ – Daniel Mathias Jan 16 '19 at 15:19
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You can consider the "average" radius $\bar r=(5.7 + 2.6)/2=4.15$ cm and simply compute $$ L=2\pi\bar r\cdot N=2\pi\cdot 4.15\cdot 41891.9=1.09234\times10^6\ \text{cm}= 1.09234\times10^4\ \text{m}. $$ If you insist on using calculus, you can of course integrate along the spiral: $$ L=\int_0^{2N\pi}\sqrt{\left({dr\over d\phi}\right)^2+r^2}\,d\phi =\int_0^{2N\pi}\sqrt{b^2+(a+b\phi)^2}\,d\phi=1.09234\times10^4\ \text{m}, $$ with: $a=2.6$ cm and $b={0.74\times10^{-6}\over2\pi}$ m. As you can see there is no difference, even keeping 6 significant digits.

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