Does anyone have any clue on how to start this question off?
The highly scary helter skelter at the fair is a cylindrical tower of height 30 metres and circumference 8 metres. The slide is wound around the tower exactly 5 times. What’s the total length of the slide?
I think it's to do with Helixes or something and we have not covered any of that in class. I've thought about 'unravelling it' as in making it a rectangle with the width being 30 and the length being 8π but do not know what to do next! Any help is much appreciated as I have no clue on what to do!!
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$\begingroup$For a single winding the height would be $6$ metres.
The circumference of the cylinder is given as $8$ so no $\pi$ required.
Unfold the cylinder and the single winding unravels as a straight line corresponding to the diagonal of a right-angle triangle of side lengths $6,8$.
By Pythagoras' theorem, the length of the diagonal is $\cdots\bullet $.
Total spiral length is $5$ times this, which is $\cdots\bullet $.
If you know a bit of calculus then you could parameterize the helix as $$ x(t) = \frac{4}{\pi}\cos(10\pi t), \quad y(t) = \frac{4}{\pi}\sin(10\pi t), \quad z(t) = 30 t, \quad t \in [0,1] $$ The arc length is then \begin{align} A &= \int_0^1 \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2}\,dt\\ &= \int_0^1 \sqrt{1600 + 900}\,dt\\ &= 50 \end{align}
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