Calculating Ribbon Length for Decorating Cylindrical Flower Containers

[vismayagn33]

A florist delivers single stemmed flowers in a sealed plastic container that is cylindrical in shape. Each container has a bade radius of 4cm and a height of 45cm. the florist wished to decorate each container with a very thin coloured ribbon. The ribbon will wind around the body of the container in a single spiral, reaching the top directly above the starting point at the bottom. The florist notes that when he draws the net of the cylinder the spiral will be diagonal of the body of the container.
a) Find the length of ribbon the florist will need to decorate this container (base radius 4cm and a height of 45 cm) with a single spiral if no ribbon is required for finishing/gluing.
b) Develop a rule that will determine the length of ribbon L, required to wrap a cylinder of radius, r cm, and height, h cm, with a single spiral. Provide a clearly labelled diagram to support your conclusion. Note both r and h are > o cm.
a regular customer has suggested that certain species of flowers would look ever more appealing with two, three or even four, spirals wrapped around the cylindrical container.
c) Calculate the total length of ribbon required for 2 evenly spaced spirals on a cylindrical container with a radius off 4cm and a height of 45 cm if the end of the ribbon reaches the top of the container directly above its starting point.
Develop a general formula, which will give the length of ribbon L cm, required to decorate a cylindrical container (with radius of r cm) and a height of h cm) with 4 evenly spaced spirals if the end of the ribbon reaches the top of the container directly above its starting point and both r and h are >0cm.
You must show and explain how the general formula is developed for full marks. Providing only an answer will earn one mark at most. Clearly define all variables used in your formula.

 

[MarkFL]

Hello and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

We also discourage multiple thread for the same problem because not only is it redundant, but it causes a potential duplication of effort on the part of our helpers, whose time is valuable.

edit: Our policy here at MHB is not to provide help for questions that appear to be part of a graded assignment:

Teachers expect graded assignments to be the work of that student and not others. Therefore, MHB policy is not knowingly to help with questions in such assignments. If you present a question in such a way as to suggest that it falls in this category, a moderator will close the thread and explain why. The original poster can always PM the moderator to discuss the situation, but if the moderator is unconvinced, the thread will stay closed. When a moderator is certain that a particular member is trying to cheat, he will ban that member and remove the offending thread. Where possible, the moderator will notify the institute at which the member studies and provide a copy of the thread. This may sound harsh, but MHB places a high premium on academic honesty and integrity.

 

[MarkFL]

I have re-opened this thread since enough time has gone by.

To determine the length of a helical spiral about a right circular cylinder, consider taking a piece of paper of base $b$ and height $h$ and drawing a diagonal line on the paper and then rolling the paper into a cylinder with the drawn diagonal on the outside. You will find you have exactly one spiral going around the cylinder. So, the length $L$ of the spiral is simply the length of the diagonal of the unrolled paper, which we know by Pythagoras is:

\(\displaystyle L=\sqrt{b^2+h^2}\)

If we are given the radius $r$ of the cylinder, then we know the base of the unrolled cylinder is its circumference, given by:

\(\displaystyle b=C=2\pi r\)

And so our formula for the spiral length becomes:

\(\displaystyle L=\sqrt{(2\pi r)^2+h^2}\)

Now, suppose we wish to know the spiral length for $n$ spirals...what we could do with our paper is divide the height into $n$ evenly spaced segments and then draw diagonals across each segment in the same direction. And thus, we find the total length of the helix would be given by:

\(\displaystyle L_{n}=n\sqrt{(2\pi r)^2+\left(\frac{h}{n}\right)^2}=\sqrt{(2n\pi r)^2+h^2}\)

And now we have a general formula with which we can answer the given questions.

a) Find the length of ribbon the florist will need to decorate this container (base radius 4cm and a height of 45 cm) with a single spiral if no ribbon is required for finishing/gluing.

Here, we identify $n=1,\,r=4\text{ cm},\,h=45\text{ cm}$ and thus:

\(\displaystyle L=\sqrt{(2(1)\pi (4\text{ cm}))^2+(45\text{ cm})^2}=\sqrt{64\pi^2+2025}\text{ cm}\approx51.54\text{ cm}\)

b) Develop a rule that will determine the length of ribbon L, required to wrap a cylinder of radius, r cm, and height, h cm, with a single spiral. Provide a clearly labelled diagram to support your conclusion. Note both r and h are > o cm.

Here, we simply identify $n=1$ and write:

\(\displaystyle L_{1}=\sqrt{(2\pi r)^2+h^2}\)

c) Calculate the total length of ribbon required for 2 evenly spaced spirals on a cylindrical container with a radius of 4cm and a height of 45 cm if the end of the ribbon reaches the top of the container directly above its starting point.

Here, we identify $n=2,\,r=4\text{ cm},\,h=45\text{ cm}$ and thus:

\(\displaystyle L=\sqrt{(2(2)\pi (4\text{ cm}))^2+(45\text{ cm})^2}=\sqrt{256\pi^2+2025}\text{ cm}\approx67.47\text{ cm}\)

d) Develop a general formula, which will give the length of ribbon L cm, required to decorate a cylindrical container (with radius of r cm) and a height of h cm) with 4 evenly spaced spirals if the end of the ribbon reaches the top of the container directly above its starting point and both r and h are >0cm.

Here, we simply identify $n=4$ and write:

\(\displaystyle L_{4}=\sqrt{(2(4)\pi r)^2+h^2}=\sqrt{(8\pi r)^2+h^2}\)