Eight turns of a wire are wrapped around a pipe with a length of 20 cm and a circumference of 6cm. What is the length of the wire?

- anonymous

- schrodinger

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- dumbcow

my guess would be to think of the pipe as being rolled out so it looks like a rectangle of length 20 and width the circumference 6.
Now each time the wire is wrapped around it travels one revolution of the pipe or 6cm but it also goes down the length of the pipe 20/8 or 2.5 cm, going back to the rectangle that would look like pulling the wire diagonally across the rectangle to a point 2.5 cm down forming a right triangle.
Find the length of the hypotenuse of that triangle which is length of the wire for first revolution and multiply by 8.
_>c^2 = 6^2+2.5^2 = 42.25
->c = 6.5
_>length of pipe = 6.5 * 8 = 52

- dumbcow

hope that made sense

- anonymous

SO IF THE RECTANGLE IS ROLLED OUT WOULD THE 6 BE THE BASE

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## More answers

- dumbcow

yes

- dumbcow

wait it depends if the pipe is standing up then its the base :)

- anonymous

What would be the equation without the the numbers only using (b) (H) or whatever, so I can try it

- dumbcow

Given length of pipe H, circumference C with wire wrapped N times.
length of wire = sqrt(C^2+(H/N)^2)*N

- anonymous

How did you get 6.5 c

- dumbcow

sqrt(C^2+(H/N)^2), C=6,H=20,N=8

- anonymous

I forgot to sqrt it, then 52 will be the length of the wire?

- dumbcow

yes

- anonymous

thanks so much

- dumbcow

welcome, i dont get your railroad problem, is there more info?

- anonymous

thats was the whole question the teacher gave me, I emailed her today hopefully she get give me a hint.

- anonymous

what number did you get in the end for the length?

- dumbcow

ok sounds like you need the initial height of the rail, im prob just missing something

- anonymous

I get about 507.5cm

- dumbcow

how do you get that? there are only 8 turns of the wire

- anonymous

8 complete turns stretched out over 20cm of length.

- anonymous

A turn is 2pi radians.

- anonymous

It's a spiral.

- anonymous

You have to find a parametric equation for the spiral in terms of the information you have and find the arc length of the spiral.

- anonymous

ok so what equation would use to slove the question?

- anonymous

The equation for a spiral is \[u(\theta) = (r \cos \theta , r \sin \theta , a \theta + b)\]

- anonymous

I've jumped a couple of steps because I don't have much time.

- anonymous

You know what the radius of the spiral should be, so you need to find a and b. This last coordinate is the height, z. When theta =0, the height of the spiral is 0, so b is 0. When the height of the spiral is 20, theta = 8 x 2pi = 16 pi (since you're told there are 8 complete turns in 20cm), so 20 = 16pi.a --> a = 5/(4pi).

- anonymous

\[u(\theta) = (r \cos \theta, r \sin \theta , \frac{5 \pi}{4}\theta)\]

- anonymous

The arc length is given by\[L=\int\limits_{\theta = 0}^{\theta = 16 \pi}\sqrt{r^2 \cos^ 2 \theta + r^2 \sin^2 \theta + \frac{25 \theta^2}{16\pi^2}}d\theta \]\[=\int\limits_{\theta = 0}^{\theta = 16 \pi}\sqrt{r^2 + \frac{25 \theta^2}{16\pi^2}}d \theta =\frac{1}{4\pi}\int\limits_{0}^{16 \pi}\sqrt{16\pi^2 r^2+25 \theta^2}d \theta\]

- anonymous

\[r=\frac{3}{\pi}\]so substitute that in and integrate.

- anonymous

You'll have to make a substitution in your integral in order to solve it. I don't have anymore time, sorry. If this isn't something due very soon, I can come back to it.

- dumbcow

hmm what class is this for?

- anonymous

Sorry if I've confused matters :s

- dumbcow

no you're correct i just doubt parametric equations is required for the class

- dumbcow

rosette what class was this problem for?

- dumbcow

anyway 507 is obviously way too much you would have to wind the wire around the pipe like 80 times

- anonymous

this is for quantitavtive literacy

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