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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 2,000

A water pipe whose diameter is 84 cm dan length is 2,4 m can contain rain water with the water height 68 cm like in the attached picture.

Determine:

a. The surface area which gets contact with the water

b. The volume of the water (in liters)

So... How do I do? What is the simple way to determine the area of a... truncated circle? (dunno what the proper term is)

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.

May his adventurous soul rest in peace at heaven.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 129

I am sure that there are several approaches to tackle the problem. However, one of them is as follows:

If we let d and r to be the water height in the pipe and the radius of the base of the pipe, respectively, then we would have:

where h is the length of the arc of the circle covered by the rain water, and hence the surface area a of the pipe contacted with the water would be given by:

where l is the length of the water pipe. To get the volume of the water in the pipe, we first calculate the area A of the base covered by the water use the integration:

where the integration as been performed using trigonometric substitutions (if you would like to have the details I can provide it to you). Hence the volume of the water in the pipe is:

Note: We get the Volume of the cylinder (the water pipe) in the case that r-d=-r (that is when the pipe is full of rain water).

*Last edited by Grantingriver (2018-07-05 03:31:54)*

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**Monox D. I-Fly****Member**- From: Indonesia
- Registered: 2015-12-02
- Posts: 2,000

Is there any approach without using trigonometry? This is supposed to be for 9 graders.

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.

May his adventurous soul rest in peace at heaven.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,829

hi Monox D. I-Fly

There is a circle theorem which will help for the surface area.

If chords AB and CD, for any circle, intersect at E then AE.EB = CE.ED You can prove this using similar triangles.

So if the width of the surface area is 2x (CD) then x^2 = 16 times 68

You can also use Pythag on triangle CEO. (x^2 = 42^2 - 26^2)

To get the volume you'd need the area of cross section . Let O be the centre of the circle.

You can work out the area of triangle CDO. Then you'd need the area of the sector CBD which requires the angle COD (obtuse). I cannot see a non trig way of getting that with 68 as a measurement. With another length it might be possible.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 129

If you would like to avoid trigonometry you can use Maclaurin series of the inverse trigonometric functions. So if we let D=r-d we would have:

and

This will give you the answer correct to the third decimal place (if you want more accuracy you can include more terms of the related Maclurin series). Therefore, in this case, the previous formulae will become:

and

So there is no trigonometry at all and I think the 9 graders can do the arithmetic operations.

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