Volume Question (1 Viewer)

[Shuuya]

Mathematics HSC 2010 Question 10b) (ii)

Could someone please explain how to approach these questions?

I'm not sure if this is actually hard to visualise or I'm just being dumb, because the sample answers solved it pretty easily xD

Thanks!

 

[kashkow]

Oh I see now.

It's actually measuring the height of the water, and not the volume as I, and perhaps you may have first thought.

The height of the water is initially r (radius of hemisphere) and you need to decrease it to half it's height when tilted. I think this question is badly designed/worded since "one half of it's original depth" may refer to when the bowl is put back horizontally but the answers show that you're measuring the water when it's tilted. I guess it makes some sense since it is asking for the angle to make the depth of water half the height r, implying it has to be tilted when the height is measured. Maybe this is just me. (Although now thinking about it, it doesn't really matter since the water will stay at half however the bowl is tilted - however tilting it helps form a triangle to solve the question easier).

Anyway so half the depth means the water is at the level of r/2 and from the center of the hemisphere perpendicular to the water (important to note) it is also a distance of r/2 SINCE the point from the center to anywhere on the surface will be a distance of r and no matter which way the bowl is, the horizontal water level will stay at r/2. However you've need to measure the heights perpendicular to the surface of the water or the water level ratio to the original height won't be the same (this is sorta self-evident). Then you can probably solve it from there yourself with guidance of the sample answers.

 

[pikachu975]

Lol the question is misleading because they put the theta in the wrong spot... it's meant to be to the horizontal (check the solutions you'll see the diagram has it in the wrong spot)

 

[InteGrand]

For the convenience of those who want to see the Board of Studies' solutions, here is the link: https://www.boardofstudies.nsw.edu....ams/pdf_doc/mathematics-sample-answers-10.pdf .

The solution for that question is on the last page (which is page 19).

 

[kashkow]

Lol the question is misleading because they put the theta in the wrong spot... it's meant to be to the horizontal (check the solutions you'll see the diagram has it in the wrong spot)

No, don't think so? In both it's between water level (horizontal line) and upper face of the hemisphere, that's how I see it...

 

[pikachu975]

No, don't think so? In both it's between water level (horizontal line) and upper face of the hemisphere, that's how I see it...

The red circle is what I thought the original diagram had theta as

 

[trecex1]

View attachment 33482

The red circle is what I thought the original diagram had theta as

The original diagram doesn't have l that vertical line.

 

[pikachu975]

The original diagram doesn't have l that vertical line.

I don't get it because the original diagram has theta in the middle or something but the answers has it in a corner

 

[trecex1]

I don't get it because the original diagram has theta in the middle or something but the answers has it in a corner

Exact same thing...couldn't be clearer which angle they mean.

 

[jathu123]

The arc kinda kinda shows which angle it is for me

 

[pikachu975]

I think I'm retarded or something lol I don't see how that angle at all relates to the angle the answers use, don't worry about it

 

[InteGrand]

Both thetas appear to be measuring the angle between the tilted diameter of the hemisphere and the horizontal.

 

[Drongoski]

This was a very clever and tricky question and I suspect at least 60% of those who attempted it got it wrong. (So Shuuya, no, you are not dumb; question is not easy) I myself was baffled when I first encountered it a few years ago. I have not looked at solutions by others, but having got another look at the question, it is not that difficult if you have the right diagrams (which, damn, I am too bumb to put up). I'll describe the diagram and one of you who is good at using software to draw the diagram can later help put it up.

This is how we can draw the diagram:

i) draw a circle, radius 'r'
ii) thru the centre, O, draw the horizontal diameter, AB, say, with B on the right
iii) draw chord, CB, to B, under this diameter, making an angle @ at B
iv) now draw another diameter, PQ, parallel to CB (this step not essential, but helps you visualise)
v) draw the radius, OE, perpendicular to PQ, cutting CB at D and meeting the circle at E

Then OD = 0.5r and OB = r so that sin@ = 0.5r/r = 1/2 ==> @ = 30 degrees


For the 2nd part, find the volume, using the integral:

 

[pikachu975]

Both thetas appear to be measuring the angle between the tilted diameter of the hemisphere and the horizontal.

Yeah but doesn't it look like it's between that curved line and the new diameter?

 

[trecex1]

Yeah but doesn't it look like it's between that curved line and the new diameter?

No. It does not.

 

[pikachu975]

I think I'll just pray this isn't in the 2016 HSC lol... I just don't see how that is showing the inclination of the bowl when the theta is next to the surface

Edit: Nvm I see it, but they should at least put an arrow pointing to where the theta actually is

 

[Drongoski]

I think I'll just pray this isn't in the 2016 HSC lol... I just don't see how that is showing the inclination of the bowl when the theta is next to the surface

Edit: Nvm I see it, but they should at least put an arrow pointing to where the theta actually is

I can guarantee it will not appear in this year's 2U paper.

 

[kashkow]

LOL pikachu, you need a tutor.. Jks

This was a very clever and tricky question and I suspect at least 60% of those who attempted it got it wrong. (So Shuuya, no, you are not dumb; question is not easy) I myself was baffled when I first encountered it a few years ago. I have not looked at solutions by others, but having got another look at the question, it is not that difficult if you have the right diagrams (which, damn, I am too bumb to put up). I'll describe the diagram and one of you who is good at using software to draw the diagram can later help put it up.

This is how we can draw the diagram:

i) draw a circle, radius 'r'
ii) thru the centre, O, draw the horizontal diameter, AB, say, with B on the right
iii) draw chord, CB, to B, under this diameter, making an angle @ at B
iv) now draw another diameter, PQ, parallel to CB (this step not essential, but helps you visualise)
v) draw the radius, ODE, perpendicular to PQ, cutting CB at D and meeting the circle at E

Then OD = 0.5r and OB = r so that sin@ = 1/2 ==> @ = 30 degrees


For the 2nd part, find the volume, using the integral:

The solutions from the Board of Studies are pretty simple for the first question, I'd check em out for an alternate soln. (see Integrand's link) It is basically the same as the diagram in the question but simplified. Your method probably works but is just a bit longer and may not be figured out by all 2U students. But thanks anyway

 

[pikachu975]

LOL pikachu, you need a tutor.. Jks




The solutions from the Board of Studies are pretty simple for the first question, I'd check em out for an alternate soln. (see Integrand's link) It is basically the same as the diagram in the question but simplified. Your method probably works but is just a bit longer and may not be figured out by all 2U students. But thanks anyway

Tutoring for 2 unit xD
Nah don't really need one tbh

 

[Drongoski]

LOL pikachu, you need a tutor.. J

The solutions from the Board of Studies are pretty simple for the first question, I'd check em out for an alternate soln. (see Integrand's link) It is basically the same as the diagram in the question but simplified. Your method probably works but is just a bit longer and may not be figured out by all 2U students. But thanks anyway

Had a quick check of the Board's solution. Mine is just the same thing. Would have helped if I displayed my diagram.