Solving a Dipstick Problem for Cone and Cylinder Volumes

[pat666]

oh yeah, don't know why I missed that.
so V_(fluid)=(\pi((h/H)*R)^2*h)/3 probably simplify down further.

Thanks

 

[Mentallic]

Theres every chance that I'm wrong

Haha fair enough

Since we've already failed a few times at understanding each other with much simpler diagrams, I'll have to draw up a picture for this more complicated one.

First of all, we can save ourselves a lot of time if we don't have to prove the area of the segment of a circle. The wiki article for it was already posted:

I found this on wikipedia... which might help http://en.wikipedia.org/wiki/Circular_segment"

If you don't understand it though just ask and I'll walk you through it.

Ok so we have that the area of the segment of the circle is A=\frac{r^2}{2}\left(\theta-\sin\theta\right)

But we don't know \theta so again we're going to have to do something like in the other question to find \theta in terms of other constants that we do know, mainly h (the height of the water level) and r (the radius of the circle).

[PLAIN]http://img17.imageshack.us/img17/1326/circleo.png

The angle \theta sweeps the entire brown angle in the circle picture, and therefore due to symmetry, the brown angle in the triangle picture is going to be \frac{\theta}{2}

Now you can easily get your relationship and plug it back into the equation. And by the way, \sin\left(\cos^{-1}\left(x\right)\right)=\sqrt{1-x^2}

 

[Mentallic]

V_(fluid)=(\pi((h/H)*R)^2*h)/3 probably simplify down further.

Sure, V=\frac{\pi h^3R}{3H^2}

By the way, you can click on my LaTeX to see how it is written, in case you're interested.

 

[pat666]

I get the circular segment part (to some degree), what I don't get is that the adjacent side of your triangle is labeled r-h, I can't see how that can be true?? wouldn't r=h and therefore your triangle be non existent or is this a misprint or an error on my part? Also just to confirm, we are simply trying to get theta in terms of the depth of the water?

P.S just to confirm the last post I made on the cone question is correct?
Thanks

 

[Mentallic]

what I don't get is that the adjacent side of your triangle is labeled r-h, I can't see how that can be true?? wouldn't r=h and therefore your triangle be non existent

Why does r need to equal h? r is the radius of the circle and h is the height of the water. First let h<r, so that the side of the triangle is r-h because it is what is left of the radius after taking away h. For r=h we have a degenerate case where the water fills up half the circle, so it would simply be A=\frac{\pi r^2}{2}. You can also take h> r and try find the formula for that, but it'll be the same as for h<r.

Also just to confirm, we are simply trying to get theta in terms of the depth of the water?

In terms of the depth of the water AND the radius of the circle. We can't do it in terms of the depth of the water alone. Imagine we had a circle radius 1 and the depth of the water was very close to it, about 0.99. The angle \theta will be very close to \pi. Now what if we kept the water depth the same, but made the circle very big now? Well the angle will get a lot smaller. So obviously \theta is dependent on more than just the depth of the water.

P.S just to confirm the last post I made on the cone question is correct?
Thanks

Well you tell me, what makes you unsure if it is correct or not?

 

[pat666]

I think it is correct but given the amount of trouble I've had getting the solution you can see why I am unsure. just the way the picture was drawn made me think r=h, which would be true if the water was at the centre. I get that \theta=2cos^-^1(r-h)/r
not sure if that's correct because you gave me some trig info that was a bit more complex.
also this will only work to the halfway point but I was thinking I would just do the reflection of the earlier dipstick points for points after the mid line.
thanks

 

[Mentallic]

I think it is correct but given the amount of trouble I've had getting the solution you can see why I am unsure.

Yes it's correct. I was just showing you that you can tidy up

V_(fluid)=(\pi((h/H)*R)^2*h)/3 probably simplify down further.

into

Sure, V=\frac{\pi h^3R}{3H^2}

just the way the picture was drawn made me think r=h,

Oh, yeah that would be my fault, sorry

which would be true if the water was at the centre.

The blue line was meant to be where the water level was at.

I get that \theta=2cos^-^1(r-h)/r

Yes that's right.

not sure if that's correct because you gave me some trig info that was a bit more complex.

My trig info was wrong, ignore it. I forgot about the 2 that was going to be in front of it. What I was meant to give you was
\sin\left(2\cos^{-1}\left(x\right)\right)=2x\sqrt{1-x^2}
It may look complex, but its purpose is simple. When you plug \theta into the area equation A=\frac{r^2}{2}\left(\theta-\sin\theta\right) you're going to be left with
edit: \sin\left(2\cos^{-1}\left(\frac{r-h}{r}\right)\right) which is where you can simplify this with the equality I gave above.
You already have the answer, but it's just if you wanted to simplify things a bit more.

Markers wouldn't give full marks if you left an answer as \sin\left(\sin^{-1}\left(x\right)\right) so I doubt they would give full marks if you left it as \sin\left(\cos^{-1}\left(x\right)\right) either.

also this will only work to the halfway point but I was thinking I would just do the reflection of the earlier dipstick points for points after the mid line.
thanks

I believe the same formula will work for the water level anywhere from 0 to 2r (the diameter of the circle) but I'll check to see.

 

[pat666]

Ok thanks a lot for all your help

 

[OmCheeto]

The cylinder on it's side problem reduces to y=x - sin(x) in it's simplest form.
If you can solve for x, then you can solve the problem.
There is a reason they put this problem in computer science textbooks and never in mathematics textbooks.

 

[Mentallic]

Ok thanks a lot for all your help

No worries

The cylinder on it's side problem reduces to y=x - sin(x) in it's simplest form.
If you can solve for x, then you can solve the problem.

Can you please elaborate?

There is a reason they put this problem in computer science textbooks and never in mathematics textbooks.

I've seen questions similar to this in maths books.

 

[pat666]

This question is for yr 11 math. My solution is I believe V=L*\sin\left(\cos^{-1}\left(2\cdot\frac{r-h}{r}\right)\right).

 

[Mentallic]

This question is for yr 11 math. My solution is I believe V=L*\sin\left(\cos^{-1}\left(2\cdot\frac{r-h}{r}\right)\right).

That's not right. The formula is A=\frac{r^2}{2}\left(\theta-\sin\theta\right) where \theta=2\cos^{-1}\left(\frac{r-h}{r}\right)

 

[pat666]

whoops I forgot the first theta, so V=L*(r^2/2(2arccos(r-h/r)-sin(2arccos(r-h/r))) pretty messy.

 

[OmCheeto]

Can you please elaborate?

Perhaps the question in the book was worded differently.
All I know is that if you have a 10 gallon tank, it is not possible to place integer gallon marks on the dipstick. Unless of course you are clever enough to solve for x.


And perhaps I should give some https://www.physicsforums.com/showthread.php?p=3042826#post3042826" regarding this particular problem before someone yells at me for playing mind games.