Solving a Dipstick Problem for Cone and Cylinder Volumes

[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.