Great Houdini's Escape

[th1nkp1nk]

Harry Houdini's favorite magic act is to have himself tied up with rope, lowered into a giant flask, and have poured into the flask at a constant rate. Harry is 6 feet tall and we can assume he is a perfect cylinder with a radius of 6 inches. The flask is 8 feet tall and has a radius of 4 feet at the bottom and a radius of 2 feet at the top. The large radius extends straight up for 2 feet before slanting in, and theneck of the flask (small radius) is also 2 feet.

Note that when the height of the water was 3 feet high, the water was rising at a rate of .058 ft/sec.

V(h) = 63 pi/4 h if 0<=h<=2
V(h) = pi/12 h^3 - 5pi/2 h^2+ 99pi/4 h -26pi/3 if 2 <= h <= 6
V(h) = 4pi h+ 263pi/6 if 6 <=h<=8

------------

We're supposed to verify these volume formulas - I got the 1st and 3rd one, but can't figure out how to do the middle one. My teacher told us to use solids of revolution, using 2 as the lower bound and h as the upper bound, but my answer didn't equal the volume equation.

 

[TchrWill]

Harry Houdini's favorite magic act is to have himself tied up with rope, lowered into a giant flask, and have poured into the flask at a constant rate. Harry is 6 feet tall and we can assume he is a perfect cylinder with a radius of 6 inches. The flask is 8 feet tall and has a radius of 4 feet at the bottom and a radius of 2 feet at the top. The large radius extends straight up for 2 feet before slanting in, and theneck of the flask (small radius) is also 2 feet.

Note that when the height of the water was 3 feet high, the water was rising at a rate of .058 ft/sec.

V(h) = 63 pi/4 h if 0<=h<=2
V(h) = pi/12 h^3 - 5pi/2 h^2+ 99pi/4 h -26pi/3 if 2 <= h <= 6
V(h) = 4pi h+ 263pi/6 if 6 <=h<=8

------------

We're supposed to verify these volume formulas - I got the 1st and 3rd one, but can't figure out how to do the middle one. My teacher told us to use solids of revolution, using 2 as the lower bound and h as the upper bound, but my answer didn't equal the volume equation.

The contained volume at any h < 2 is V = Pi(r^2)h = 50.265h.

The contained volume at any h >2 but <6 is as follows.
The volume of a right circular cone is V = Pih'(r^2 + rr' + r'^2)/3 where r = the radius at the bottom of the cone, 4 ft, and r' = the radius at the top of the conical frustrum, 2 ft. and h' = the height from the bottom of the conical section.
The slope of the conical side is arctan(2/4) = µ = 26.565º the tangent being 2/4.
The radius at any given h' is given by r' = (4 - h')tan µ = (4 - h')(2/4) = (4 - h')/2
Therefore, the volume of any conical portion above 2 feet in flask height is given by Vcon = Pih'[r^2 + r(4 - h')/2 + ((4 - h)/2)^2]

The volume of the upper cylindrical portion f the flask is given by Vu = Pi(2^2)h" where h" = the height from the top of the conical frustrum top to the height in question.

I;ll let you add them together and simplify to derive the three equations for volume within each height constraint.

 

[galactus]

I think I found a nice solution. The cylinder being lowered into the flask has volume \(\displaystyle \L\\{\pi}(\frac{1}{2})^{2}h=\frac{1}{4}{\pi}h\)

The volume of the lower portion form 0 to 2 is \(\displaystyle \L\\32{\pi}\)

Now, the trickiest part is the portion from 2 to 6, where the shoulders are slanted.

Find the equation of the line which makes up a shoulder. It passes through
(4,2) and (2,6). This gives a line equation of \(\displaystyle \L\\y=-2x+10\)
Solve for x and get \(\displaystyle \L\\x=5-\frac{y}{2}\)

Make the solid of revolution as such:

\(\displaystyle \L\\{\pi}\int_{2}^{h}(5-\frac{y}{2})^{2}dy=\frac{\pi}{12}h^{3}-\frac{5\pi}{2}h^{2}+25{\pi}h-\frac{122{\pi}}{3}\)

Now, add the bottom portion and subtract the cylinder submerged into the flask:

\(\displaystyle \L\\\frac{\pi}{12}h^{3}-\frac{5\pi}{2}h^{2}+25{\pi}h-\frac{122{\pi}}{3}+32{\pi}-\frac{1}{4}{\pi}h=\fbox{\frac{\pi}{12}h^{3}-\frac{5\pi}{2}h^{2}+\frac{99\pi}{4}h-\frac{26\pi}{3}}\).....there she is!.

 

[morson]

Nice work galactus, elegant and concise. I initially thought the shoulders were not straight lines (curves) so I was having a bit of trouble.