My Math Forum water in a cup

 Calculus Calculus Math Forum

 March 8th, 2010, 12:11 PM #1 Senior Member   Joined: Dec 2006 Posts: 167 Thanks: 3 water in a cup A cup ( as a frustum of a cone), with top radius R, bottom radius r and height H, filled full of water initially. Find the remaining water volume function $v(\theta)$ where $\theta$ is the inclination of its axis that initially was vertical. Ignore anything make the problem more difficult like surface tension of the water etc. Assume no differences of the dimensions of the cup measured inside or outside.
 March 9th, 2010, 02:46 PM #2 Senior Member   Joined: Dec 2006 Posts: 167 Thanks: 3 Re: water in a cup I tried to attach a gif picture but for some reason it says "Could not upload attachment to ./files/..." Anyway I got a link for the picture below. To calculate the volumn of the cone above x-y plane, it'll involve a integral $\int \frac{1+ \cos (2\beta+2\theta)}{(\cos 2\beta + \cos 2\theta)^{3/2}} d\theta$ that leads to ellipse integral, and is complicated to me.
 March 9th, 2010, 04:40 PM #3 Senior Member   Joined: Mar 2007 Posts: 428 Thanks: 0 Re: water in a cup Just surmising, as it's been far too long since I did any of this, but if you project the ellipse onto a plane perpendicular to the axis of the cone, you'd have a circle again for the liquid surface. [I.E. straighten out the cone.] Have that plane pass through the center of the ellipse. The volume of liquid would be the same. I'd think that symmetry about the center of the surface would account for the same amount lost as that gained in the tipping. That is, you can subtract the volume of the straight cone of liquid now from the full volume to get the volume spilled. Perhaps not well described, but I know what I mean.
 March 10th, 2010, 10:48 AM #4 Senior Member   Joined: Dec 2006 Posts: 167 Thanks: 3 Re: water in a cup My calculation like this (see the figure above for notations): $h/r=\frac{h+H}{R}, \quad h=\frac{rH}{R-r}$ $H_{0}=h+H=\frac{RH}{R-r}, \quad L=\sqrt{R^{2}+H_{0}^{2}}$ $d=-L\sin \gamma =-L\sin \left( \pi /2-\theta -\beta \right) =-L\cos \left(\theta +\beta \right)$ The axis of the cone $\mathbf{k}=\left( 0,\sin \theta ,\cos \theta \right)$ cone equation $\mathbf{k}\cdot \left( x,y,z-d\right)=\left\vert \left( x,y,z-d\right) \right\vert \cos \beta$ ellipse equation on x-y plan: $\mathbf{k}\cdot \left( x,y,-d\right)=\left\vert \left( x,y,-d\right) \right\vert \cos \beta$ $y\sin \theta -d\cos \theta=\left( \cos \beta \right) \sqrt{x^{2}+y^{2}+d^{2}}$ $d^{2}\cos ^{2}\theta -2dy\cos \theta \sin \theta +y^{2}\sin ^{2}\theta =\left( x^{2}+y^{2}+d^{2}\right) \cos ^{2}\beta$ $\left( \cos ^{2}\theta -\cos ^{2}\beta \right) d^{2}-2dy\cos \theta \sin \theta +y^{2}\sin ^{2}\theta =\left( x^{2}+y^{2}\right) \cos ^{2}\beta$ $x^{2}\cos ^{2}\beta +y^{2}\left( \cos ^{2}\beta -\sin ^{2}\theta \right) +2dy\cos \theta \sin \theta =\left( \cos ^{2}\theta -\cos ^{2}\beta \right) d^{2}$ [color=#FF4000](*)[/color] $\quad x^{2}\cos ^{2}\beta +\left( \cos ^{2}\beta -\sin ^{2}\theta \right) \left( y+\frac{d\cos \theta \sin \theta }{\cos ^{2}\beta -\sin ^{2}\theta }\right)^{2}= \frac{1}{4}\frac{\left( 1-\cos 4\beta \right) d^{2}}{\cos 2\theta +\cos 2\beta }$ the area of the ellipse $\frac{\frac{\pi }{4}\frac{\left( 1-\cos 4\beta \right) d^{2}}{\cos 2\theta +\cos 2\beta }}{\left( \cos \beta \right) \sqrt{\cos ^{2}\beta -\sin ^{2}\theta }}=\frac{\sqrt{2}\pi \left( 1-\cos 4\beta \right) L^{2}\cos ^{2}\left( \theta +\beta \right) }{4\left( \cos \beta \right) \left( \cos 2\theta +\cos 2\beta \right) ^{\frac{3}{2}}} = \left(\frac{\sqrt{2}\pi \left( 1-\cos 4\beta \right) L^{2}}{8 \cos \beta }\right)\frac{1+\cos \left( 2\theta +2\beta \right) }{\left( \cos 2\theta+\cos 2\beta \right) ^{\frac{3}{2}}}$ Form the ellipse equation I don't think your projection will produce a circle.
 March 24th, 2010, 11:26 AM #5 Senior Member   Joined: Dec 2006 Posts: 167 Thanks: 3 Re: water in a cup Actually this problem can be solved without involving ellipse integrals. Notice that any cone-kind volume equals height x bottom_area x (1/3):

 Tags cup, water

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post r-soy Physics 1 December 15th, 2013 02:54 PM Chikis Algebra 2 May 29th, 2013 09:02 PM Chikis Elementary Math 2 May 12th, 2013 02:44 PM Joe_65 Physics 0 January 23rd, 2013 04:21 AM johnny Physics 7 October 28th, 2007 03:29 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top