Tilted conical cup - new water depth

This is a follow-up of my solution of the previous problem . Refer to my previous solution to understand better the calculations in this solution.

First we find the volume of the half-filled water tumbler, $V_\frac 12 = \frac {2\sqrt 6\pi}3(6^3-2^3) \approx 97.47378709$ . Now we align the axis of the tumbler along the $x$ -axis with its bottom along the $z$ -axis. We note that the radius of the cross-section circle of the tumbler is given by $r(x) = mx + 2$ , where $m=\frac 1{\sqrt{24}}$ . The volume of the water $V(x)$ is the sum of the volume of a frustrum $V_1(x)$ and the volume of a "wedge" $V_2(x)$ . The volume of the frustrum is given by $V_1(x) = \frac {2\sqrt 6\pi}3(r_1(x)^3 - 2^3)$ , where $r_1(x)$ is the base radius of the frustrum.

For the volume of the wedge $V_2(x)$ , we can make use of the result of the previous problem. Using the farthest point of the wedge from the $z$ -axis as the reference for $x$ , at $x=4\sqrt 6$ or at the brim of the tumbler, the water poured out when the tumbler is tilted at $30^\circ$ is found to be $V_{\rm out}\approx 98.1614004197936$ . The width of the wedge $l(4\sqrt 6) = \frac {16\sqrt 6}{1+6\sqrt 2}$ . This means that $V_2(4\sqrt 6) = V_1(4\sqrt 6) - V_1(4\sqrt 6 - l(4\sqrt 6)) - V_{\rm out} \approx 68.81430502$ (volume of a frustrum with height $l$ minus the volume of water poured out). We note that the shape of the wedge does not change with $x$ , therefore $V_2(x)$ is directly proportional to $r(x)^3$ . Therefore we can find $V_2(x) = \left(\frac {r(x)}4\right)^3 V_2(4\sqrt 6)$ .

Using the $x$ -coordinate of the wedge's farthest end from the $z$ -axis, the base radius of the frustrum $r_1(x) = m(x-l(x))+2$ , where $l(x) = \frac {4r(x)\sqrt 6}{1+6\sqrt 2}$ . And the final formula is

$V(x) = \frac {2\sqrt 6\pi (x-l(x))^3 - 8)}3 + \frac {r(x)^3 V_2(4\sqrt 6)}{64}$

We have to find the value of $x$ , when $V(x) = V_\frac 12$ . As you can see the computation is very complex, I used a Microsoft Excel spreadsheet and manual iterations. The required $x_1 \approx 6.746320373$ and the required $d = \frac 5{2\sqrt 6} \left(\frac {2\sqrt 6}5 \times \frac {\sqrt 3}2 - \frac 15 \times \frac 12 \right)x_1 \approx 5.153941385$ . Therefore $\lfloor 10^3 d \rfloor = \boxed{5153}$ .

We are given that the bottom radius is $r_1 = 2$ and the top radius is $r_2 = 4$ and the slant height is $s = 10$ , hence the height $h = \sqrt{s^2 - (r_2 - r_1)^2 } = \sqrt{96}$

The radius of the water surface before tilting is $r_0 = \frac{1}{2} (r_1 + r_2) = 3$ , And from similar triangles, we deduce that height of the cone underneath the water surface is $\frac{3}{2} h$ .

Hence, the original volume of the whole cone underneath the water plane is

$V_1 = \frac{1}{3} \pi r_0^2 \frac{3}{2} h = \frac{1}{2} \pi r_0^2 h$

If we take the origin at the apex of the cone, then its equation is

$z = a \sqrt{x^2 + y^2} = a r$

since $r_1 = 2$ at the bottom and $r_2 = 4$ at the top and the difference in $z$ is $h =\sqrt{96}$ , then $\sqrt{96} = a (4 - 2)$ , so that $a = \sqrt{96}/2 = \sqrt{24}$

Next, the water surface in the tilted cup will cross the axis of the cup at an elevation of $z = z_0$ . This is shown in the attached image which shows

the cup after untilting. We need the find the volume under the blue line (plane). If we assume the x-axis is pointing towards the reader, then the y-axis is pointing to the right and the z-axis is pointing vertically up.

The equation of the plane of the surface of the water is $z = \tan(\dfrac{\pi}{6}) y + z_0$ where $z_0$ is yet to be determined.

To find the volume of the cone under this plane, we'll find its intersection

with the plane. Let $t = \tan \dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}$ , then

$t y + z_0 = a \sqrt{x^2 + y^2}$

$t^2 y^2 + z_0^2 + 2 t y z_0 =a^2( x^2 + y^2)$

Re-arranging,

$y^2 (a^2 - t^2) - 2 t y z_0 + a^2 x^2 = z_0^2$

Completing the square,

$(a^2 - t^2) (y - t z_0 /(a^2 - t^2) )^2 + a^2 x^2 = z_0^2 + t^2 z_0^2/ (a^2 - t^2) = z_0^2 ( a^2 )/(a^2 - t^2)$

This can be written as

$x^2 / A^2 + (y - y_0)^2 / B^2 = 1$

where

$y_0 = t z_0 / (a^2 - t^2)$

$A = z_0 / (a^2 - t^2)^{1/2}$

$B = z_0 a / (a^2 - t^2)$

This is the equation of the projection of the elliptical surface of water onto the xy plane.

The area of this region is

$A_1 = \pi A B = \pi z_0^2 \dfrac{a}{ (a^2 - t^2)^{3/2} }$

Hence the area of the cone underneath the water surface is

$A_2 = \dfrac{A_1}{ \sin \theta } = 5 A_1$

And finally the volume is $V_2 = \frac{1}{3} d A_2 = \frac{1}{3} z_0 \sin \theta (5 A_1) = \frac{1}{3} \pi z_0^3 \dfrac{a} { (a^2 - t^2)^{3/2} }$

Setting $V_2$ equal to $V_1$ , gives us $z_0$ . Solving for $z_0$ , we get $z_0 = 14.59451952$

Hence the altitude of water at $r = r_1$ is $z^* = z_0 + t r - h = 15.74922006 - \sqrt{96} = 5.951261087$

And finally, the depth of water $= z^* \cos \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2} (5.951261087) = 5.153943286$

which makes the answer $\lfloor 5153.943286 \rfloor = \boxed{5153}$