I'm a high school math teacher trying to stay engaged with the subject, and I've started wondering about this question:
"Given a cylindrical ice cream container with radius r and height h, determine the best way to scoop out a given volume s of ice cream in order to minimize the surface area that you expose."
If we can solve this problem, we can minimize freezer burn, which is inversely proportional to ice cream enjoyment. The impact could be huge.
My Question:
While investigating a certain scooping method that I believe to be optimal for many containers and scoop sizes, I eventually stumbled upon this equation as a solution to a relevant indefinite integral: $$ \frac{s}{h}-\frac{\pi r^2}{2}=r^2\arcsin \left( \frac{x}{r} \right) + x\sqrt{r^2-x^2 } $$ My goal is to solve this for x, but I'm really struggling. A bit of research has presented me with the idea that this is likely to be a "transcendental equation", which is likely to be unsolvable. But I'll be honest -- the idea of transcendental equations is new to me, and I don't exactly know where to go from here.
I'm wondering if you all can tell me whether this can be solved, and if so, how, and if not, why not?
I should mention that I already solved the integral with a Riemann sum, and that works perfectly fine for my purposes. But still... I'd like a perfect solution if there is one.
Thanks!
Assuming that the equation to be solved for $x$ is $$\frac{s}{h}-\frac{\pi r^2}{2}=r^2\arcsin \left( \frac{x}{r} \right) + x\sqrt{r^2-x^2}$$ let $$x=r t \qquad \text{and}\qquad k= \frac 1{r^2} \left(\frac{s}{h}-\frac{\pi r^2}{2}\right)$$ to make $$k=\arcsin(t)+t\sqrt{1-t^2}$$
Expanded as a series $$\arcsin(t)+t\sqrt{1-t^2}=2t-\sum_{n=1}^\infty \frac{|a_n|}{b_n} t^{2n+1}$$ where the $a_n$ and $b_n$ form sequences $A091154$ and $A143582$ in $OEIS$.
Using series reversion, we have $$t=\frac{k}{2}+\frac{k^3}{48}+\frac{13 k^5}{3840}+\frac{493 k^7}{645120}+\frac{37369 k^9}{185794560}+\frac{4732249 k^{11}}{81749606400}+$$ $$\frac{901188997 k^{13}}{51011754393600}+\frac{240798388357 k^{15}}{42849873690624000}+\frac{6611433892597 k^{17}}{3586204813492224000}+O\left(k^{19}\right)$$
Before trying to make simpler, check the results for e few values of $k$
$$\left( \begin{array}{ccc} k & \text{estimate} & \text{solution} \\ 0.1 & 0.050021 & 0.050021 \\ 0.2 & 0.100168 & 0.100168 \\ 0.3 & 0.150571 & 0.150571 \\ 0.4 & 0.201369 & 0.201369 \\ 0.5 & 0.252716 & 0.252716 \\ 0.6 & 0.304787 & 0.304787 \\ 0.7 & 0.357787 & 0.357787 \\ 0.8 & 0.411969 & 0.411969 \\ 0.9 & 0.467654 & 0.467654 \\ 1.0 & 0.525267 & 0.525268 \\ 1.1 & 0.585404 & 0.585410 \\ 1.2 & 0.648947 & 0.648986 \\ 1.3 & 0.717288 & 0.717509 \\ 1.4 & 0.792768 & 0.793961 \\ 1.5 & 0.879532 & 0.886571 \end{array} \right)$$
which seems to be more than decent up to $k \sim 1.4$.
Let $t_0$ to be the above estimate. Polish the root using a couple of Newton iterations;
Edit
Using Padé approximants, we have
$$\arcsin(t)+t\sqrt{1-t^2}=\frac{2 t \left(60-19 t^2\right)}{60-9 t^2}+O\left(t^7\right)$$
In fact, compared to the series, this is in an error of $\frac{29 t^7}{2800}$
Solving the cubic equation $$38 t^3-9 k t^2-120 t+60k=0$$ which shows three real roots but the one to be retained is $$t=\frac{1}{38} \left(3 k-2 \sqrt{9 k^2+1520} \cos \left(\frac{1}{3} \left(\cos ^{-1}\left(-\frac{3 k \left(12160-9 k^2\right)}{\left(9 k^2+1520\right)^{3/2}}\right)+\pi \right)\right)\right)$$
$$\left( \begin{array}{ccc} k & \text{estimate} & \text{solution} \\ 0.1 & 0.050021 & 0.050021 \\ 0.2 & 0.100168 & 0.100168 \\ 0.3 & 0.150571 & 0.150571 \\ 0.4 & 0.201369 & 0.201369 \\ 0.5 & 0.252716 & 0.252716 \\ 0.6 & 0.304785 & 0.304787 \\ 0.7 & 0.357783 & 0.357787 \\ 0.8 & 0.411956 & 0.411969 \\ 0.9 & 0.467620 & 0.467654 \\ 1.0 & 0.525185 & 0.525268 \\ 1.1 & 0.585212 & 0.585410 \\ 1.2 & 0.648520 & 0.648986 \\ 1.3 & 0.716387 & 0.717509 \\ 1.4 & 0.791046 & 0.793961 \\ 1.5 & 0.877156 & 0.886571 \end{array} \right)$$
Same quality but nicer formula.