Solving a cubic equation in exact terms which is the key to solving the question in picture below

Question

I have arrived at the formula below, I need help with solving this cubic equation for h please.

Answer 1

Since at a height $z \ge 0$ the area of a horizontal section of the vase is equal to:

$$ A(z) = \left(6+2\,z\tan(30°)\right)^2 $$

at time $t \ge 0$ the volume of water in the vase is equal to:

$$ V(t) = \int_0^{h(t)} A(z)\,\text{d}z = \frac{4}{9}\,h^3(t) + 4\sqrt{3}\,h^2(t) + 36\,h(t) $$

from which:

$$ V'(t) = \frac{4}{3}\,h^2(t)\,h'(t) + 8\sqrt{3}\,h(t)\,h'(t) + 36\,h'(t) $$

i.e.:

$$ h'(t) = \frac{3\,V'(t)}{4\left(3\sqrt{3}+h(t)\right)^2} = \frac{9}{\left(3\sqrt{3}+h(t)\right)^2}\,. $$

To calculate $h'(3)$ we need to know $h(3)$ which can be determined by solving the equation:

$$ V(3) = V'(3)\cdot 3 = 36 \quad \quad \Rightarrow \quad \quad \frac{4}{9}\,h^3(3) + 4\sqrt{3}\,h^2(3) + 36\,h(3) = 36 $$

that is, setting for simplicity $h(3) \equiv x$, we have to solve the cubic equation:

$$ x^3 + 9\sqrt{3}\,x^2 + 81\,x - 81 = 0. $$

Although the roots of polynomials up to the fourth degree can always be expressed by radicals, at the application level it makes more sense to immediately determine a numerical approximation.

In particular, being:

$$ f(x) = x^3 + 9\sqrt{3}\,x^2 + 81\,x - 81 \quad \quad \Rightarrow \quad \quad f'(x) = 3\,x^2 + 18\,\sqrt{3}\,x + 81 $$

if $x_0 = 1$ is reasonably the root of the first attempt, for the Newton-Raphson method we have:

$$ \begin{aligned} & x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \approx 0.855974\,; \\ & x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \approx 0.852492\,; \\ & x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} \approx \underbrace{0.852490}_{h(3)}\,; \\ \end{aligned} $$

and here we stop, as at least the first three decimal digits have stabilized, as required!

In conclusion, we have:

$$ h'(3) = \frac{9}{\left(3\sqrt{3}+h(3)\right)^2} \approx \boxed{0.246\,cm/s}. $$

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ADD BONUS - Wanting to avoid numerical approximations at all costs:

$$ x^3 + 9\sqrt{3}\,x^2 + 81\,x - 81 = 0 $$

the general idea is to reduce to a depressed cubic equation of the quadratic term:

$$ x \mapsto y - \frac{9\sqrt{3}}{3} $$

that is, by substitution and simplifying we arrive at the equation:

$$ y^3 - 81\sqrt{3} - 81 = 0 $$

where, unexpectedly, the linear term has also disappeared; very lucky! ^_^

It's thanks to this stroke of luck that we can do without complex numbers, in fact:

$$ y = 3\sqrt[3]{3\left(1+\sqrt{3}\right)} \quad \quad \Rightarrow \quad \quad x = 3\sqrt[3]{3\left(1+\sqrt{3}\right)}-3\sqrt{3} $$

so the exact answer to the question is:

$$ h'(3) = \frac{9}{\left(3\sqrt{3}+h(3)\right)^2} = \boxed{\frac{1}{\sqrt[3]{\left(3+3\sqrt{3}\right)^2}} \approx 0.246\,cm/s}. $$

Obviously this happens because wanted by the author, otherwise not even in dreams! ^_*