Flow Rate Problem

I don't know if my solution is right or I just luckily arrived at the correct answer. hahahahahaha

$V = \frac{1}{3} \pi r^2 h$

$r = \frac{4}{10} h$ or $\frac{2}{5} h$

$V = \frac{1}{3} \pi (\frac{2}{5} h)^2 h$

$V = \frac{4}{75} \pi h^3$

$\frac{dV}{dt} = 2.15$

$V = 2.15t$

$2.15t = \frac{4}{75} \pi h^3$

$h = (\frac{161.25}{4\pi} t)^\frac{1}{3}$

$\frac{dh}{dt} = \frac{\frac{161.25}{4\pi}}{3 \sqrt[3]{((\frac{161.25}{4\pi} t)^2)} }$

$t = \frac{4\pi}{161.25} h^3$

$\frac{dh}{dt} = \frac{\frac{161.25}{4\pi}}{3 \sqrt[3]{((\frac{161.25}{4\pi} (\frac{4\pi}{161.25} h^3))^2)} }$

$\frac{dh}{dt} = \frac{161.25}{12\pi h^2}$

$\frac{161.25}{12\pi (3.5^2)} = 0.349166457... \approx \boxed{0.3492}$

The volume of water in the conical tank is given by:

$\begin{aligned} V & = \frac {\pi r^2 h}3 & \small \color{#3D99F6} \text{Note that } \frac rh = \frac 4{10} \implies r = 0.4h \\ & = \frac {0.16\pi h^3}3 \\ \color{#3D99F6}\frac {dV}{dt} & = 0.16\pi h^2 \cdot \frac {dh}{dt} & \small \color{#3D99F6} \text{Note that } \frac {dV}{dt} = 2.15 \\ \implies \frac {dh}{dt} & = \frac {\color{#3D99F6}2.15}{0.16\pi h^2} \\ \frac {dh}{dt}\bigg|_{h=3.5} & = \frac {2.15}{0.16\pi \cdot 3.5^2} \\ & \approx \boxed {0.3492} \end{aligned}$