Coffee is draining from a conical filter into a cylindrical

We know that the volume of a cone of tadius r and height h is:
${\displaystyle V=\frac{1}{3}\pi r^{2}h}$
We know that the volume of cylinder of radius r and height h is:
${\displaystyle V=\pi r^{2}h}$
From the exercise, we know that the rate of change of both of volume of the filter and the volume of coffee pot is.
${\displaystyle \frac{dV}{dt}=10\frac{i{n}^3}{min}}$
We can also see that the diameter and the height of the filter is 6, which means h=2r (for the cone).
The radius of the cylinder is always r=3
a. The radius of the pot is r=3. We want to find ${\displaystyle \frac{dh}{dt}}$
${\displaystyle V=\pi r^{2}h=\pi \cdot 9\cdot hmid\frac{d}{dt}}$
${\displaystyle \frac{dV}{dt}=9\pi \cdot \frac{dh}{dt}}$
${\displaystyle \frac{dh}{dt}=\frac{dt}{9\pi }=\frac{10}{9\pi }\frac{in}{min}}$
b. The height of the full cone is h=5 in, which also means that r = 2.5 in.
We want to find ${\displaystyle \frac{dh}{dt}}$
$$V=\frac{1}{3}\pi r^{2}h=\frac{\pi }{3}(\begin{array}{c}\frac{h}{2}\end{array}{)}^2\cdot h=\frac{\pi }{12}{h}^3\mid \frac{d}{dt}$$$$V=\frac{1}{3}\pi r^{2}h=\frac{\pi }{3}(\begin{array}{c}\frac{h}{2}\end{array}{)}^2\cdot h=\frac{\pi }{12}{h}^3\mid \frac{d}{dt}$$
${\displaystyle \frac{dV}{dt}=\frac{\pi }{12}\cdot 3h^2\cdot \frac{dh}{dt}=\frac{\pi }{4}\cdot h^2\cdot \frac{dh}{dt}}$
${\displaystyle -10=\frac{\pi }{12}\cdot 3h^2\cdot \frac{dh}{dt}=\frac{\pi }{4}\cdot h^2\cdot \frac{dh}{dt}}$
${\displaystyle \frac{dh}{dt}=-\frac{10}{\frac{25\pi }{4}}=-\frac{8}{5\pi }\frac{in}{min}}$
(The sign of the rate of change of volume is negative, because the volume is decreasing)

(a) How fast is the level in the coffee pot rissing when the coffee in the cone is 5 in
deep?
The volume of a cylinder is ${\displaystyle V=\pi r^{2}h}$, where r is the radius of thee cylinder and h is the height. In
a cylinder, the radius is a constant (in this case 3), so we have
${\displaystyle V_{cyl}=\pi {\left(3\right)}^{2}h}$
${\displaystyle =9\pi h}$
${\displaystyle \frac{d{V}_{cyl}}{dt}=9\pi \frac{dh}{dt}}$
The rate of change of volume going into the cylinder is the same as the rate of change of volume of
coffee leaving the cone (since there is nowhere else for it to go). So ${\displaystyle \frac{dV}{dt}=10}$ for the cylinder. Thus
we have
${\displaystyle \frac{d{V}_{cyl}}{dt}=9\pi \frac{dh}{dt}}$
${\displaystyle 10=9\pi \frac{dh}{dt}}$
${\displaystyle \frac{dh}{dt}=\frac{10}{9\pi }}$
(b) How fast is the level in the cone falling at the same instant?
For the cone, if we let y equal the current height of the coffee in the cone and x equal the radius
of the cone at the top of the coffee in the filter, the current volume of coffee ${\displaystyle V_{cone}=\frac{1}{3}\pi x^{2}y}$. Since
the radius of the base of the cone is 3 inches and the height of the cone is 6 inches, we know from
smilar triangles that ${\displaystyle \frac{x}{y}=\frac{3}{6}=\frac{1}{2}}$, or ${\displaystyle x=\frac{1}{2}y}$. Differentiating this formula implicitly also gives us
${\displaystyle \frac{dx}{dt}=\frac{1}{2}\frac{dy}{dt}}$ . Therefore, we have:
${\displaystyle V_{cone}=\frac{1}{3}\pi x^{2}y}$
${\displaystyle =\frac{1}{3}\pi {\left(\frac{1}{2}y\right)}^{2}y}$
${\displaystyle =\frac{1}{12}\pi y^3}$
${\displaystyle \frac{d{V}_{cone}}{dt}=\frac{3}{12}\pi y^2\frac{dy}{dt}}$
${\displaystyle =\frac{1}{4}\pi y^2\frac{dy}{dt}}$
At the given instant, we then have
${\displaystyle -10=\frac{1}{4}\pi {\left(5\right)}^2\frac{dy}{dt}}$
${\displaystyle \frac{-10\left(4\right)}{25\pi }=\frac{dy}{dt}}$