Hello, Super Mallow!

3) Coffee is draining from a conical filter nto a cylindrical coffee pot at the rate of 10 in³/min.

a) How fast is the level in the pot rising when the coffee in the cone is 5 inches deep?

b) How fast is the level in the cone falling then?

[A diagram on the page shows the height and diameter of the cone is 6 inches;

diameter of pot is also 6 inches] Code:

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6 \:::|:::/ |-------+-------|
: \::|h:/ |:::::::|:::::::|
: \:|:/ |:::::::|:::::::|y
: \|/ |:::::::|:::::::|
- * *-------+-------*
: - 3 - : - 3 - :

(a) The pot has radius of 3 inches.

The coffee in the pot has radius 3 and height $\displaystyle y$.

The volume of the coffee in the pot is: .$\displaystyle V \:=\:\pi r^2h$

Hence, the volume of coffee is: .$\displaystyle V \:=\:\pi(3^2)y \:=\:9\pi y$

Differentiate with respect to time: .$\displaystyle \frac{dV}{dt} \:=\:9\pi\left(\frac{dy}{dt}\right)$

We are told that: $\displaystyle \frac{dV}{dt} = 10$ in³/min.

So we have: .$\displaystyle 10 \:=\:9\pi\left(\frac{dy}{dt}\right)\quad\Rightarro w\quad\frac{dy}{dt} \:=\:\frac{10}{9\pi}$

The coffee in the pot is rising at a rate of about $\displaystyle \boxed{0.35\text{ in/min}}$

(b) The cone has radius 3 and height 6.

The coffee in the cone has radius $\displaystyle r$ and height $\displaystyle h$.

The volume of the coffee is: .$\displaystyle V \:=\:\frac{\pi}{3}r^2h $ .**[1]**

From the similar right triangles, we have: .$\displaystyle \frac{r}{h} \:=\:\frac{3}{6}\quad\Rightarrow\quad r \,=\,\frac{h}{2}$

Substitute into [1]: .$\displaystyle V \:=\:\frac{\pi}{3}\left(\frac{h}{2}\right)^2h \:=\:\frac{\pi}{12}h^3$

Differentiate with respect to time: .$\displaystyle \frac{dV}{dt} \:=\:\frac{\pi}{4}h^2\left(\frac{dh}{dt}\right)$ .**[2]**

We are given: .$\displaystyle h =5,\;\frac{dV}{dt} =-10$ in³/min

Substitute into [2]: .$\displaystyle -10 \:=\:\frac{\pi}{4}(5^2)\left(\frac{dh}{dt}\right)\ quad\Rightarrow\quad\frac{dh}{dt}\:=\:-\frac{8}{5\pi}$

The coffee in the cone is falling at about $\displaystyle \boxed{0.51\text{ in/min}}$