Math Help - Related Rates 3

1. Related Rates 3

3) Coffee is draining from a conical filter nto a cylindrical coffee pot at the rate of 10 in^3/min.
a) How fast is the level in the pot rising when the coffee in the cone is 5 inches deep?
b) How hast 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]

Huge thanks in advance to whoever helps

2. So there are two figures to analyze here.

The cylindrical pot:
---radius, r1 = 6/2 = 3 in.
---dV2 /dt = 10 cu.in/min

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

Whatever is the height of the coffee in the cone, the volume of coffee pouring into the pot is 10 cu.in/min always.

V2 = pi(r2)^2 *h2

r2 does not change. It is constant at 3 inches, so,
V2 = pi(3^2)h2 = (9pi)h2
So,
dV2 /dt = (9pi)(dh2/dt)
10 = (9pi)(dh2/dt)
dh2/dt = 10/(9pi) in/sec

Therefore, the level of the coffee in the pot is rising at the rate of 10/(9pi) ft/min. ------------------answer.

The conical filter.
---total height = 6 in.
---radius at top = 6/2 = 3 in.
---dV1 /dt = -10 cu.in/min

V1 = (1/3)[pi(r1)^2]h1 = (pi/3)(r1)^2 *h1 -----(i)

Since we are interested on the height of the coffee only, we will express r1 in terms of h1
Imagine, or draw the figure on paper. It is an inverted isosceles triangle that is 6in at the top, 0 at the bottom, and 6in high.
Draw the level of the coffee inside. Another inverted isosceles triangle is formed, whose top is 2r1 wide, 0 at the bottom, and h1 high.
By proportion,
top/height: 6/6 = 2r1/h1
Cross multiply,
6*h1 = 6*2r1
h1 = 2r1
r1 = h1 /2

Plug that into (i),
V1 = (pi/3)(h1 /2)^2 *h1
V1 = (pi/12)(h1)^3

dV1 /dt = (pi/12)[3(h1)^2 dh1/dt]
dV1/dt = (pi/4)(h1)^2 dh1/dt
Plug in the givens,
-10 = (pi/4)(5^2) dh1/dt
-10 = 25pi/4 dh1/dt
dh1/dt = -10 / (25pi/4) = -40/(25pi) = -8/(5pi) in/min

Therefore, at that instant, the depth of the coffee in the cone is decreasing at the rate of 8/(5pi) in/min. --------------answer.

--------------------------------------------------------------
Uh, I may be wrong in my understanding of
Coffee is draining from a conical filter nto a cylindrical coffee pot at the rate of 10 in^3/min.

My understanding is the rate is always 10 cu.in/min always, whatever is the depth of the coffee in the conical filter.

If the 10 cu.in/min is only when the conical filter is full, then my answers above are wrong.

3. 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
:          \|/            |:::::::|:::::::|
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: - 3 - : - 3 - :
(a) The pot has radius of 3 inches.
The coffee in the pot has radius 3 and height $y$.
The volume of the coffee in the pot is: . $V \:=\:\pi r^2h$
Hence, the volume of coffee is: . $V \:=\:\pi(3^2)y \:=\:9\pi y$
Differentiate with respect to time: . $\frac{dV}{dt} \:=\:9\pi\left(\frac{dy}{dt}\right)$
We are told that: $\frac{dV}{dt} = 10$ in³/min.
So we have: . $10 \:=\:9\pi\left(\frac{dy}{dt}\right)\quad\Rightarro w\quad\frac{dy}{dt} \:=\:\frac{10}{9\pi}$

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

(b) The cone has radius 3 and height 6.
The coffee in the cone has radius $r$ and height $h$.
The volume of the coffee is: . $V \:=\:\frac{\pi}{3}r^2h$ .[1]

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

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

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

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

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

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

4. Ah, I start to see it now. The Conical part was a bit confusing; I tried the problem and actually got part A! Part B makes sense now.

Thanks so much for hte help guys!