# Math Help - Rate Of Change of Cone

1. ## Rate Of Change of Cone

A cone-shaped coffee filter of radius 6cm and depth 10cm contains water, which drips out through a hole at the bottom at a constant rate of 1.5cm^3 per second.

a) If the filter starts out full, how long does it take to empty?

v = (1/3)(pi)(r^2)h - 1.5t
0 = (1/3)(pi)(6^2)(10) - 1.5t
0 = 376.99 - 1.5t
376.99 = 1.5t
t = 251.33 sec

b) Find the volume of water in the filter when the depth of the water is h cm.

v = (1/3)(pi)(r^2)h - 1.5t
v + 1.5t = (1/3)(pi)(r^2)h
h = (v + 1.5t) / ((1/3)(pi)r^2)

c) How fast is the water level falling when the depth is 8cm?

8 = (v + 1.5t) / ((1/3)(pi)r^2)

How do I solve this with so many unknown variables (volume, time and radius)? My suspicion is that I did part (b) incorrectly. Can anyone please show me where I went wrong? Thanks!

2. I apologize, I really hate to bounce but the problem seems pretty basic and I would really like to get it figured out so I can get some sleep. I'm still trying to figure it out but can't get anywhere. Any help would be greatly appreciated! Thanks!

3. (a)

$Rate=1.5cm^{3}/s$
$Volume=\frac{1}{3}\pi(6)^{2}(10)=1200\pi$
So $Time\;taken=\frac{1200\pi}{1.5}=2513.27s$

(b)

By similar triangles, $\frac{h}{10}=\frac{r}{6}\Rightarrow r=\frac{3}{5}h$
So $\displaystyle V=\frac{1}{3}\pi\left(\frac{3}{5}h\right)^{2}h=\fr ac{3}{25}\pi h^{3}$

(c)

By Chain Rule, $\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}$
Given $\frac{dV}{dt}=-1.5cm^{3}/s$
Differentiate V with restpect to h, $\frac{dV}{dh}=\frac{9}{25}\pi h^{2}$
At $h=8$, $\frac{dV}{dh}=\frac{9}{25}\pi (9)^{2}=\frac{729}{25}\pi$
Thus, $\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}\Rig htarrow -1.5=\frac{729}{25}\pi \times\frac{dh}{dt}\Rightarrow\frac{dh}{dt}=?$

4. Originally Posted by acc100jt
(a)

$Rate=1.5cm^{3}/s$
$Volume=\frac{1}{3}\pi(6)^{2}(10)=1200\pi$
So $Time\;taken=\frac{1200\pi}{1.5}=2513.27s$
I do believe that (1/3)(6^2)(10) = 120, not 1200. Which means that t=251.33s

Originally Posted by acc100jt
(b)

By similar triangles, $\frac{h}{10}=\frac{r}{6}\Rightarrow r=\frac{3}{5}h$
So $\displaystyle V=\frac{1}{3}\pi\left(\frac{3}{5}h\right)^{2}h=\fr ac{3}{25}\pi h^{3}$
I've never learned this 'similar triangles' thing, but I see that you're just finding the ratio between r and h of the cone so we can express r in terms of h. Sounds good, this is what I was missing.

Originally Posted by acc100jt
(c)

By Chain Rule, $\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}$
Given $\frac{dV}{dt}=-1.5cm^{3}/s$
Differentiate V with restpect to h, $\frac{dV}{dh}=\frac{9}{25}\pi h^{2}$
At $h=8$, $\frac{dV}{dh}=\frac{9}{25}\pi (9)^{2}=\frac{729}{25}\pi$
Thus, $\frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}\Rig htarrow -1.5=\frac{729}{25}\pi \times\frac{dh}{dt}\Rightarrow\frac{dh}{dt}=?$
The height of the water should be 8cm, but this was also very helpful. I got -0.0164 = dh/dt for yours (using h = 9cm), and I got dh/dt = -0.021 (using h = 8cm).

Thanks for all your help!