# Conical tank calc help

1. Mar 16, 2005

### punjabi_monster

I have a calculus question and was wondering if some could help me see what I am doing wrong in this question. Thank you

A conical tank with an altitude of 10m and whose base has a radius of 4m is mounted with its vertex down. The tank is full of water which is draining through the vertex at the rate of 5 m^3/min. how fast is the level of the water dropping when the radius is 3?

this is how i did it:

r/h=4/10
h=5pi/2

V=[(pi)(r^2)(h)]/3
V=[(pi)(r^2)(5r/2)]/3
V=[pi5r^3]/6
dv/dt=[15pi(r^2)]/6] * dr/dt
5=[(15pi(3^2)]/6] * dr/dt
dr/dt= (2pi)/9

2. Mar 16, 2005

### Data

Firstly, the answer is actually $$\frac{5}{9\pi}\frac{\mbox{m}}{\mbox{min}}$$, not $$\frac{5\pi}{9}\frac{\mbox{m}}{\mbox{min}}$$.

Secondly, are you sure $$\frac{d r}{d t}$$ is what the question is looking for? Re-read it.

Last edited: Mar 16, 2005
3. Mar 17, 2005

### erik05

Notice how the question asks for how fast the level of the water is dropping. This would mean a change in height with respect to time, not the radius. Since it's asking for the rate of change when the radius is 3 then:

$$r=\frac{3h}{10}$$
$$V=\frac{1}{3} \pi (\frac{3h}{10})^2 h$$
$$V= \frac {3}{100} \pi h^3$$
$$\frac {dV}{dt}= \frac{9}{100} \pi h^2 \frac{dh}{dt}$$

Solving for dh/dt should get you the answer $$\frac{5}{9\pi}\frac{\mbox{m}}{\mbox{min}}$$ as stated above by Data.

4. Mar 17, 2005

### Data

You actually need to use $$r = \frac{2h}{5}$$. But other than that you're completely right.

5. Mar 18, 2005