volume of cone

• May 27th 2008, 12:51 AM
doktoro
volume of cone
1. the level of a leaking oil-storage tank drops by 0.1% of the height above a leak every hour. The leak(hole) starts at a height of 4.5m above the ground at 2 am. The tank is a large cylinder of diameter 20m and total height 10m, but is only three-quarters full when it starts leaking.
A spillage of more than 5000L must be notified to the environment protection Agency. By what time should the leak be fixed to avoid notification?

(I don't know how to set up the formula of oil decreasing)

Thank you
• May 27th 2008, 01:48 AM
earboth
Quote:

Originally Posted by doktoro
1. the level of a leaking oil-storage tank drops by 0.1% of the height above a leak every hour. The leak(hole) starts at a height of 4.5m above the ground at 2 am. The tank is a large cylinder of diameter 20m and total height 10m, but is only three-quarters full when it starts leaking.

2. A spillage of more than 5000L must be notified to the environment protection Agency. By what time should the leak be fixed to avoid notification?
...

The volume of a cylinder is calculated by:

$V_{cyl}=\pi \cdot r^2 \cdot h$

With your problem you have r = 10, h = 10. That means the maximum volume is:

$V_{max}= \pi \cdot (10\ m)^2 \cdot (10\ m)=1000 \pi \ m^3$

The actual volume is three quarter of the maxium volume:

$V_{actual} = \frac34 \cdot V_{max} = 750 \pi \ m^3$ that means the surface of the oil is 7.5 m above ground. So the height above the leak is 3 m:

$h(0)= 3\ m$
$h(1)=3 \cdot 0.9 \ m$
$h(2)=(3 \cdot 0.9 \ m)\cdot 0.9 = 3 \cdot 0.9^2 \ m$
...
$h(t)=3 \cdot 0.9^t \ m$

2. Calculate the volume which has already leaked from the tank:

$V_{leak}=\pi \cdot 10^2 \cdot (3 - h(t)) = \pi \cdot 10^2 \cdot (3 - 3 \cdot 0.9^t)$

Since 5000 L = 5 m³ you have to solve for t:

$5 = \pi \cdot 10^2 \cdot (3 - 3 \cdot 0.9^t)$

$5=300\pi - 300\pi \cdot 0.9^t$

$\frac{5-300 \pi}{-300 \pi} = 0.9^t$

Use logarithms to calculate t. I've got t ≈ 0.0505 h and that are nearly 3 min (This result seems a little funny to me :confused:

I assume that I've made a silly mistake with my calculations. So check!
• May 27th 2008, 09:27 AM
doktoro
Thank you
Quote:

Originally Posted by earboth
The volume of a cylinder is calculated by:

$V_{cyl}=\pi \cdot r^2 \cdot h$

With your problem you have r = 10, h = 10. That means the maximum volume is:

$V_{max}= \pi \cdot (10\ m)^2 \cdot (10\ m)=1000 \pi \ m^3$

The actual volume is three quarter of the maxium volume:

$V_{actual} = \frac34 \cdot V_{max} = 750 \pi \ m^3$ that means the surface of the oil is 7.5 m above ground. So the height above the leak is 3 m:

$h(0)= 3\ m$
$h(1)=3 \cdot 0.9 \ m$
$h(2)=(3 \cdot 0.9 \ m)\cdot 0.9 = 3 \cdot 0.9^2 \ m$
...
$h(t)=3 \cdot 0.9^t \ m$

2. Calculate the volume which has already leaked from the tank:

$V_{leak}=\pi \cdot 10^2 \cdot (3 - h(t)) = \pi \cdot 10^2 \cdot (3 - 3 \cdot 0.9^t)$

Since 5000 L = 5 m³ you have to solve for t:

$5 = \pi \cdot 10^2 \cdot (3 - 3 \cdot 0.9^t)$

$5=300\pi - 300\pi \cdot 0.9^t$

$\frac{5-300 \pi}{-300 \pi} = 0.9^t$

Use logarithms to calculate t. I've got t ≈ 0.0505 h and that are nearly 3 min (This result seems a little funny to me :confused:

I assume that I've made a silly mistake with my calculations. So check!

Thank you so much! (Clapping)I got the answer now, it is after 5 hours and 19 minutes, because it's $0.999^t$,