# Word problem involving the weather and a leaking barrel

• Nov 10th 2010, 06:36 PM
Zamzen
Word problem involving the weather and a leaking barrel
The question states
During a rain weather, a barrel with the height of 90cm is filled to the top. When it stops to rain the barrel starts to leak so that the waterlevel is declining proportional to the squareroot of the water depth. How long time does it take for the barrel to be empty of it drops from 90cm to 85cm in 1h?

My attempt at the solution
y'=-k(Y)^1/2
After solving for Y by using separaration technique i receive Y=(K^2)(t^2)/4 -ktD +D^2
Solving for D i get (90)^1/2
solving for K i get (4(90)^1/2 +-1360^(1/2)) / 2
When i solve for y=0 i recive either 5h our 177h. And its supposed to be 35h. I cant see where im doing wrong, i think that K^2 is weird. along with that t^2 is weird as well.
• Nov 10th 2010, 07:08 PM
skeeter
Quote:

Originally Posted by Zamzen
The question states
During a rain weather, a barrel with the height of 90cm is filled to the top. When it stops to rain the barrel starts to leak so that the waterlevel is declining proportional to the squareroot of the water depth. How long time does it take for the barrel to be empty of it drops from 90cm to 85cm in 1h?

My attempt at the solution
y'=-k(Y)^1/2
After solving for Y by using separaration technique i receive Y=(K^2)(t^2)/4 -ktD +D^2
Solving for D i get (90)^1/2
solving for K i get (4(90)^1/2 +-1360^(1/2)) / 2
When i solve for y=0 i recive either 5h our 177h. And its supposed to be 35h. I cant see where im doing wrong, i think that K^2 is weird. along with that t^2 is weird as well.

$\displaystyle \frac{dy}{dt} = -k\sqrt{y}$

$\displaystyle \frac{dy}{\sqrt{y}} = -k \, dt$

$2\sqrt{y} = -kt + C$

$\displaystyle y = \frac{(-kt+C)^2}{4}$

when $t = 0$ , $y = 90$ ...

$\displaystyle 90 = \frac{C^2}{4}$ ... $C = 6\sqrt{10}$

$\displaystyle y = \frac{(-kt+ 6\sqrt{10})^2}{4}$

when $t = 1$ , $y = 85$

$\displaystyle 85 = \frac{(-k + 6\sqrt{10})^2}{4}$

$340 = (-k + 6\sqrt{10})^2
$

$\sqrt{340} = -k + 6\sqrt{10}
$

$k = 6\sqrt{10} - \sqrt{340} \approx 0.5346$

set $y = 0$ ...

$\displaystyle 0 = \frac{(-kt+ 6\sqrt{10})^2}{4}
$

$\displaystyle t = \frac{6\sqrt{10}}{k} \approx 35.5 \, hrs$
• Nov 10th 2010, 07:17 PM
Zamzen
thank you very much