1. ## again..im lost!

A tank is in the shape of 1m x 1m x 1m cube and is positioned with its base in a horizontal position.There is a amall hole in the side of the tank at a height of 80cm above the base.At 12.00 water starts to enter the empty tank , and the rate at which the water is entering t minutes later is given in litres per minute by the formula 20+2t.
At what time does the water start to exit the tank through the small hole?

I don't even know where to start!

2. Originally Posted by ail101
A tank is in the shape of 1m x 1m x 1m cube and is positioned with its base in a horizontal position.There is a amall hole in the side of the tank at a height of 80cm above the base.At 12.00 water starts to enter the empty tank , and the rate at which the water is entering t minutes later is given in litres per minute by the formula 20+2t.
At what time does the water start to exit the tank through the small hole?

I don't even know where to start!
This looks like a remarkably trivial Calculus problem- there no calculus involved at all.

Water will start to exit the tank when the height of the water in the tank equals the height of the hole, .8 m. As the water rises to, say, height h, the volume of water in the tank will be 1*1*h= h so we want to determine when the volume of water is .8 cubic meters. Since a liter the same volume as a cube 10 cm= .1 m on a side, a liter is $\displaystyle (.1)^3= .001$ cubic meters. Water is coming in at 20+2t liters= .001(20+ 2t) cubic meters per minute, water will start draining out the hole when .001(20+ 2t)= .8.

Solve that equation.

3. Originally Posted by ail101
A tank is in the shape of 1m x 1m x 1m cube and is positioned with its base in a horizontal position.There is a amall hole in the side of the tank at a height of 80cm above the base.At 12.00 water starts to enter the empty tank , and the rate at which the water is entering t minutes later is given in litres per minute by the formula 20+2t.
At what time does the water start to exit the tank through the small hole?

I don't even know where to start!
As the rate at which the water is entering is dependent on t,
then the rate increases with time linearly.

Then, if we integrate the rate, we get the amount of water in the tank,

which is rectangle area + triangle area = $\displaystyle 20t+\frac{1}{2}2t(t)$

or $\displaystyle \int_{0}^t(20+2t)dt=20t+t^2$

$\displaystyle t^2+20t=800\ \Rightarrow\ t^2+20t-800=0\ \Rightarrow\ (t-20)(t+40)=0$

$\displaystyle t=20\ minutes.$

Water starts to leak at 12:20