Precal app problem

**bilbobaggins** · October 11th 2007, 01:54 PM

There are 2 tanks of water. The first tank holds 200 liters and will be drained by pump at a constant rate completely in 5 hours. The second tank holds 900 liters and will be drained by opening a plug on the bottom, this tank will also drain in 5 hours. The second tanks volume will decrease according to toricellis law which is V(t)=Vo(1-t/T)^2

When will the volumes of the 2 tanks be equal?

**galactus** · October 11th 2007, 03:03 PM

The amount of water in the first tank at time t will be 200-40t

The amount of water in the second at time t is $900(1-\frac{t}{5})^{2}$

Set them equal and solve for t. You will get two solutions.

**bilbobaggins** · October 11th 2007, 04:00 PM

thanks, so far I've got to -45t+t^2/25=676
does that sound right to you? I dont know where to go from here any pointers?

**galactus** · October 11th 2007, 04:36 PM

Just a quadratic.

$200-40t=900(1-\frac{t}{5})^{2}$

$200-4t=36t^{2}-360t+900$

$-36t^{2}+320t+700=>36t^{2}-3200t+747=>4(9t^{2}-80t+175)=>9t^{2}-80t+175=0$

Now, solve the quad. Use the quadratic formula. May be the best bet.

But you could factor.

Divide by 9 and rewrite:

$t^{2}-\frac{35}{9}t-5t+\frac{175}{9}$

$(t^{2}-\frac{35}{9}t)-(5t-\frac{175}{9})$

$t(t-\frac{35}{9})-5(t-\frac{35}{9})=0$

$\boxed{(t-5)(t-\frac{35}{9})=0}$

**bilbobaggins** · October 11th 2007, 04:52 PM

Originally Posted by galactus

Just a quadratic.

$200-40t=900(1-\frac{t}{5})^{2}$

$200-4t=36t^{2}-360t+900$

$-36t^{2}+320t+700=>36t^{2}-3200t+747=>4(9t^{2}-80t+175)=>9t^{2}-80t+175=0$

Now, solve the quad. Use the quadratic formula. May be the best bet.

But you could factor.

Divide by 9 and rewrite:

$t^{2}-\frac{35}{9}t-5t+\frac{175}{9}$

$(t^{2}-\frac{35}{9}t)-(5t-\frac{175}{9})$

$t(t-\frac{35}{9})-5(t-\frac{35}{9})=0$

$\boxed{(t-5)(t-\frac{35}{9})=0}$

how did you get from here

to here?

**topsquark** · October 11th 2007, 06:29 PM

Originally Posted by bilbobaggins

how did you get from here

to here?

It's just a typo:
$200 - 40t = 900 \left ( 1 - \frac{t}{5} \right )^2$

$200 - 40t = 900 \left ( 1 - \frac{2t}{5} + \frac{t^2}{25} \right )$

$200 - 40t = 900 - 900 \cdot \frac{2t}{5} + 900 \cdot \frac{t^2}{25}$

$200 - 40t = 36t^2 - 360t + 900$

He used the correct equation to proceed to the next line.

-Dan

**bilbobaggins** · October 13th 2007, 08:43 PM

Originally Posted by topsquark

It's just a typo:
$200 - 40t = 900 \left ( 1 - \frac{t}{5} \right )^2$

$200 - 40t = 900 \left ( 1 - \frac{2t}{5} + \frac{t^2}{25} \right )$

$200 - 40t = 900 - 900 \cdot \frac{2t}{5} + 900 \cdot \frac{t^2}{25}$

$200 - 40t = 36t^2 - 360t + 900$

He used the correct equation to proceed to the next line.

-Dan

k thanks

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