Question

A 200 liter tank initially full of water develops a leak at the bottom. Given that 20% of the water leaks out in the first 10 minutes, find the amount of water left in the tank t minutes after the leak develops if the water drains off at a rate that is proportional to the amount of water present.

Answer 1

@Laynalove

Answer 2

@zzr0ck3r

Answer 3

@jim_thompson5910

Answer 4

@thephysicsman

Answer 5

I think we have to use related rates with this problem

Answer 6

nah, I wish. This is for my chapter on differential equations. I need to use the formula for exponential growth and decay

Answer 7

woah, lol what class is that for?

Answer 8

just Cal II

Answer 9

its not actually differential equations lol

Answer 10

You said something about exponential growth and decay

\[\frac{ dA }{ dt } = e^{-kt}*A_{0}\]

Answer 11

\[P=A _{0}e ^{-kt}\]

Yea thats the one

Answer 12

@legomyego180 what else were you thinking?

Answer 13

I have a hunch on the equation? I think I need to solve for k, the decay constant first.

Answer 14

\[160=200e^{-k10}\]

Answer 15

how does that look to you?

Answer 16

Im getting \[k=-\frac{ \ln \frac{ 4 }{ 5 } }{ 10 }\]

Which we can use to plug into our original equation to solve for t?

Answer 17

\[\frac{ 4 }{ 5 } = e^{-k*10}\]

\[\ln(\frac{ 4 }{ 5 }) = -10k \]

\[\frac{ \ln(\frac{ 4 }{ 5 }) }{ -10 } = k \]

Answer 18

yeah I agree :) but I feel there should be more to this problem?

Answer 19

Yea we need to plug k back into the original equation to solve for t but Im not sure what I need to do with the inital values variable and end value variable

Answer 20

@zepdrix @UnkleRhaukus you guys around?

Answer 21

@mww

Answer 22

this is VERY similar to a Q u posted a few days ago: http://openstudy.com/study#/updates/577474a5e4b0140fe65398af

u could easily re-use the same equation and solve for the different constants in this case.

Answer 23

they just need you to put back the k into the original equation (find the amount at t minutes)

Answer 24

The rate of change is proportional to the amount at t time,
\(\large\rm A'=kA\)
and solving for A gives us the equation for growth/decay that we're used to seeing,
\(\large\rm A=A_o e^{kt}\)

Ya looks like you guys got everything figured out :)
\(\large\rm k=\dfrac{\ln.8}{10}\)

Plugging it in gives you,\[\large\rm A(t)=200e^{\left(\frac{\ln.8}{10}~t\right)}\]

Answer 25

That's all they wanted.

Answer 26

You don't need a negative on the k when you setup the equation.
The negative will come out naturally.

But I guess it's ok if you do.

Answer 27

your initial amount is already included as P(0)= A = 200 when t = 0.

Answer 28

Gotcha, I think I actually have a fighting chance on this subject on my test tomorrow

Answer 29

In fact if they say that a proportion of the original was left at time t,, then you can always write:

\[\frac{ P(t) }{ A } = C = e^{-kt}\] immediately on LHS for some constant C being the proportion.