Question

A 1000-gallon tank, initially full of water, develops a leak at the bottom. Given that 500
gallons of water leak out in the first 30 minutes, find the amount of water left in the tank t
minutes after the leak develops given that the water drains off a rate proportional to the
amount of water present.

Answer 1

calculus right?

let Q be the quantity of water in tank
dQ/dt will be the rate of change of water in tank (drain rate)

given that the water drains off a rate proportional to the amount of water present

dQ/dt = kQ, where k is a constant

Integrate n use the given "500 gallons of water leak out in the first 30 minutes" to solve for k.

Good luck!

Answer 2

\[\frac{ dQ }{ dt }=kQ\]
separating the variables and integrating
\[\int\limits \frac{ dQ }{ Q }=\int\limits k~dt+c\]
\[\ln Q=kt+c\]
\[Q=e ^{kt+c}=e^c~e ^{kt}=a e ^{kt}\]
when t=0,Q=1000 g
\[1000=e^ce^0=e^c\]
\[Q=1000~e ^{kt}\]
when t=30 min.
Q=1000-500=500g
\[500=1000e^{30k}\]
\[\frac{ 500 }{ 1000 }=e^{30k},\frac{ 1 }{ 2 }=e^{30k},30k=\ln \frac{ 1 }{ 2 }=\ln 1-\ln 2=0-\ln 2=-\ln 2\]
\[k=\frac{ -\ln~2 }{ 30 }\]
replace the value of k and findQ

Answer 3

Thanks for the help guys, sorry my computer died earlier and I didnt have the charger. Ended up getting the correct answer with your help so thank you