Respuesta :
Answer:
Therefore the tank will be half empty after 10.10 days.
Therefore there will be 2735.53 liters water in the tank after 4 days.
Step-by-step explanation:
Given that,the square root of the volume of remaining water in the tank is proportional to the rate of water leakage.
Let V be volume of water at any instant time t.
[tex]\therefore \frac{dV}{dt} \propto \sqrt V[/tex]
[tex]\Rightarrow \frac{dV}{dt}= k \sqrt V[/tex]
where k is constant of proportionality.
[tex]\Rightarrow \frac{dV}{\sqrt V}= k \ dt[/tex]
Integrating both sides
[tex]\Rightarrow \int \frac{dV}{\sqrt V}= \int k \ dt[/tex]
[tex]\Rightarrow 2\sqrt V=kt+C[/tex] [ C is integrating constant]
At t=0, the volume of water is 350 liters
[tex]2\sqrt{350}=k.0+C[/tex]
[tex]\Rightarrow C=2\sqrt{350}[/tex]
The equation becomes
[tex]\Rightarrow 2\sqrt V=kt+2\sqrt{350}[/tex]
Again at t=1, the volume of water is(350-20)liters=330 liters
[tex]2\sqrt {330}=k.1+2\sqrt{350}[/tex]
[tex]\Rightarrow k=2\sqrt {330}-2\sqrt{350}[/tex]
The equation becomes
[tex]2\sqrt V=2(\sqrt{330}-\sqrt{350} ) t+2\sqrt{350}[/tex]
[tex]\Rightarrow \sqrt V=(\sqrt{330}-\sqrt{350} ) t+\sqrt{350}[/tex]
Now when the tank is half empty,then V= (350÷2) liters = 175 liters
[tex]\sqrt {175}=(\sqrt{330}-\sqrt{350} ) t+\sqrt{350}[/tex]
[tex]\Rightarrow (\sqrt{330}-\sqrt{350} ) t=\sqrt{175}-\sqrt{350}[/tex]
[tex]\Rightarrow t=\frac{\sqrt{175}-\sqrt{350}}{ (\sqrt{330}-\sqrt{350} )}[/tex]
[tex]\Rightarrow t=10.10[/tex] days
Therefore the tank will be half empty after 10.10 days.
After 4 days,
[tex]\sqrt V=(\sqrt{330}-\sqrt{350} ) (4)+\sqrt{350}[/tex]
[tex]\Rightarrow \sqrt V= 16.54[/tex]
[tex]\Rightarrow V= 373.53[/tex] liters
Therefore there will be 2735.53 liters water in the tank after 4 days.
