A tank with a capacity of 2000 L is full of a mixture of water and chlorine with a concentration of 0.01 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 20 L/s. The mixture is kept stirred and is pumped out at a rate of 50 L/s. Find the amount of chlorine in the tank as a function of time.(Let y be the amount of chlorine in grams and t be the time in seconds.)

You can represent the unknown amounts by the use of variables. Follow whatever the description is and convert it one by one mathematically. For example if it is asked to increase some item by 4 , then you can add 4 in that item to increase it by 4. If something is for example, doubled, then you can multiply that thing by 2 and so on methods can be used to convert description to mathematical expressions.

For the given case, we can model the situation by assuming that things are done in discrete way rather than continuous, such that in 1 second, the inputted water(instantaneously coming inside at the start of second) get mixed with the whole water available in the tank, and then going outside instantaneously at the end of the considered second.

We also need to work with variable amount of concentration.

Let us suppose that at time t, the concentration of chlorine in the water present in the tank be [tex]\rm y_t \: \rm grams/Liter[/tex], and the amount of water present in tank be [tex]V_t[/tex] L,

then total amount of chlorine present in tank = [tex]V_t \times y_t[/tex] grams

Then, after 1 second(water not still outputted), the time becomes t+1, the volume of water in tank is [tex]V_t[/tex] +20 L (20L is pure water)(assuming capacity of tank doesn't contradict at this moment).

then the concentration of chlorine becomes:

[tex]y_{t+1} = \dfrac{V_t \times y_t}{V_t + 20} = \dfrac{V_t \times y_t}{V_{t+1}}\: \rm grams/liter[/tex] Now 50L is outputted at the end of that considered second of time. Thus remaining water in tank is V - 30 L.

Thus,

[tex]y_{t+2} = \dfrac{V_{t+1} \times y_{t+1}}{V_{t+1} + 20} = \dfrac{V_{t+1} \times y_{t+1}}{V_{t+1}} = y_{t+1} = \dfrac{V_{t+1}}{V_{t+2}}\dfrac{V_t \times y_t}{V_{t+1}}\: \rm grams/liter\\\\y_{t+2} = \dfrac{V_t\times y_t}{V_{t+2}}\\\\[/tex]

Similarly, we get:

[tex]y_{t+k} = \dfrac{V_t\times y_t}{V_{t+k}}[/tex]

At t = 0, we had: [tex]y_0 = 0.01\\V_0 = 2000[/tex], taking t =0, and k = t, we get:

[tex]y_t = \dfrac{V_0\times y_0}{V_{t}} = \dfrac{0.01 \times 2000}{V_t} = \dfrac{20}{V_t}[/tex]

After each second , the volume is net decreasing 30 L, so after t seconds, the water will decrease as V - 30t

Thus, after t seconds from starting, we get [tex]V_t = V_0 - 30t = 2000 - 30t[/tex]

Using this value, and the value of [tex]y_t[/tex] we had obtained above, we get:

[tex]y_t = \dfrac{20}{2000-30t} \: \rm grams/liter[/tex]

Thus, if we take [tex]y_t[/tex] as y simply, representing the amount of chlorine in grams and t be the time in seconds, then

[tex]y = \dfrac{20}{2000-30t}[/tex]

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