How do you find the amount of sugar in the tank after t minutes if a tank contains 1640 liters of pure water and a solution that contains 0.09 kg of sugar per liter enters a tank at the rate 5 l/min the solution is mixed and drains from the tank at the same rate?

Answer 1

This needs a bit of calculus and some thought.

Let's start by defining a few useful quantities: #\sigma# the rate at which solution enters and drains (l/min); #\rho# the sugar solution concentration entering the tank (kg/l); #m(t)# the total mass of sugar in the tank at time #t# (kg); #V# the volume of the tank (l)
The rate of sugar mass gained from the solution entering is simply #\sigma \rho#. And since the sugar solution concentration at time #t# inside the tank is #m(t)#/#V#, the mass-loss rate by draining is #\sigma m(t)#/#V#

The overall gain rate is just the difference between above gain and loss rates:

#{dm(t)}/{dt} = \sigma \rho - \sigma {m(t)}/V#.

Rearranging this to integrate, we have

#\int_0^{m(t)}{dm'}/{\sigma \rho - \sigma m'\mbox{/}V} = \int_0^t dt'#,

which leads to

#V/\sigma\ln({V\rho}/{V\rho-m}) = t#.
And solving for #m# we finally obtain
#m(t) = V\rho(1-e^{-\sigma t\mbox{/}V})#.
To check this makes sense, note that at the start #(t=0)# the mass of sugar is zero and then slowly rises, reaching an aymptotic value (i.e. after an infinite time) corresponding to the concentration of the solution entering: #m(t\mbox{=}\infty)\mbox{/}V = \rho#.
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Answer 2

To find the amount of sugar in the tank after t minutes, you can use the formula:

Amount of sugar = Initial amount of sugar + (Rate of sugar solution in - Rate of sugar solution out) * t

Given: Initial amount of sugar = 1640 liters * 0.09 kg/liter Rate of sugar solution in = 5 liters/minute * 0.09 kg/liter Rate of sugar solution out = 5 liters/minute * (Amount of sugar in tank / Total volume of tank)

Substitute the values into the formula and solve for the amount of sugar in the tank after t minutes.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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