Question

At time t0 a really big tank contains 10 kg of salt dissolved in 100 litres of water. We pour water containing 1/5 kg of salt per litre of water into the tank at a rate of 2 litre/min. At the same time, the mixture is drained from the tank at the rate of 1 litre/min. (a) Let x(t) be the quantity of salt in the tank at time t ? t0. Find a differential equation for x(t), valid until the water overflows the tank. You should get dx(t)/dt = 2/5 - x(t)/(100 + (t - t0)), t ? t0, where t is the time in minutes and x(t) is in units of kg. Explain in your own words how this equation was obtained, carefully justifying each step (you may draw a figure illustrating how x(t) is determined if it helps). (b) Solve this equation for x(t), accounting for the initial condition, assuming that the starting time is t0 (not t = 0). Please include all intermediate steps. (c) Find the value for the concentration of salt (mass of salt divided by volume of water) as a function of time. Assuming that the tank is infinitely big, what is the limiting value for the concentration of salt as t ? ?? Explain in your own words why this makes sense.

          At time t0 a really big tank contains 10 kg of salt dissolved in 100 litres of water. We pour water containing 1/5 kg of salt per litre of water into the tank at a rate of 2 litre/min. At the same time, the mixture is drained from the tank at the rate of 1 litre/min.

(a) Let x(t) be the quantity of salt in the tank at time t ? t0. Find a differential equation for x(t), valid until the water overflows the tank. You should get

dx(t)/dt = 2/5 - x(t)/(100 + (t - t0)), t ? t0,

where t is the time in minutes and x(t) is in units of kg. Explain in your own words how this equation was obtained, carefully justifying each step (you may draw a figure illustrating how x(t) is determined if it helps).

(b) Solve this equation for x(t), accounting for the initial condition, assuming that the starting time is t0 (not t = 0). Please include all intermediate steps.

(c) Find the value for the concentration of salt (mass of salt divided by volume of water) as a function of time. Assuming that the tank is infinitely big, what is the limiting value for the concentration of salt as t ? ?? Explain in your own words why this makes sense.

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At time t0 a really big tank contains 10 kg of salt dissolved in 100 litres of water. We pour water containing 1/5 kg of salt per litre of water into the tank at a rate of 2 litre/min. At the same time, the mixture is drained from the tank at the rate of 1 litre/min.

(a) Let x(t) be the quantity of salt in the tank at time t ? t0. Find a differential equation for x(t), valid until the water overflows the tank. You should get

dx(t)/dt = 2/5 - x(t)/(100 + (t - t0)), t ? t0,

where t is the time in minutes and x(t) is in units of kg. Explain in your own words how this equation was obtained, carefully justifying each step (you may draw a figure illustrating how x(t) is determined if it helps).

(b) Solve this equation for x(t), accounting for the initial condition, assuming that the starting time is t0 (not t = 0). Please include all intermediate steps.

(c) Find the value for the concentration of salt (mass of salt divided by volume of water) as a function of time. Assuming that the tank is infinitely big, what is the limiting value for the concentration of salt as t ? ?? Explain in your own words why this makes sense.

Added by Carrie D.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sam Stansfield

07/18/2023

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The tank initially contains 10 kg of salt in 100 litres of water. Show more…

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At time t0 a really big tank contains 10 kg of salt dissolved in 100 litres of water. We pour water containing 1/5 kg of salt per litre of water into the tank at a rate of 2 litre/min. At the same time, the mixture is drained from the tank at the rate of 1 litre/min. (a) Let x(t) be the quantity of salt in the tank at time t ≥ t0. Find a differential equation for x(t), valid until the water overflows the tank. You should get dx(t)/dt = 2/5 - x(t)/(100 + (t - t0)), t ≥ t0, where t is the time in minutes and x(t) is in units of kg. Explain in your own words how this equation was obtained, carefully justifying each step (you may draw a figure illustrating how x(t) is determined if it helps). (b) Solve this equation for x(t), accounting for the initial condition, assuming that the starting time is t0 (not t = 0). Please include all intermediate steps. (c) Find the value for the concentration of salt (mass of salt divided by volume of water) as a function of time. Assuming that the tank is infinitely big, what is the limiting value for the concentration of salt as t → ∞? Explain in your own words why this makes sense.

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Transcript

00:01 Okay, we have a tank with not a fixed volume or a fixed amount of salt.

00:07 The initial volume was 100 liters and started out with 10 kilograms of salt at some initial time t zero.

00:15 We're gonna bring salt water in at two liters per minute.

00:19 We're gonna pump it out at one liter per minute.

00:22 The concentration coming in is 0 .2 kilograms per liter.

00:31 So first of all, let's figure out the volume at time t.

00:34 So the volume at time t is the volume at t zero plus the rate in minus the rate out times the elapsed time since t zero.

00:49 So we can write that as 100 plus two minus one times t minus t zero, which is 100 plus t minus t zero.

01:04 So the volume's not constant.

01:06 And eventually, if we wait long enough, the tank will overflow.

01:11 So the formula or the equation for the amount of salt in the tank is that the rate of change of that is the stuff coming in minus the stuff going out.

01:29 And so we have the rate coming in times the concentration coming in minus the rate going out times the concentration going out.

01:41 And we have numbers for r in and c in and r out.

01:45 C out is the current concentration.

01:48 So it's x of t divided by v of t.

01:55 And i'm gonna rearrange that, put the x's on the left and substitute in for my v from up above.

02:12 And that's gonna be equal to 0 .4, which is 2 5ths.

02:18 So it is the same formula that they get in the problem statement, but it's written a little differently.

02:31 So the next step is to solve it.

02:33 So what i'm gonna do is i'm gonna multiply through by this denominator.

02:58 Then the left -hand side is a perfect derivative.

03:02 100 and t zero are both constants, but we have t times dx dt plus x is the only stuff that kinda depends on anything.

03:14 And so what we have is we have a total time derivative of this quantity, 100 plus t minus t zero times x.

03:30 If you take that derivative, you'll get the same thing up here.

03:33 And so then the right -hand side, 0 .4 times 100 plus t minus t zero...

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