Question 5cii
So we want to verify that the function
 = \frac{600}{t+1} - \frac{100e^{-0.2t}(t+6)}{t+1})
satisfies the original differential equation (i.e. that it actually is a solution to that particular differential equation), and in addition to that, verify that it is the solution for the differential equation for that particular initial condition. We'll do it by substitution, since that's what the question tells us to do.
The differential equation is
. We could of course, find the general solution for this differential equation (i.e. find a solution).
Note that you find what
was in 5ci. We also know what x is (the function x(t)). So basically take the LHS of the differential equation, sub in dx/dt and x, simplify all the algebra and see that it equals
. It's fairly hairy, I'll write out some of it:
-600}{(t+1)^2} + \frac{ \frac{600}{t+1} - \frac{100e^{-0.2t}(t+6)}{t+1}}{t+1} <br />\end{align*}<br />)
The rest is just doing the algebra. I've done the rest of the working here, but you should be able to do the rest of it quite quickly by hand. Note that this is a 'show that' question, so you're expected to show all steps of working.
So that shows that x(t) is a solution to the differential equation. Now to show that it is the solution for that particular initial condition. We're told in the question that initially there is no chemical. So the initial condition is t = 0, x = 0. This is simply a matter of evaluating
)
and checking to see if it equals 0.
Question 5e
So we want to find the amount of chemical (x) that has flowed out of the tank in the first 10 minutes (t = 0 to t = 10).
We know that the rate of outflow is
, as we figured out in question 5b. Since you asked
For instance does the integral mean rate of outflow? Why?
I'm going to guess you didn't completely get your head around question 5b. I'll go ahead and explain that one too. In the question we're told that "the solution flows out at the rate of 10 litres per minute".
The rate of outflow of chemical is well... it's exactly what the name suggests. It's a rate, so the measure of amount of chemical that's flowing out per whatever unit of time. Given the units we're told in the question, we're measuring it in grams per minute.
The latter term in that is of course
.
The first term is what we figured out in Question 5a. The concentration of the solution inside the tank is the concentration of the solution that flows out. We measure the concentration in grams per litre (mass over volume). We know that x grams is the amount of chemical in the tank at any time. We know that at any time t, we have
t + 10 = 10(t+1))
litres. The +10 is because there's 10 litres of solution already in there at the start. 20-10 because with each minute we have 20 litres of chemical coming in, and then 10 litres flowing out of the tank every minute. So what remains in the tank is 20-10 litres.
Hence that's where we get
}{(t+1)^2} )
(after subbing in x(t)).
So since the rate of outflow is the change in amount flowing out per minute, we can integrate that to find the amount flowing out. We'll integrate from t = 0 to t = 10 to find the amount that flows out in the first 10 minutes.
Which gives
}{(t+1)^2} dt)
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