Mathematics, 02.06.2020 23:01 igtguith

Consider a vat that at time t contains a volume V (t) of salt solution containing an amount Q(t) of salt, evenly distributed throughout the vat with a concentration c(t), where c(t) = Q(t) / V(t). Assume that water containing a concentration k of salt enters the vat at a rate rin, and that water is drained from the vat at a rate rout > rin.
(a) If V (0) = V0, find an expression for the amount of solution in the vat at time t. At what time T will the vat become empty? Find an initial value problem that describes the amount of salt in the vat at time t <= T. You may assume that Q(0) = Q0.
(b) Solve the initial value problem in part (a). What is the amount of salt in the vat at time t? What is the concentration of the last drop that leaves the vat at time t = T

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a.i $\frac{dQ(t)}{dt} = kr_{in} - \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t}\\\frac{dQ(t)}{dt} + \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t} = kr_{in}$

ii. time at which vat will be empty

$T = \frac{Q(T)r_{out} }{[V_{0} + (r_{in} - r_{out} )]kr_{in}}$ when dQ/dt = 0

b. i Amount of salt at time t

Q(t) = [[exp(rout/(rin - rout)]krin[V(0)t + [exp(rout/(rin - rout)](rin - rout)t²/2 + Q(0)V(0)]/[V(0) + [exp(rout/(rin - rout)](rin - rout)t]

ii. Concentration at t = T

c(T) = [krin[(V(0)T + (rin - rout)T²/2)exp(rout/(rin - rout) + Q(0)V(0)]/([V(0) + exp(rout/(rin - rout)[(rin - rout)T]][V₀ + (rin - rout)T])

Step-by-step explanation:

i. We determine the differential equation for the solution in the vat at time,t.

Let Q be the quantity of salt in the vat at any time,t.

So, dQ/dt = rate of change of quantity of salt in the vat. = rate of change of quantity of salt into the vat - rate of change of quantity of salt out of the vat.

Now, since a concentration of salt, k enter the vat at a rate of rin, rate of change of quantity of salt into the vat = krin.

Let V(0) = V₀ be the volume of water int the vat at time t = 0. Now, the volume of water in the vat increases at a rate of (rin - rout). The increase in volume after a time t is (rin - rout)t. So the volume after a time, t is V(t) = V₀ + (rin - rout)t. The concentration of this liquid is thus Q(t)/V(t) = Q(t)/V₀ + (rin - rout)t. Now, the rate of change of quantity of salt out of the vat is thus [Q(t)/V₀ + (rin - rout)t]rin = Q(t)rout/[V₀ + (rin - rout)t].

So, dQ(t)/dt = krin - Q(t)rout/[V₀ + (rin - rout)t].

$\frac{dQ(t)}{dt} = kr_{in} - \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t}\\\frac{dQ(t)}{dt} + \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t} = kr_{in}$

ii Time T at which vat will be empty

At time T, when the vat is empty, dQ(T)/dt = 0.

$\\\frac{dQ(T)}{dt} + \frac{Q(T)r_{out} }{V_{0} + (r_{in} - r_{out} )T} = kr_{in}\\0 + \frac{Q(T)r_{out} }{V_{0} + (r_{in} - r_{out} )T} = kr_{in}\\T = \frac{Q(T)r_{out} }{[V_{0} + (r_{in} - r_{out} )]kr_{in}}$

b. i

i. We solve the differential equation to find the amount of salt at time,t

The integrating factor is ex

$exp(\int\limits {\frac{r_{out} }{V_{0} + (r_{in} - r_{out})t} } \, dt ) = exp(\frac{r_{out}}{(r_{in} - r_{out})} \int\limits {\frac{ (r_{in} - r_{out})}{V_{0} + (r_{in} - r_{out})t} } \, dt )\\= exp(\frac{r_{out}}{(r_{in} - r_{out})} ln [V_{0} + (r_{in} - r_{out})t} ]) \\\= [V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})$

Multiplying both side of the equation by the integrating factor, we have

$[V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})\frac{dQ(t)}{dt} + [V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})\frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t} = kr_{in}[V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})$ dQ(t)[V(0) + (rin - rout)texp(rout/(rin - rout))]/dt = krin[V(0) + (rin - rout)texp(rout/(rin - rout))]

Integrating both sides we have

∫(dQ(t)[V(0) + (rin - rout)texp(rout/(rin - rout))]/dt)dt = ∫(krin[V(0) + (rin - rout)texp(rout/(rin - rout))])dt

Let exp(rout/(rin - rout) = A

Q(t)[V(0) + A(rin - rout)t] = Akrin[V(0)t + A(rin - rout)t²/2 + C

At t = 0, Q(t) = Q(0),

So,

Q(0)[V(0) + A(rin - rout)×0] = AkrinV(0)t + A(rin - rout)×0²/2 + C

C = Q(0)V(0)

Q(t)[V(0) + A(rin - rout)t] = Akrin[V(0)t + A(rin - rout)t²/2 + Q(0)V(0)

Q(t) = [[exp(rout/(rin - rout)]krin[V(0)t + [exp(rout/(rin - rout)](rin - rout)t²/2 + Q(0)V(0)]/[V(0) + [exp(rout/(rin - rout)](rin - rout)t]

The above is the amount of salt at time ,t.

ii. The concentration of the last drop of salt at time, t = T

To find the concentration of the last drop of salt at time t = T, we insert T into Q(t) to find its quantity and insert t = T into V(t) = V₀ + (rin - rout)t.

Q(T) = [Akrin[V(0)T + A(rin - rout)T²/2 + Q(0)V(0)]/[V(0) + A(rin - rout)T]

Q(T) = [[exp(rout/(rin - rout)]krin[V(0)T + [exp(rout/(rin - rout)](rin - rout)T²/2 + Q(0)V(0)]/[V(0) + [exp(rout/(rin - rout)](rin - rout)T]

the volume at t = T is

V(T) = V₀ + (rin - rout)T.

The concentration at t = T is c(T) = Q(T)/V(T) = [Akrin[V(0)T + A(rin - rout)T²/2 + Q(0)V(0)]/[V(0) + A(rin - rout)T]/V₀ + (rin - rout)T.

= [Akrin[V(0)T + A(rin - rout)T²/2 + Q(0)V(0)]/([V(0) + A(rin - rout)T][V₀ + (rin - rout)T])

= [krin[(V(0)T + (rin - rout)T²/2)exp(rout/(rin - rout) + Q(0)V(0)]/([V(0) + exp(rout/(rin - rout)[(rin - rout)T]][V₀ + (rin - rout)T])

Answer from: Quest

It doesn’t show the question

Answer from: Quest

false

step-by-step explanation:

false : it can not go to 9 the highest that part can go is 7

here is your proof

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Consider a vat that at time t contains a volume V (t) of salt solution containing an amount Q(t) of...

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