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## Answered question

2022-01-03

Brine containing 1.5lb/gal of salt enters a tank at the rate of 2gal/min and the mixture well stirred leaves at the same rate. The tank initially contains 100 gal of brine with10 lbs of dissolved salt. When will the amount of salt in the tank be 20 lbs?

### Answer & Explanation

Gerald Lopez

Beginner2022-01-04Added 29 answers

$y=$ amount of salt at time t
${r}_{i}=2$
${c}_{i}=1.5$
${r}_{o}=2$
$v\left(t\right)={v}_{0}+\left({r}_{i}-{r}_{o}\right)t$
$v\left(t\right)=100+\left(2-2\right)t$
$v\left(t\right)=100+0t$
$v\left(t\right)=100$
${c}_{o}=\frac{y}{v\left(t\right)}$
${c}_{o}=\frac{y}{100}$
DE is given by
$\frac{dy}{dt}={r}_{i}{c}_{i}-{r}_{o}{c}_{o}$
$\frac{dy}{dt}=2\cdot 1.5-2\cdot \frac{y}{100}$
$\frac{dy}{dt}=3-\frac{y}{50}$
$\frac{dy}{dt}=-\frac{y}{50}+3$
$\frac{dy}{dt}=-\left(\frac{y}{50}-3\right)$
$\frac{dy}{dt}=-\frac{1}{50}\left(y-150\right)$
$\frac{dy}{dt}=-0.02\left(y-150\right)$
$\frac{dy}{\left(y-150\right)}=-0.02dt$
$\int \frac{dy}{\left(y-150\right)}=\int -0.02dt$
$\mathrm{ln}\left(y-150\right)=-0.02t+c$
$y-150={e}^{-0.02t+c}$
$y-150={e}^{-0.02t}\cdot {e}^{c}$
$y-150={e}^{-0.02t}\cdot C$
$y-150=C{e}^{-0.02t}$

Matthew Rodriguez

Beginner2022-01-05Added 32 answers

$\frac{dy}{dt}=\left(\text{rate in}\right)-\left(\text{rate out}\right)$
$\frac{dy}{dt}=2\left(1\cdot 5\right)-\left(\frac{y}{100}\right)$
$\frac{dy}{dt}+\frac{y}{100}=3-\left(1\right)$
$\frac{dy}{dt}+p\left(x\right)y=q\left(x\right)$
Which is a first order linear ode.
$I.f={e}^{\int pdx}={e}^{\int \frac{1}{100}dx}={e}^{\frac{x}{100}}$
$y×IF=\int q×IF.dx$
$⇒y.{e}^{\frac{x}{100}}=\int 3\cdot {e}^{\frac{x}{100}}\cdot dx$
$⇒y\cdot {e}^{\frac{x}{100}}=300{e}^{\frac{x}{100}}+c$
$⇒y=300+c{e}^{\frac{x}{100}}$
So $y=300+c{e}^{\frac{t}{100}}-\left(2\right)$
$y\left(0\right)=\frac{10}{100}=\frac{1}{10}$
(2) $⇒300+c{e}^{0}=\frac{1}{10}⇒C=\frac{1}{10}-300$
$⇒C=\frac{2.999}{10}=-299.9$
So (2) $⇒300-299.9\cdot {e}^{-t/100}=100$
$⇒299.9\cdot {e}^{-\frac{t}{100}}=200$
$⇒{e}^{-\frac{t}{100}}=0.66688$
$⇒-\frac{t}{100}=\mathrm{log}\left(0.66688\right)$
$⇒t=100\mathrm{log}\left(0.66688\right)$
$t=100×0.4051$
$t=40.51$

karton

Expert2022-01-09Added 613 answers

Let x(t)= The amount of salt in the tank at time t.
Rate of increase of x,
$\frac{dx}{dt}=\text{Rate of salt in}-\text{Rate of salt out}$
$\left(\text{Rate}\right)in=3lb/gal×4gal/min$

The amount of salt in the tank after 5 min is 50lb.

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