# Brine containing 1.5lb/gal of salt enters a tank at the

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?

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Gerald Lopez
$$\displaystyle{y}=$$ amount of salt at time t
$$\displaystyle{r}_{{{i}}}={2}$$
$$\displaystyle{c}_{{{i}}}={1.5}$$
$$\displaystyle{r}_{{{o}}}={2}$$
$$\displaystyle{v}{\left({t}\right)}={v}_{{{0}}}+{\left({r}_{{{i}}}-{r}_{{{o}}}\right)}{t}$$
$$\displaystyle{v}{\left({t}\right)}={100}+{\left({2}-{2}\right)}{t}$$
$$\displaystyle{v}{\left({t}\right)}={100}+{0}{t}$$
$$\displaystyle{v}{\left({t}\right)}={100}$$
$$\displaystyle{c}_{{{o}}}={\frac{{{y}}}{{{v}{\left({t}\right)}}}}$$
$$\displaystyle{c}_{{{o}}}={\frac{{{y}}}{{{100}}}}$$
DE is given by
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={r}_{{{i}}}{c}_{{{i}}}-{r}_{{{o}}}{c}_{{{o}}}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={2}\cdot{1.5}-{2}\cdot{\frac{{{y}}}{{{100}}}}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={3}-{\frac{{{y}}}{{{50}}}}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=-{\frac{{{y}}}{{{50}}}}+{3}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=-{\left({\frac{{{y}}}{{{50}}}}-{3}\right)}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=-{\frac{{{1}}}{{{50}}}}{\left({y}-{150}\right)}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=-{0.02}{\left({y}-{150}\right)}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left({y}-{150}\right)}}}}=-{0.02}{\left.{d}{t}\right.}$$
$$\displaystyle\int{\frac{{{\left.{d}{y}\right.}}}{{{\left({y}-{150}\right)}}}}=\int-{0.02}{\left.{d}{t}\right.}$$
$$\displaystyle{\ln{{\left({y}-{150}\right)}}}=-{0.02}{t}+{c}$$
$$\displaystyle{y}-{150}={e}^{{-{0.02}{t}+{c}}}$$
$$\displaystyle{y}-{150}={e}^{{-{0.02}{t}}}\cdot{e}^{{{c}}}$$
$$\displaystyle{y}-{150}={e}^{{-{0.02}{t}}}\cdot{C}$$
$$\displaystyle{y}-{150}={C}{e}^{{-{0.02}{t}}}$$
$$\displaystyle{y}={150}+{C}{e}^{{-{0.02}{t}}}$$ (1)
The tank initially contains initially 10 lb of salt
$$\displaystyle{10}={150}+{C}{e}^{{-{0.02}\cdot{0}}}$$
$$\displaystyle{10}={150}+{C}\cdot{1}$$
$$\displaystyle{10}={150}+{C}$$
$$\displaystyle{C}={10}-{150}$$
$$\displaystyle{C}=-{140}$$
$$\displaystyle{y}={150}-{140}{e}^{{-{0.02}{t}}}$$
find t when $$\displaystyle{y}={20}$$
$$\displaystyle{20}={150}-{140}{e}^{{-{0.02}{t}}}$$
$$\displaystyle{140}{e}^{{-{0.02}{t}}}={150}-{20}$$
$$\displaystyle{140}{e}^{{-{0.02}{t}}}={130}$$
$$\displaystyle{e}^{{-{0.02}{t}}}={\frac{{{130}}}{{{140}}}}$$
$$\displaystyle{e}^{{-{0.02}{t}}}={\frac{{{13}}}{{{14}}}}$$
$$\displaystyle-{0.02}{t}={\ln{{\left({\frac{{{13}}}{{{14}}}}\right)}}}$$
$$\displaystyle{t}=-{\frac{{{\ln{{\left({\frac{{{13}}}{{{14}}}}\right)}}}}}{{{0.02}}}}$$
$$\displaystyle{t}={3.705398}$$
$$\displaystyle{t}\approx{3.71}$$ minutes
the amount of salt in the tank be 20 lbs after 3.71 minutes
###### Not exactly what you’re looking for?
Matthew Rodriguez

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

karton

Let x(t)= The amount of salt in the tank at time t.
Rate of increase of x,
$$\frac{dx}{dt}=\text{Rate of salt in}-\text{Rate of salt out}$$
$$(\text{Rate})in=3lb/gal \times 4gal/min$$
$$\begin{array}{}=12 lb/min \\(\text{Rate})out=2gal/min \times \frac{x(t)}{2t+20} lb/gal \\=\frac{2x}{2t+20} lb/min \\\frac{dx}{dt}=12-\frac{2x}{2t+20} \\\Rightarrow \frac{dx}{dt}=+\frac{2x}{2t+20}=12 \\\text{This is linear first order differential equation of the form} \\\frac{dx}{dt}+px=q \\p=\frac{2}{2t+20}\ q=12 \\\text{Integrating factor}\ (I.F)=e^{\int pdt} \\=e^{\int \frac{2}{2t+20}dt} \\\text{Put}\ 2t+20=z \\2dt=dz \\\therefore IF=e^{\int \frac{dz}{z}} \\=e^{\ln |z|} \\=e^{\ln |2t+20|} \\IF=2t+20 \\\text{The solution is} \\x \times I.F=\int q \times IF dt \\\Rightarrow x(2t+20)=\int 12(2t+20)dt \\\Rightarrow x(2t+20)=12 \int 2(t+10)dt \\=24 \int (t+10)dt \\=24 [\frac{t^{2}}{2}+10t]+c \\\Rightarrow x(t)=\frac{12t^{2}+240t+c}{2t+20} \\\text{Given},\ x(0)=0 \\0=\frac{12(0)^{2}+240 \times 0+c}{2 \times 0+20} \\\Rightarrow 0=\frac{c}{20} \\\text{After}\ t=5min \\x(5)=\frac{12(5)^{2}+240 \times 5+20}{(2 \times 5)+20} \\=\frac{(12 \times 25)+1200+20}{10+20} \\=\frac{300+1200+20}{300} \\=\frac{1520}{30} \\=50.67lb \\\Rightarrow c=0 \\x(t)=\frac{12t^{2}+240t}{2t+20} \\=\frac{\not{2}(6t^{2}+120t)}{\not{2}(t+10)} \\x(t)=\frac{6t^{2}+120t}{t+10} \\\text{After}\ t=5min \\x(5)=\frac{6(5)^{2}+120(5)}{5+10} \\=\frac{(6 \times 25)+600}{15}=\frac{150+600}{15} \\=\frac{750}{15} \\=50lb \end{array}$$
The amount of salt in the tank after 5 min is 50lb.