Question

A tank consists of 50 liters of freshwater. Two liters of brine each containing 5 grams of dissolved salt is running into the tank per minute; the mixture is kept uniform by stirring and runs out at the rate of one liter per minute. If m grams of salt is present in the tank after t minutes then expression of m in terms of t is given by $m=kt\left( r+\dfrac{2p+t}{p+t} \right)$ grams. Find $\dfrac{p}{k}+r$

Answer 1

Now we have 2L of water is added and 1L of water is removed every minute hence we can form an equation of volume of water at time t. Now at time t salt contained is m. hence we can calculate the concentration at time t by the formula $\dfrac{\text{amount of salt present}}{\text{Total volume}}$.

Complete step-by-step solution:
Now we know the total salt added each minute. The total output can be calculated by concentration × volume of water that is removed. Hence we can form a differential equation for salt at time t. Solving this differential equation we will get the equation in the form of $m=kt\left( r+\dfrac{2p+t}{p+t} \right)$ . Now that we know k, p, and r we can find the value of $\dfrac{p}{k}+r$
Now we first know that initially, 50 liters of water are present.
Now, 2L of water is added and 1L of water is removed every minute.
Hence we have that after t minutes the Volume of the tank is 50 + (2 – 1)t
Hence the Volume after t minutes is given by 50 + t.
Now we know that at time t m grams of salt is present.
So concentration at time t is given by $\dfrac{\text{amount of salt present}}{\text{Total volume}}$ gm per liters
Hence we have at time t the concentration is given by $\dfrac{m}{50+t}gm/L$
Now we have that 1L water flows out hence output rate is $\begin{aligned} & \dfrac{m}{50+t}\times 1 \\\ & =\dfrac{m}{50+t}gm/\min ...............................(1) \\\ \end{aligned}$
Now 2L of water each containing 5 grams of salt is put in tank per minute hence per minute 10grams of salt is added
Input rate = 10 grams ……………………….(2)
Now total change is given by input rate – output rate.
Hence we have

& \dfrac{dm}{dt}=10-\dfrac{m}{50+t} \\\ & \Rightarrow \dfrac{dm}{dt}=\dfrac{10(50+t)-m}{(50+t)} \\\ \end{aligned}$$ $$\Rightarrow \dfrac{dm\left( 50+t \right)}{dt}=10\left( 50+t \right)-m$$ $$\Rightarrow \dfrac{dm\left( 50+t \right)}{dt}+m=10(50+t)................(3)$$ Now we know that $\left( f.g \right)'=f'g+g'f$ Hence if we see $\begin{aligned} & \dfrac{d\left( \left( 50+t \right).m \right)}{dt}=\dfrac{dm\left( 50+t \right)}{dt}+m\dfrac{d\left( 50+t \right)}{dt} \\\ & \dfrac{d\left( \left( 50+t \right).m \right)}{dt}=\dfrac{dm\left( 50+t \right)}{dt}+m \\\ \end{aligned}$ Substituting this in equation(3) We get $$\begin{aligned} & \dfrac{d\left( m\left( 50+t \right) \right)}{dt}=10(50+t). \\\ & \Rightarrow d\left( m(50+t) \right)=\left( 500+10t \right)dt \\\ \end{aligned}$$ Now integrating on both sides we get. $$\begin{aligned} & m.\left( 50+t \right)=\int{\left( 500+10t \right)dt} \\\ & \Rightarrow m.\left( 50+t \right)=\int{\left( 500dt \right)}+\int{10tdt} \\\ & \Rightarrow m\left( 50+t \right)=500t+\dfrac{10{{t}^{2}}}{2}+C \\\ & \Rightarrow m\left( 50+t \right)=500t+5{{t}^{2}}+C \\\ & \Rightarrow m\left( 50+t \right)=5t\left( 100+t \right)+C \\\ & \Rightarrow m=\dfrac{5t\left( 100+t \right)+C}{50+t} \\\ \end{aligned}$$ Now we have at t = 0 m = 0 and hence c = 0 $0=\dfrac{C}{50}\Rightarrow C=0$ Hence we have $m=5t\times \left( \dfrac{(100+t)}{50+t} \right)$ Comparing this with $m=kt\left( r+\dfrac{2p+t}{p+t} \right)$ we get K = 5, p = 50 and r = 0 Hence $\dfrac{p}{k}+r=\dfrac{50}{5}+0=10$ **Hence we have the value of $\dfrac{p}{k}+r$ is 10.** **Note:** Now in this solution while using formulas we must always check if the units are the same. Also the condition at t = 0 m = 0 and hence c = 0 is not given but understood. Since at t = 0 no water was salt was there in the tank we say m = 0.