Question

Water with a small salt content (5 lb in 1000 gal) is flowing into a very salty lake at the rate of \(4 \cdot 10^{5}\) gal per hr. The salty water is flowing out at the rate of \(10^{5}\) gal per hr. If at some time (say \(t=0\) ) the volume of the lake is \(10^{9}\) gal, and its salt content is \(10^{7}\) lb, find the salt content at time \(t\). Assume that the salt is mixed uniformly with the water in the lake at all times.

Short answer

The salt content in the lake at time t is given by: \[ S(t) = \frac{2 (3 \cdot 10^5 t + 10^9)}{5} \]

Step-by-step solution

Define the variables

Let the volume of the lake at time t be V(t) and the amount of salt in the lake at time t be S(t). Given that V(t) = 10^9 gallons initially, and S(t) = 10^7 pounds initially.

Determine the rate of change of the volume

Since water flows in at the rate of 4x10^5 gallons per hour and flows out at the rate of 10^5 gallons per hour, the net change in volume per hour is: \[ \frac{dV}{dt} = 4 \cdot 10^5 - 1 \cdot 10^5 = 3 \cdot 10^5 \text{ gal/hr} \]

Express the volume as a function of time

Integrate the rate of change from Step 2 to find V(t): \[ V(t) = 3 \cdot 10^5 t + 10^9 \] This represents the volume of the lake at any time t.

Determine the rate of change of the salt content

Salt flows into the lake with the incoming water at a concentration of 5 lb per 1000 gallons, which is 0.005 lb per gallon. Therefore, the rate of salt inflow is: \[ \text{Salt inflow rate} = 0.005 \times 4 \cdot 10^5 = 2000 \text{ lb/hr} \] Since the water leaving the lake is uniformly mixed, its salt concentration is \( \frac{S(t)}{V(t)} \), leading to a salt outflow rate of: \[ \text{Salt outflow rate} = 10^5 \times \frac{S(t)}{V(t)} \]

Write the differential equation for S(t)

Use the rates of salt inflow and outflow to express the change in salt content: \[ \frac{dS}{dt} = 2000 - 10^5 \times \frac{S(t)}{V(t)} = 2000 - 10^5 \times \frac{S(t)}{3 \cdot 10^5 t + 10^9} \] Simplify the equation: \[ \frac{dS}{dt} = 2000 - \frac{10^5 S(t)}{10^9} \times \left( \frac{10^9}{3 \cdot 10^5 t + 10^9 } \right) \]

Solve the differential equation

This is a linear first-order differential equation. To solve it, use an integrating factor or substitution. Let's denote: \[ \frac{dS}{dt} + \frac{10^5}{3 \cdot 10^5 t + 10^9} S = 2000 \] The integrating factor, \( \mu(t) \), is: \[ \mu(t) = e^{\int \frac{10^5}{3 \cdot 10^5 t + 10^9} dt} = e^{ \frac{1}{3} \ln (3 \cdot 10^5 t + 10^9) } = (3 \cdot 10^5 t + 10^9)^{\frac{1}{3}} \] Substituting and solving gives: \[ S(t) = \frac{2000 (3 \cdot 10^5 t + 10^9)}{10^5} \] Solve for S(t) to get: \[ S(t) = \frac{2 (3 \cdot 10^5 t + 10^9)}{5} + C(3 \cdot 10^5 t + 10^9)^{-\frac{1}{3}} \]

Use initial conditions

Given S(0) = 10^7 lbs, use this to solve for the constant C: \[ 10^7 = \frac{2 \times 10^9}{5} + C(10^9)^{-\frac{1}{3}} \] Solve for C.

Write the final solution

Substitute the value of C back into S(t): \[ S(t) = \frac{2 (3 \cdot 10^5 t + 10^9)}{5}\] This is the salt content in the lake at time t.

Key concepts

Differential Equations

Differential equations involve equations that relate functions with their derivatives, describing how a quantity changes over time. In this exercise, we are dealing with a first-order linear differential equation to find the salt content in a lake over time. The equation connects the rate of change of the lake's salt content with its current volume and inflow/outflow rates. Understanding the initial conditions and the rates of change is crucial to forming the differential equation.

Salt Concentration

Salt concentration tells us how much salt is present in a certain amount of water. It's a crucial aspect of environmental science and water treatment. In the lake problem, we start with a known salt concentration and need to account for salt entering and leaving the lake over time. We can calculate salt inflow because the concentration of incoming water is given. Similarly, understanding outflow is just as important. Outflow concentration depends on how uniformly the salt is mixed with water, which in this exercise is assumed to be perfectly uniform.

Rate of Change

The rate of change represents how quickly something is increasing or decreasing. In our problem, we have two crucial rates of change to consider: the rate at which the volume of the lake changes, and the rate at which the lake's salt content changes. We started by determining the net change in the lake’s volume, found by subtracting the outflow rate from the inflow rate. Next, we looked at how fast salt is entering and exiting the lake. This involves both inflow concentration and the lake's current salt concentration as water exits.

First-Order Linear Differential Equations

First-order linear differential equations involve the first derivative of the unknown function and the function itself. In this exercise, the unknown function is the salt content, S(t), of the lake. The general form of the equation is \(\frac{dS}{dt} + P(t)S = Q(t)\), where P(t) and Q(t) are given functions. To solve the equation, we employed the method of integrating factors, which transforms the equation into a simpler form that is easier to solve.

Integrating Factor

To solve the first-order linear differential equation, we used an integrating factor. The integrating factor is a function used to multiply the original differential equation to make it more solvable. For our problem, the integrating factor was \(\mu(t) = e^{\frac{1}{3} \ln(3 \cdot 10^5 t + 10^9)}\), which allows us to rewrite and simplify the equation. Applying the integrating factor enabled us to integrate both sides of the equation and, ultimately, find the function that represents the salt content in the lake over time.