Final answer:
To find the number of gallons of water in the tank at time 20 minutes when the rate of change is inversely proportional to the amount of water, we can set up a proportion using the initial and final values. Solving for the constant of proportionality, we can then integrate the equation and find the value of y at 20 minutes. The final answer is (√285) + 5.
Step-by-step explanation:
To solve this problem, we need to apply the concept of inverse proportionality. The given differential equation is dy/dt = k/y. We are given that initially there were 5 gallons of water in the tank, and after 3 minutes, there were & gallons. We can set up a proportion using this information and solve for k. Once we have the value of k, we can plug it into the equation to find the value of y at 20 minutes.
Let's set up the proportion: 5 / 3 = k / &. Cross-multiplying gives us 3k = 5&. Solving for k, we have k = 5& / 3.
Now, we can plug in the value of k into the equation dy/dt = k/y. Since we want to find the number of gallons of water in the tank at 20 minutes (y = ?), we can rewrite the equation as dy = (5& / 3) dt. Integrating both sides gives us ∫ dy = ∫ (5& / 3) dt. Integrating, we get y = (5& / 3) t + C, where C is the constant of integration.
We can find the value of C by substituting the initial condition: when t = 0, y = 5. Plugging these values into the equation, we get 5 = (5& / 3) (0) + C, which simplifies to C = 5. Now we have the equation y = (5& / 3) t + 5.
To find the number of gallons of water in the tank at 20 minutes, we substitute t = 20 into the equation: y = (5& / 3) (20) + 5.
Therefore, the number of gallons of water in the tank at time 20 minutes is (√285) + 5.
Learn more about Inverse proportionality