Question

A large tank is filled with water when an outflow valve is opened at \(t=0 .\) Water flows out at a rate, in gal/min, given by \(Q^{\prime}(t)=0.1\left(100-t^{2}\right),\) for \(0 \leq t \leq 10\) a. Find the amount of water \(Q(t)\) that has flowed out of the tank after \(t\) minutes, given the initial condition \(Q(0)=0\) b. Graph the flow function \(Q,\) for \(0 \leq t \leq 10\) c. How much water flows out of the tank in 10 min?

Short answer

Answer: Approximately 66.67 gallons of water flow out of the tank within 10 minutes.

Step-by-step solution

Integrate the flow rate function

To find the flow function \(Q(t)\), we will integrate the flow rate function \(Q^{\prime}(t) = 0.1(100-t^2)\). The antiderivative of this function, based on the initial condition \(Q(0)=0\), will give us the flow function. \(\int Q^{\prime}(t) dt = \int 0.1(100 - t^2) dt\) Apply the constant multiple rule and apply power rule to find the antiderivative. \(Q(t) = 0.1 \int (100 - t^2) dt = 0.1 (100t - \frac{t^3}{3}) + C\) Now, use the initial condition \(Q(0)=0\) to find the constant C. \(Q(0) = 0.1 (100(0) - \frac{0}{3}) + C = 0\) \(C=0\) Thus, the flow function \(Q(t)\) is: \(Q(t) = 0.1 (100t - \frac{t^3}{3})\) b. Graphing the flow function \(Q\), for \(0 \leq t \leq 10\)

Find critical points

To graph the flow function, we first need to find its critical points. Critical points occur where the derivative of the function is either zero or undefined. In this case, our derivative is the flow rate function \(Q^{\prime}(t) = 0.1(100-t^2)\), which is continuous, so we just need to find where it is equal to zero. \(0.1(100-t^2) = 0\) \(100-t^2 = 0\) \(t^2 = 100\) \(t = \pm 10\) Since the domain of \(t\) is restricted to \(0 \leq t \leq 10\), we only need to consider \(t=10\) as a critical point.

Graph the flow function

Graph the given function using \(Q(t) = 0.1 (100t - \frac{t^3}{3})\), with the critical point at \(t=10\). You'll notice that the function starts at \(Q(0)=0\) and initially increases as water flows out of the tank and then decreases as the valve closes, reaching \(0\) at \(t=10\). c. Finding the amount of water that goes out of the tank in 10 min

Compute the flow function at t=10

To find the amount of water that flows out of the tank in \(10\) minutes, simply compute \(Q(10)\) using the flow function: \(Q(10) = 0.1 (100(10) - \frac{10^3}{3})\) \(Q(10) = 0.1 (1000 - \frac{1000}{3})\) \(Q(10) = 0.1 (2000/3)\) \(Q(10) = 200/3 \approx 66.67\) gallons So, after 10 minutes, approximately 66.67 gallons of water flow out of the tank.

Key concepts

Rate of Change

Rate of change in a calculus context typically refers to how a quantity, represented by a function, changes with respect to change in another quantity, often time or position. In our exercise, the rate of change is described by the function \(Q'(t) = 0.1(100 - t^2)\), which represents the water flow rate out of a tank in gallons per minute.

In real-world terms, this function tells us how fast the water is being emptied from the tank at any given time \(t\). When \(t=0\), the rate of change is at its maximum, indicating the valve is fully open. As time progresses, the rate decreases because the \(t^2\) term grows, reflecting the closing of the valve until it's completely shut at \(t=10\) minutes. Understanding rate of change is crucial to predicting the behavior of dynamic systems, be it water flow, car speed, or stock prices in finance.

Definite Integrals

Definite integrals are foundational in calculus for finding the exact accumulation of quantities, such as area, volume, distance, and, in our case, the amount of water flowing out of a tank over a period. We calculate the definite integral of the rate function over an interval to find the total change in quantity.

For example, we determine how much water has flowed out by integrating the flow rate function \(Q'(t)\) from \(t=0\) to \(t=10\). The integral essentially sums up all the tiny slices of water flow rate over each instant within the 10 minutes, which in this scenario, yields us the total volume of water that has left the tank. To connect back to our exercise, finding \(Q(t)\) via integrating \(Q'(t)\) is precisely using a definite integral to sum up the continuous flow rates over the given time.

Initial Value Problems

Initial value problems (IVPs) are a type of differential equation along with a specified value, called the initial condition, which the solution must satisfy at a certain point. Solving an IVP helps determine the specific function that not only matches the general behavior described by the differential equation but also passes through the given starting point or condition.

In the context of our integral problem, the initial condition is \(Q(0)=0\), asserting that no water has flowed out at the beginning (\(t=0\)). With this information, we find the constant of integration \(C\) when solving the indefinite integral of the flow rate, ensuring that our solution for \(Q(t)\) is tailored to the specific scenario of the water tank. This application illustrates how initial conditions shape the solutions of dynamic processes to fit real-world situations, whether in engineering, physics, or finance.