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: The total amount of water that flows out of the tank in 10 minutes is \(\frac{2000}{3}\) gallons.

Step-by-step solution

Determine the function for the water flow

Given the rate of water flow function \(Q^{\prime}(t)=0.1(100-t^{2})\), we need to find the function \(Q(t)\) that represents the amount of water that has flowed out of the tank after \(t\) minutes. To do this, we need to integrate the given function with respect to \(t\).

Integrate the water flow rate

To find \(Q(t)\), we will integrate the water flow rate function, \(Q^{\prime}(t)\), with respect to time \(t\): $$Q(t)=\int{Q^{\prime}(t) \, dt}=\int{\left(0.1\left(100-t^{2}\right)\right) \, dt}$$

Evaluate the integral

Now, we solve the integral, by applying power rule, and take into account the initial condition \(Q(0) = 0\): $$Q(t)=0.1\int{(100-t^2) \, dt} = 0.1(100t-\frac{1}{3}t^3) + C$$ To find the constant \(C\), we plug in the initial condition \(Q(0) = 0\): $$0 = 0.1(100\times0-\frac{1}{3}0^3) + C \Rightarrow C = 0$$ So, the final expression for \(Q(t)\) is: $$Q(t) = 0.1(100t-\frac{1}{3}t^3)$$

Graph the flow function

To graph the flow function \(Q(t)\) for \(0 \leq t \leq 10\), we can plot the equation \(Q(t) = 0.1(100t-\frac{1}{3}t^3)\) for \(t\) in the interval [0, 10]. Since this is a cubic function, you should see that the graph first grows and later on decreases as \(t\) increases.

Total water flow in 10 minutes

To find the total amount of water that flows out of the tank in 10 minutes, we evaluate \(Q(t)\) at \(t=10\): $$Q(10) = 0.1(100\cdot10-\frac{1}{3}10^3) = 0.1(1000-\frac{10^3}{3}) = \frac{2000}{3}$$ So, \(\frac{2000}{3}\) gallons of water flow out of the tank in 10 minutes. To summarize: a. The amount of water, \(Q(t)\), that has flowed out of the tank after \(t\) minutes is given by \(Q(t) = 0.1(100t-\frac{1}{3}t^3)\). b. Graph the flow function \(Q(t)\) for \(0 \leq t \leq 10\) by plotting the equation. c. In 10 minutes, \(Q(10) = \frac{2000}{3}\) gallons of water flow out of the tank.