a large tank is filled with water when an outflow valve is opened at t = 0. water flows out at a rate in…

Explanation:

Step1: Recall the relationship between rate and amount

The rate of change of the amount of water flowing out is $Q'(t)=0.5(16 - t^{2})$. To find $Q(t)$, we integrate $Q'(t)$ with respect to $t$. [Q(t)=\int Q'(t)dt=\int0.5(16 - t^{2})dt]

Step2: Expand and integrate

[Q(t)=0.5\int(16 - t^{2})dt=0.5\left(16t-\frac{t^{3}}{3}\right)+C = 8t-\frac{t^{3}}{6}+C]

Step3: Use the initial - condition

Given $Q(0) = 0$, substitute $t = 0$ and $Q(0)=0$ into $Q(t)=8t-\frac{t^{3}}{6}+C$. We get $0=8\times0-\frac{0^{3}}{6}+C$, so $C = 0$. Thus, $Q(t)=8t-\frac{t^{3}}{6}$.

To choose the correct graph: First, find the derivative of $Q(t)$ which is $Q'(t)=8 - 0.5t^{2}$. The critical points of $Q(t)$ are found by setting $Q'(t)=0$, so $8 - 0.5t^{2}=0$, which gives $t^{2}=16$ and $t = 4$ (since $0\leq t\leq4$). The second - derivative $Q''(t)=-t$. When $t = 4$, $Q''(4)=- 4<0$, so $Q(t)$ has a maximum at $t = 4$. Also, $Q(0)=0$ and $Q(4)=8\times4-\frac{4^{3}}{6}=32-\frac{64}{6}=\frac{192 - 64}{6}=\frac{128}{6}\approx21.33$. The function $Q(t)$ is a cubic function with a negative leading coefficient for the cubic term. It starts at $(0,0)$, increases, and then reaches a maximum at $t = 4$.

The correct graph is A.

Answer:

a. $Q(t)=8t-\frac{t^{3}}{6}$ b. A.