antiderivatives\na large tank is filled with water when an outflow valve is opened at t = 0. water flows out…
antiderivatives\na 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(t)=0.5(16 - t²), for 0≤t≤4.\n 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.\n b. graph the function q for 0≤t≤4.\n c. how much water flows out of the tank in 4 min?\n a. q(t)=
Answer
Explanation:
Step1: Integrate the rate - function
We know that $Q(t)$ is the antiderivative of $Q^{\prime}(t)$. Given $Q^{\prime}(t)=0.5(16 - t^{2})=8-0.5t^{2}$. Integrating term - by - term using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int(8 - 0.5t^{2})dt=\int8dt-\int0.5t^{2}dt$. $Q(t)=8t-0.5\times\frac{t^{3}}{3}+C=8t-\frac{t^{3}}{6}+C$.
Step2: Use the initial condition
Since $Q(0) = 0$, substituting $t = 0$ 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, $Q(t)=8t-\frac{t^{3}}{6}$.
Answer:
$8t-\frac{t^{3}}{6}$