## Calculus: Early Transcendentals (2nd Edition)

$a.$ Putting $t=0$ into the equation we get $d(0)=100$. $b.$ $t_e=\frac{50}{11}\approx 4.55$. $c.$ $\mathcal{D}=\left[0,\frac{50}{11}\right]$.
$a.$ Indeed, $$d(0)=(10-2.2\cdot0)^2=10^2=100.$$ $b.$ Tank is empty when the depth of water is equal to zero. So we have to solve for $t_e$ such that $d(t_e)=0$. This gives $$d(t_e)=(10-2.2t_e)^2=0\Longrightarrow 10-2.2 t_e=0\\2.2 t_e=10\Longrightarrow t_e=\frac{10}{2.2}=\frac{50}{11}\approx 4.55.$$ $c.$ Since $d$ is depth, and we stard at $t=0$, the appropriate domain is any $t>0$ such that $d\in[100,0]$. In this case, the boundaries for the time interval that is the domain should be the initial moment $t=0$ and the moment when the tank is empty $t_e=\frac{50}{11}$, so the domain is $$\mathcal{D}=\left[0,\frac{50}{11}\right].$$