Answer
a. $\frac{dy}{dt}=\frac{t}{12}-1$ m/h
b. fastest $t=0$ $\frac{dy}{dt}|_{t=0}=-1$ m/h, slowest $t=12$, $\frac{dy}{dt}|_{t=12}=0$
c. See graph and explanations.
Work Step by Step
a. Given the equation $y=6(1-\frac{t}{12})^2$, we have $\frac{dy}{dt}=(2)6(1-\frac{t}{12})(-\frac{1}{12})=-(1-\frac{t}{12})=\frac{t}{12}-1$ m/h
b. As $0\leq t\leq 12$, the fluid level in the tank is falling fastest when $t=0$ which gives $\frac{dy}{dt}=\frac{0}{12}-1=-1$ m/h (here negative sign means falling). The fluid level in the tank is falling slowest when $t=12$, which gives $\frac{dy}{dt}=\frac{12}{12}-1=0$
c. See graph; the derivative $dy/dt$ is below the x-axis -- that is, all negative -- and this is because the fluid level keeps dropping all the time. The value of the derivative is more negative at the beginning ($t=0$). This is because the fluid level drops faster at the beginning and approaches zero at the end ($t=12$) as reflected in the $dy/dt$ approaching zero toward the end.
