Answer
(a) At $t=0$, rate $=\frac{4800\times 0}{(0^2+2)^3}=0$ bacteria per day
(b) At $t=1$, rate $=177.78$ bacteria per day
(c) At $t=2$, rate $=44.44$ bacteria per day
(d) At $t=3$, rate $=10.82$ bacteria per day
(e) At $t=4$, rate $=3.29$ bacteria per day
(f) The rate of change of the bacteria population is decreasing as time passes.
Work Step by Step
Step-1: Differentiate the following equation with respect to $t$,
$$N=400\bigg[ 1-\frac{3}{(t^2+2)^2} \bigg]$$
$$\frac{dN}{dt}=(-400\times3)\cdot(-2)\cdot(t^2+2)^{-3}\cdot(2t)$$
$$\frac{dN}{dt}=\frac{4800t}{(t^2+2)^3}$$
Step-2: At $t=0$, rate $=\frac{4800\times 0}{(0^2+2)^3}=0$ bacteria per day
Step-3: At $t=1$, rate $=177.78$ bacteria per day
Step-4: At $t=2$, rate $=44.44$ bacteria per day
Step-5: At $t=3$, rate $=10.82$ bacteria per day
Step-6: At $t=4$, rate $=3.29$ bacteria per day
Step-7: The rate of change of the bacteria population is decreasing as time passes.
