Volume of a container is $\frac{4 \pi}{3}$. Water can flow in and out of container.The volume of water in container is given by: $g(t), 0 \leq t \leq 4$, where $t$ is time in hours and $g(t)$ is measured in $m^3$. The rate of change of the volume of water in the container is: $g'(t)= 0.9 – 2.5 cos(0.4 t^2)$.
a) The volume of water is increasing when $p < t < q$. Find $p$ and $q$. During that interval the volume of water increases by k $ m^3$. Find k.
When $t =0$, the volume of water in the container is $2.3 m^3$. The container is never completely full of water during 4 hour period.
b) Find the minimum volume of empty space in the container during the 4 hour period.
I did part a) and found that p= 1.733874 and q = 3.56393, so volume increased by $ g(q) – g(p)= \int_{p}^{q} g'(t) dt = 3.7454$ $m^3$ during that period.
I don't know how to do part b). Any help would be appreciated.
Best Answer
Part b is just asking to find the volume of air in the container at a certain time. Let $h(t)$ be the volume occupied by air such that $$ h(t) = \frac{4\pi}{3} - g(t). $$ Notice that the extrema of $h$ occur at the extrema of $g$. Can you take it from here?