This is a problem I am stuck with seems like a rate of change problem but stuck, how can I solve this?
the volume of water in the container is given by the function $v(t)$ for $0\le t \le 4$ where $t$ is given in hours. the rate of change of volume is given by:
$$v'(t)=0.9-2.5\cos(0.4t^2)$$
i) the volume of the water is increasing when $s<t<r$, find $r, s.$ Find $r, s.$
ii) for the interval $s<t<r$,the volume of water increased by $V \text{m}^3$, find the increase volume $V \text{m}^3$.
iii) at $t = 0$, the volume of the water in the container is $2.4\text{m}^3$, we are also given that the water tank will not be entirely full during the entire $4$ hour period. What is the minimum volume of the empty space in the tank for the $4$ hour period.
I will appreciate the help as it will help me understand rates of change problems more.
I won't give away any full answers, but will provide some hints on how to interpret the questions.
$(i)$ If the volume is increasing, then the rate of change of volume clearly has to be positive. For what values of $t$ is $v'(t)>0$?
$(ii)$ At time $s$, the volume was $v(s)$. At time $r$, the volume is $v(r)$. The increase in volume is the difference between these, i.e. $v(r)-v(s)$. You don't know the formula for $v(t)$, but you do know the formula for $v'(t)$. How can you write the quantity $v(r)-v(s)$ in terms of $v'(t)$? How do you get from $v'(t)$ to $v(t)$?
$(iii)$ When the tank reaches a point where it has minimum volume of empty space, this is when the volume of water is at a maximum. I.e. $v(t)$ is at a maximum. For what value of $t$ does this happen? Can you use this value of $t$ to find the volume of empty space at time $t$?