# Rates of change problem involving volume

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 $$s, find $$r, s.$$ Find $$r, s.$$

ii) for the interval $$s,the volume of water increased by $$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.

$$(i)$$ 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 $$t$$?
• @AuroraBorealis yes, that's it. That should give you the required value of $t$. – John Doe Jan 7 '19 at 4:51
• i still do not get this question part iii), how can we integrate it if the integral $\cos(t^2)$ cannot be integrated using standard functions, given that this is a high school question? – Aurora Borealis Jan 7 '19 at 14:08