# Chapter 3 - Section 3.4 - The Derivative as a Rate of Change - Exercises - Page 146: 27

a) $dy/dt=-1+\frac{t}{12}(m/h)$ b) The fluid level was falling the fastest at $t=0h$, or the start of the draining, when $dy/dt=-1m/h$ It was falling the slowest at $t=12h$, or the end of the draining, when $dy/dt=0m/h$ c) See down for explanations. $$y=6\Big(1-\frac{t}{12}\Big)^2m$$ a) First, we need to rewrite the formula of $y$: $$y=6\Big(1-\frac{t}{12}\Big)^2=6\Big(1-\frac{t}{6}+\frac{t^2}{144}\Big)=6-t+\frac{t^2}{24}$$ Find $dy/dt$: $$\frac{dy}{dt}=\Big(6-t+\frac{t^2}{24}\Big)'=-1+\frac{t}{12}(m/h)$$ $(t\in[0,12])$ b) Here we know that the tank is completely drained after $12$ hours, so we limit $t$ to the interval $[0,12]$, which means that $$0\le t\le12$$ $$-1\le\frac{dy}{dt}=\frac{t}{12}-1\le0$$ This means $\min(dy/dt)=-1$ when $t=0$ and $\max(dy/dt)=0$ when $t=12$. Fixing the new found information into the speed of falling of the fluid level, we have - The fluid level was falling the fastest at $t=0h$, or the start of the draining, when $dy/dt=-1m/h$, but the speed of falling is $|dy/dt|=1m/h$. - The fluid level was falling the slowest at $t=12h$, or the end of the draining, when $dy/dt=0m/h$, or basically there is no more water to drain. c) The graphs are enclosed below. The graph of $dy/dt$ (blue line) is negative all in the interval $[0,12]$, and this corresponds to the fact that the graph of $y$ (red line) decreases continuously throughout this whole interval, an understandable fact since $y$ represents the fluid level of a tank being drained. However, we see the movement trend of $dy/dt$ is upward, and this fact is represented in the graph of $y$ as the steepness of decrease of $y$ gets less and less as $t\to12$ before almost lying flat to reach $0$ at $t=12$. As the fluid level went down, the speed of drainage became slower and slower and stopped after $12$ hours when the drainage finished. 