Question

Any calculus nerds want to help me on this ?

Answer 1

What is thisssss stuff?????

Answer 2

Water flows in with rate $R(t)$ and out with rate $D(t)$. The net rate - call it $N(t)$ - at which it flows through the pipe is $N(t)=R(t)-D(t)$.

a. This part is asking how much water flows into the pipe, regardless of how much flows out of it. To find this, we integrate $R(t)$:

$\displaystyle\int_0^8R(t)\,\mathrm dt\approx76.57\,\mathrm{ft}^3$

(I assume you know how to find this value with a graphing calculator?)

b. The amount of water in the pipe increases when $N(t)>0$. For $t=3$, we have $N(3)\approx-0.313$, which means the amount of water is decreasing at this time.

c. You can use the first derivative test here. $t_0$ is a critical point if $N(t_0)=0$, and a minimum occurs at this $t_0$ if $N(t)$ for $t$, and $N(t)>0$ for $t>t_0$. Refer to a plot - you'll see that this is the case for $t_0$ between $t=2$ and $t=4$. Use your calculator to get a reasonable approximation (by solving $N(t)=0$ and picking the value between 2 and 4.)

d. The pipe starts with 30 ft^3 of water, so the amount of water in the pipe at any time $w\ge0$ is

$30+\displaystyle\int_0^wN(t)\,\mathrm dt$

The pipe will overflow for the first time when this is equal to 50.