Question

Any calculus nerds want to help me on this ?

Answer 1

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<0`, 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)<0` for
`t<t_0`, 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.