Hello, ihmth!

1) A swimming pool is 40 ft long, 20 ft wide, 8 ft deep at the deep end

and 3 ft deep at the shallow end. The bottom is rectangular.

If the pool is filled with by pumping water into it at the rate of 100 ft³/min,

how fast is the water level rising when it is 3 ft deep at the deep end? Code:

40
- * - - - - - - - - - - - - - - - - - *
: | |
3 | | 3
: | 40 |
_ * - - - - - - - - - - - - - - - - - *
: | x *
: + - - - - - - - - - - - *
5 |:::::::::::::::::*
: |h::::::::::*
: |:::::*
- *

The pool is filled to a height of $\displaystyle h$ feet.

The volume of the water is: .$\displaystyle \text{(area of triangle) } \times \text{ (width of pool)}$

So we have: .$\displaystyle V \;=\;\left(\tfrac{1}{2}xh\right)(20) \quad\Rightarrow\quad V \;=\;10xh$ .[1]

From similar right triangles, we have: .$\displaystyle \frac{x}{h} \:=\:\frac{40}{5} \quad\Rightarrow\quad x \:=\:8h$ .[2]

Substitute [2] into [1]: .$\displaystyle V \:=\:10(8h)h\quad\Rightarrow\quad V \:=\:80h^2$

Differentiate with respect to time: .$\displaystyle \frac{dV}{dt} \:=\:160h\cdot \frac{dh}{dt}$ .[3]

We are given: .$\displaystyle h = 3,\;\frac{dV}{dt} = 100$

Substitute into [3]: .$\displaystyle 100 \:=\:160(3)\frac{dh}{dt} \quad\Rightarrow\quad \frac{dh}{dt} \:=\:\frac{100}{480}\:=\:\frac{5}{24}$

Therefore, the water is rising at: .$\displaystyle \tfrac{5}{24}\text{ ft} = 2\tfrac{1}{2}\text{ inches per minute.}$