Answer:
When the water level is 3 cm, the rate at which the water level is rising is 30.86 cm/s.
Step-by-step explanation:
We first have to relate the height H of the pyramid with the volume V.
The volume of the pyramid is V=Sh/3, being S: area of the base square, and h: height of the pyramid.
The base is changing as the pyramid is filled with water.
Then, the volume for every height is:
![V(t)=[L(t)]^2h(t)/3](https://tex.z-dn.net/?f=V%28t%29%3D%5BL%28t%29%5D%5E2h%28t%29%2F3)
The side of the square will grow linearly with the height. They start at H=0 and L=0 at the bottom of the pyramid, and they end at H=10 and L=3.
So we have:
![L=(3/10)h=0.3h](https://tex.z-dn.net/?f=L%3D%283%2F10%29h%3D0.3h)
If we replace in the volume equation we have:
![V=L^2h/3=(0.3h)^2h/3=0.09h^3/3=0.03h^3](https://tex.z-dn.net/?f=V%3DL%5E2h%2F3%3D%280.3h%29%5E2h%2F3%3D0.09h%5E3%2F3%3D0.03h%5E3)
We know that the rate of variation of the volume in time is contant and equal to 25 cm3:
![dV/dt=25](https://tex.z-dn.net/?f=dV%2Fdt%3D25)
We can derive the volume equation to calculate the variation of h in time:
![dV/dt=3(0.03)h^2*(dh/dt)=0.09h^2*(dh/dt)=25\\\\dh/dt=(25/0.09)h^{-2}=278h^{-2}](https://tex.z-dn.net/?f=dV%2Fdt%3D3%280.03%29h%5E2%2A%28dh%2Fdt%29%3D0.09h%5E2%2A%28dh%2Fdt%29%3D25%5C%5C%5C%5Cdh%2Fdt%3D%2825%2F0.09%29h%5E%7B-2%7D%3D278h%5E%7B-2%7D)
The rate of variation of the height in time is dh/dt=278h^(-2). The units are cm/s.
When the water level is 3 cm, the rate at which the water level is rising is:
![h=3\\\\dh/dt=278*3^{-2}=278*(1/9)=30.86](https://tex.z-dn.net/?f=h%3D3%5C%5C%5C%5Cdh%2Fdt%3D278%2A3%5E%7B-2%7D%3D278%2A%281%2F9%29%3D30.86)