Answer:
Therefore the rate at which water level is dropping is
in per minute.
Step-by-step explanation:
Given that,
The diameter of cylindrical bucket = 12 in.
Depth is increasing at the rate of = 4.0 in per minutes.
i.e
is depth of the bucket.
The volume of the bucket is V = ![\pi r^2 h](https://tex.z-dn.net/?f=%5Cpi%20r%5E2%20h)
![=\pi \times 6^2\itimes h_1](https://tex.z-dn.net/?f=%3D%5Cpi%20%5Ctimes%206%5E2%5Citimes%20h_1)
![\therefore V=36\pi h_1](https://tex.z-dn.net/?f=%5Ctherefore%20V%3D36%5Cpi%20h_1)
Differentiating with respect yo t,
![\frac{dV}{dt}=36\pi \frac{dh_1}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D36%5Cpi%20%5Cfrac%7Bdh_1%7D%7Bdt%7D)
Putting ![\frac{dh_1}{dt}=4](https://tex.z-dn.net/?f=%5Cfrac%7Bdh_1%7D%7Bdt%7D%3D4)
![\therefore\frac{dV}{dt}=36\pi\times 4](https://tex.z-dn.net/?f=%5Ctherefore%5Cfrac%7BdV%7D%7Bdt%7D%3D36%5Cpi%5Ctimes%204)
The rate of volume change of the bucket = The rate of volume change of the aquarium .
Given that,The aquarium measures 24 in × 36 in × 18 in.
When the water pumped out from the aquarium, the depth of the aquarium only changed.
Consider h be height of the aquarium.
The volume of the aquarium is V= ( 24× 36 ×h)
V= 24× 36 ×h
Differentiating with respect to t
![\frac{dV}{dt}=24\times 36 \times \frac{dh}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D24%5Ctimes%2036%20%5Ctimes%20%5Cfrac%7Bdh%7D%7Bdt%7D)
Putting ![\frac{dV}{dt}=36\pi\times 4](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D36%5Cpi%5Ctimes%204)
![36\pi\times 4= 24\times 36\times \frac{dh}{dt}](https://tex.z-dn.net/?f=36%5Cpi%5Ctimes%204%3D%2024%5Ctimes%2036%5Ctimes%20%5Cfrac%7Bdh%7D%7Bdt%7D)
![\Rightarrow \frac{dh}{dt}=\frac{36\pi \times 4}{24\times 36}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7Bdh%7D%7Bdt%7D%3D%5Cfrac%7B36%5Cpi%20%5Ctimes%204%7D%7B24%5Ctimes%2036%7D)
![\Rightarrow \frac{dh}{dt}=\frac{11}{21}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7Bdh%7D%7Bdt%7D%3D%5Cfrac%7B11%7D%7B21%7D)
Therefore the rate at which water level is dropping is
in per minute.