Answer:
![\displaystyle \frac{dh}{dt}=-\frac{4}{\pi}\approx-1.2732\text{ centimeters per minute}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdh%7D%7Bdt%7D%3D-%5Cfrac%7B4%7D%7B%5Cpi%7D%5Capprox-1.2732%5Ctext%7B%20centimeters%20per%20minute%7D)
The water level is dropping by approximately 1.27 centimeters per minute.
Step-by-step explanation:
Please refer to the attached diagram.
The height of the conical container is 6 cm, and its radius is 1 cm.
The container is leaking water at a rate of 1 cubic centimeter per minute.
And we want to find the rate at which the water level <em>h</em> is dropping when the water height is 3 cm.
Since we are relating the water leaked to the height of the water level, we will consider the volume formula for a cone, given by:
![\displaystyle V=\frac{1}{3}\pi r^2h](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2h)
Now, we can establish the relationship between the radius <em>r</em> and the height <em>h</em>. At any given point, we will have two similar triangles as shown below. Therefore, we can write:
![\displaystyle \frac{1}{6}=\frac{r}{h}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B6%7D%3D%5Cfrac%7Br%7D%7Bh%7D)
Solving for <em>r</em> yields:
![\displaystyle r=\frac{1}{6}h](https://tex.z-dn.net/?f=%5Cdisplaystyle%20r%3D%5Cfrac%7B1%7D%7B6%7Dh)
So, we will substitute this into our volume formula. This yields:
![\displaystyle \begin{aligned} V&=\frac{1}{3}\pi \Big(\frac{1}{6}h\Big)^2h\\ &=\frac{1}{108}\pi h^3\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20V%26%3D%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%5CBig%28%5Cfrac%7B1%7D%7B6%7Dh%5CBig%29%5E2h%5C%5C%20%26%3D%5Cfrac%7B1%7D%7B108%7D%5Cpi%20h%5E3%5Cend%7Baligned%7D)
Now, we will differentiate both sides with respect to time <em>t</em>. Hence:
![\displaystyle \frac{d}{dt}[V]=\frac{d}{dt}\Big[\frac{1}{108}\pi h^3\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdt%7D%5BV%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5CBig%5B%5Cfrac%7B1%7D%7B108%7D%5Cpi%20h%5E3%5CBig%5D)
The left is simply dV/dt. We can move the coefficient from the right:
![\displaystyle \frac{dV}{dt}=\frac{1}{108}\pi\frac{d}{dt}\big[h^3\big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7BdV%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B108%7D%5Cpi%5Cfrac%7Bd%7D%7Bdt%7D%5Cbig%5Bh%5E3%5Cbig%5D)
Implicitly differentiate:
![\displaystyle\begin{aligned} \frac{dV}{dt}&=\frac{1}{108}\pi(3h^2\frac{dh}{dt})\\ &=\frac{1}{36}\pi h^2\frac{dh}{dt}\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cbegin%7Baligned%7D%20%5Cfrac%7BdV%7D%7Bdt%7D%26%3D%5Cfrac%7B1%7D%7B108%7D%5Cpi%283h%5E2%5Cfrac%7Bdh%7D%7Bdt%7D%29%5C%5C%20%26%3D%5Cfrac%7B1%7D%7B36%7D%5Cpi%20h%5E2%5Cfrac%7Bdh%7D%7Bdt%7D%5Cend%7Baligned%7D)
Since the water is leaking at a rate of 1 cubic centimeter per minute, dV/dt=-1.
We want to find the rate at which the water level h is dropping when the height of the water is 3 cm.. So, we want to find dh/dt when h=3.
So, by substitution, we acquire:
![\displaystyle -1=\frac{1}{36}\pi(3)^2\frac{dh}{dt}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20-1%3D%5Cfrac%7B1%7D%7B36%7D%5Cpi%283%29%5E2%5Cfrac%7Bdh%7D%7Bdt%7D)
Therefore:
![\displaystyle -1=\frac{1}{4}\pi\frac{dh}{dt}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20-1%3D%5Cfrac%7B1%7D%7B4%7D%5Cpi%5Cfrac%7Bdh%7D%7Bdt%7D)
Hence:
![\displaystyle \frac{dh}{dt}=-\frac{4}{\pi}\approx-1.2732\text{ centimeters per minute}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdh%7D%7Bdt%7D%3D-%5Cfrac%7B4%7D%7B%5Cpi%7D%5Capprox-1.2732%5Ctext%7B%20centimeters%20per%20minute%7D)
The water level is dropping at a rate of approximately 1.27 centimeters per minute.