Question

A conical container, with vertex down, has a height of 6 cm and a diameter of 2 cm. It is leaking water at

the rate of 1 cubic centimeter per minute. Find the rate at which the water level h is dropping when h = 3 cm.

Answer 1

`\displaystyle (dh)/(dt)=-(4)/(\pi)\approx-1.2732\text{ centimeters per minute}`

The water level is dropping by approximately 1.27 centimeters per minute.

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 h 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=(1)/(3)\pi r^2h`

Now, we can establish the relationship between the radius r and the height h. At any given point, we will have two similar triangles as shown below. Therefore, we can write:

`\displaystyle (1)/(6)=(r)/(h)`

Solving for r yields:

`\displaystyle r=(1)/(6)h`

So, we will substitute this into our volume formula. This yields:

`\displaystyle \begin{aligned} V&=(1)/(3)\pi \Big((1)/(6)h\Big)^2h\\ &=(1)/(108)\pi h^3\end{aligned}`

Now, we will differentiate both sides with respect to time t. Hence:

`\displaystyle (d)/(dt)[V]=(d)/(dt)\Big[(1)/(108)\pi h^3\Big]`

The left is simply dV/dt. We can move the coefficient from the right:

`\displaystyle (dV)/(dt)=(1)/(108)\pi(d)/(dt)\big[h^3\big]`

Implicitly differentiate:

`\displaystyle\begin{aligned} (dV)/(dt)&=(1)/(108)\pi(3h^2(dh)/(dt))\\ &=(1)/(36)\pi h^2(dh)/(dt)\end{aligned}`

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=(1)/(36)\pi(3)^2(dh)/(dt)`

Therefore:

`\displaystyle -1=(1)/(4)\pi(dh)/(dt)`

Hence:

`\displaystyle (dh)/(dt)=-(4)/(\pi)\approx-1.2732\text{ centimeters per minute}`

The water level is dropping at a rate of approximately 1.27 centimeters per minute.