Question

Water is slowly leaking from a bird bath at the rate of 8 cm3/min. The birdbath is in the shape of a cone with the vertex down and the radius is double the height. How fast is the water level changing when the water is 10 cm deep

Answer 1

The water level is changing at - 0.016 centimeters per minute when the water is 10 centimeters deep.

Explanation:

From Geometry, the volume of the cone (
`V`), measured in cubic centimeters, is defined by:

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

Where:

`r` - Radius, measured in centimeters.

`h` - Height, measured in centimeters.

Then, we find a formula for the rate of change of the volume of the birth bath (
`\dot V`), measured in cubic centimeters per minute, by means of differentiation:

`\dot V = (\pi)/(3)\cdot \left(2\cdot r\cdot h\cdot \dot r + r^(2)\cdot \dot h \right)` (2)

Where:

`\dot r` - Rate of change of radius, measured in centimeters per minute.

`\dot h` - Rate of change of height, measured in centimeters per minute.

In addition, we have the following relationship:

`r = 2\cdot h` (3)

And by Differential Calculus:

`\dot r = 2\cdot \dot h` (4)

By applying (3) and (4) in (2), we find the following expanded formula:

`\dot V = (\pi)/(3)\cdot \left[2\cdot (2\cdot h)\cdot (2\cdot \dot h)+4\cdot h^(2)\cdot \dot h\right]`

`\dot V = (\pi)/(3)\cdot (8\cdot h+4\cdot h^(2))\cdot \dot h` (5)

If
`\dot V = -8\,(cm^(3))/(min)` and
`h = 10\,cm`, then the rate of change of the water level is:

`\dot h = (\dot V)/((\pi)/(3)\cdot (8\cdot h + 4\cdot h^(2)) )`

`\dot h = (-8\,(cm^(3))/(min) )/((\pi)/(3)\cdot [8\cdot (10\,cm)+4\cdot (10\,cm)^(2)] )`

`\dot h \approx -0.016\,(cm)/(min)`

The water level is changing at - 0.016 centimeters per minute when the water is 10 centimeters deep.