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

Answer:

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

Step-by-step explanation:

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

$V = \frac{\pi\cdot r^{2}\cdot h}{3}$ (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:

$\dot V = \frac{\pi}{3}\cdot \left(2\cdot r\cdot h\cdot \dot r + r^{2}\cdot \dot h \right)$ (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:

$\dot V = \frac{\pi}{3}\cdot \left[2\cdot (2\cdot h)\cdot (2\cdot \dot h)+4\cdot h^{2}\cdot \dot h\right]$

$\dot V = \frac{\pi}{3}\cdot (8\cdot h+4\cdot h^{2})\cdot \dot h$ (5)

If $\dot V = -8\,\frac{cm^{3}}{min}$ and $h = 10\,cm$, then the rate of change of the water level is:

$\dot h = \frac{\dot V}{\frac{\pi}{3}\cdot (8\cdot h + 4\cdot h^{2}) }$

$\dot h = \frac{-8\,\frac{cm^{3}}{min} }{\frac{\pi}{3}\cdot [8\cdot (10\,cm)+4\cdot (10\,cm)^{2}] }$

$\dot h \approx -0.016\,\frac{cm}{min}$

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