Question

A cup has the shape of a right circular cone. The height of the cup is 12 cm, and the radius of the opening is 3 cm. Water is poured into the cup at a constant rate of 2cm3/sec What is the rate at which the water level is rising when the depth of the water in the cup is 5 cm?

Answer 1

Answer:

The rate at which the water level is rising when the depth of the water in the cup is 5 centimeters is approximately 0.407 centimeters per second.

Step-by-step explanation:

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

$V = \frac{\pi\cdot \cdor r^{2}\cdot h}{3}$ (Eq. 1)

All right circular cone satisfies the following relationship:

$\frac{h}{r} = \frac{h_{max}}{r_{max}}$

$r = \left(\frac{r_{max}}{h_{max}} \right)\cdot h$ (Eq. 2)

Where:

$r$ - Radius of the right circular cone, measured in centimeters.

$h$ - Height of the right circular cone, measured in centimeters.

By applying (Eq. 2) in (Eq. 1), we get the following formula:

$V = \frac{\pi}{3} \cdot \left(\frac{r_{max}}{h_{max}} \right)^{2}\cdot h^{3}$ (Eq. 1b)

Given that $r_{max}$ and $h_{max}$ are constant, we get the rate of change for the volume of the right circular cone ($\dot V$), measured in cubic centimeters per second:

$\dot V = \pi\cdot \left(\frac{r_{max}}{h_{max}} \right)^{2}\cdot h^{2}\cdot \dot h$ (Eq. 2)

Where $\dot h$ is the rate of change of the water level, measured in centimeters per second.

If we know that $\dot V = 2\,\frac{cm^{3}}{s}$, $r_{max} = 3\,cm$, $h_{max} = 12\,cm$ and $h = 5\,cm$, the rate of change of the water level is:

$\dot h = \frac{\dot V}{\pi\cdot h^{2}}\cdot \left(\frac{h_{max}}{r_{max}} \right)^{2}$

$\dot h =\left[\frac{2\,\frac{cm^{3}}{s} }{\pi\cdot (5\,cm)^{2}}\right] \cdot \left(\frac{12\,cm}{3\,cm} \right)^{2}$

$\dot h \approx 0.407\,\frac{cm}{s}$

The rate at which the water level is rising when the depth of the water in the cup is 5 centimeters is approximately 0.407 centimeters per second.