Mathematics, 04.11.2020 19:20 21marmong

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?

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

Given the height h of the cup is 12 cm, and the radius r of the opening is 3 cm.

Given, A cup has the shape of a right circular cone.

We know that the volume of cone, V $=\frac{\pi r^{2}h }{3}$ .......equation 1.

here h is the height and r is the radius of the cone respectively.

All right circular cone satisfies the relationship, $\frac{h}{r} =\frac{h_{max} }{r_{max} }$

Now, $r=\frac{r_{max} }{h_{max} } h$ ...equation 2.

From equation 1 and equation 2, we get $V=\frac{\pi }{3} . (\frac{r_{max} }{h_{max} }h) ^{2} .h\\$

$V=\frac{\pi }{3} .(\frac{r_{max} }{h_{max} } )^{2}. h^{3}$

Given that $r_{max}$ and $h_{max}$ are constant, we get the rate of change for the volume of the right circular cone $V_{1}$ , measured in cubic centimetre per second.

$V_{1} =\pi (\frac{r_{max} }{h_{max} })^{2} h^{2} .h_{1}$ here $h_{1}$ is the rate of change of the water level, measured in centimetres per second.

Here $V_{1} =2cm^{3} per sec$ , $r_{max} =3$ , $h_{max}=12$ and h=5 putting these values in above formulae, we get $2 =\pi (\frac{3}{12} )^{2} .5^{2} .h_{1}$

on calculating we get, $h_{1} = 0.407 cm per sec$

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

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Answer from: guapotaco8102

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 (), 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:

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

- 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.

Answer from: Quest

need more information

step-by-step explanation:

Answer from: Quest

what i would say the best answer is a. 40.5

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