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 ([tex]V[/tex]), measured in cubic centimeters, is defined by the formula:
[tex]V = \frac{\pi\cdot \cdor r^{2}\cdot h}{3}[/tex] (Eq. 1)
All right circular cone satisfies the following relationship:
[tex]\frac{h}{r} = \frac{h_{max}}{r_{max}}[/tex]
[tex]r = \left(\frac{r_{max}}{h_{max}} \right)\cdot h[/tex] (Eq. 2)
Where:
[tex]r[/tex] - Radius of the right circular cone, measured in centimeters.
[tex]h[/tex] - Height of the right circular cone, measured in centimeters.
By applying (Eq. 2) in (Eq. 1), we get the following formula:
[tex]V = \frac{\pi}{3} \cdot \left(\frac{r_{max}}{h_{max}} \right)^{2}\cdot h^{3}[/tex] (Eq. 1b)
Given that [tex]r_{max}[/tex] and [tex]h_{max}[/tex] are constant, we get the rate of change for the volume of the right circular cone ([tex]\dot V[/tex]), measured in cubic centimeters per second:
[tex]\dot V = \pi\cdot \left(\frac{r_{max}}{h_{max}} \right)^{2}\cdot h^{2}\cdot \dot h[/tex] (Eq. 2)
Where [tex]\dot h[/tex] is the rate of change of the water level, measured in centimeters per second.
If we know that [tex]\dot V = 2\,\frac{cm^{3}}{s}[/tex], [tex]r_{max} = 3\,cm[/tex], [tex]h_{max} = 12\,cm[/tex] and [tex]h = 5\,cm[/tex], the rate of change of the water level is:
[tex]\dot h = \frac{\dot V}{\pi\cdot h^{2}}\cdot \left(\frac{h_{max}}{r_{max}} \right)^{2}[/tex]
[tex]\dot h =\left[\frac{2\,\frac{cm^{3}}{s} }{\pi\cdot (5\,cm)^{2}}\right] \cdot \left(\frac{12\,cm}{3\,cm} \right)^{2}[/tex]
[tex]\dot h \approx 0.407\,\frac{cm}{s}[/tex]
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.
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 [tex]=\frac{\pi r^{2}h }{3}[/tex].......equation 1.
here h is the height and r is the radius of the cone respectively.
All right circular cone satisfies the relationship, [tex]\frac{h}{r} =\frac{h_{max} }{r_{max} }[/tex]
Now, [tex]r=\frac{r_{max} }{h_{max} } h[/tex]...equation 2.
From equation 1 and equation 2, we get[tex]V=\frac{\pi }{3} . (\frac{r_{max} }{h_{max} }h) ^{2} .h\\[/tex]
[tex]V=\frac{\pi }{3} .(\frac{r_{max} }{h_{max} } )^{2}. h^{3}[/tex]
Given that [tex]r_{max}[/tex] and [tex]h_{max}[/tex] are constant, we get the rate of change for the volume of the right circular cone [tex]V_{1}[/tex] , measured in cubic centimetre per second.
[tex]V_{1} =\pi (\frac{r_{max} }{h_{max} })^{2} h^{2} .h_{1}[/tex] here [tex]h_{1}[/tex] is the rate of change of the water level, measured in centimetres per second.
Here [tex]V_{1} =2cm^{3} per sec[/tex], [tex]r_{max} =3[/tex], [tex]h_{max}=12[/tex] and [tex]h=5[/tex] putting these values in above formulae, we get [tex]2 =\pi (\frac{3}{12} )^{2} .5^{2} .h_{1}[/tex]
on calculating we get, [tex]h_{1} = 0.407 cm per sec[/tex]
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.
For more details on volume of cone follow the link:
https://brainly.com/question/1315822
