Option A is correct. The rate at which the water level is rising when the depth of the water in the cup is 5 cm is [tex]\frac{32 \pi}{25}cm/sec[/tex]
The formula for calculating the volume of the right circular cone is expressed as [tex]V=\frac{1}{3}\pi r^2h[/tex] where:
- r is the radius
- h is the height of the cone
Using similar triangles:
[tex]\frac{h}{r}= \frac{12}{3}\\\frac{h}{r}= 4\\r = \frac{1}{4}h[/tex]
The volume of the expression becomes:
[tex]V=\frac{1}{3}\pi (\frac{1}{4} h)^2h\\V=\frac{1}{3} \pi \frac{h^3}{16}\\V=\frac{\pi h^3}{48}[/tex]
The formula for expressing the rate of change is volume is expressed as:
[tex]\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}\\2 = \frac{dV}{dh} \times \frac{dh}{dt}\\2=\frac{3 \pi h^2}{48} \frac{dh}{dt}[/tex]
Given the following parameters
h = 5cm
Substitute:
[tex]2=\frac{3\pi (5)^2}{48}\frac{dh}{dt}\\2=\frac{75 \pi}{48} \frac{dh}{dt}\\96=75 \pi \frac{dh}{dt}\\ \frac{dh}{dt} = \frac{96\pi}{75}\\ \frac{dh}{dt} = \frac{32 \pi}{25} cm/sec[/tex]
Hence the rate at which the water level is rising when the depth of the water in the cup is 5 cm is [tex]\frac{32 \pi}{25}cm/sec[/tex]
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