To find the rate at which the level of water is rising in a right circular cylinder, we need to consider a few key pieces of information:
1. The radius [tex]\( r \)[/tex] of the cylinder is given as 8 cm.
2. The rate at which the volume of water is increasing (dV/dt) is given as 18 cubic cm per minute.
Let's denote:
- [tex]\( h \)[/tex] as the height of the water level in the cylinder.
- [tex]\( \frac{dh}{dt} \)[/tex] as the rate at which the height of the water level is rising, which is what we need to find.
We know the volume [tex]\( V \)[/tex] of a cylinder can be calculated with the formula:
[tex]\[ V = \pi r^2 h \][/tex]
Taking the derivative with respect to time [tex]\( t \)[/tex], we get:
[tex]\[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \][/tex]
We are given:
[tex]\[ \frac{dV}{dt} = 18 \, \text{cubic cm/min} \][/tex]
The radius [tex]\( r \)[/tex] is 8 cm.
First, let's calculate the area of the base of the cylinder:
[tex]\[ \text{Area of base} = \pi r^2 \][/tex]
[tex]\[ \text{Area of base} = \pi (8)^2 \][/tex]
[tex]\[ \text{Area of base} = 64 \pi \, \text{square cm} \][/tex]
Now, we substitute the known values into the rate of change formula:
[tex]\[ 18 = \pi (8)^2 \frac{dh}{dt} \][/tex]
[tex]\[ 18 = 64 \pi \frac{dh}{dt} \][/tex]
Solving for [tex]\( \frac{dh}{dt} \)[/tex]:
[tex]\[ \frac{dh}{dt} = \frac{18}{64 \pi} \][/tex]
[tex]\[ \frac{dh}{dt} \approx \frac{18}{201.06192982974676} \][/tex]
[tex]\[ \frac{dh}{dt} \approx 0.08952465548919113 \][/tex]
Thus, the rate at which the level of water is rising in the cylinder is approximately [tex]\( 0.0895 \, \text{cm per minute} \)[/tex].
So, the area of the base of the cylinder is approximately [tex]\( 201.062 \, \text{square cm} \)[/tex] and the rate at which the water level is rising is approximately [tex]\( 0.0895 \, \text{cm per minute} \)[/tex].
