Answer:
The radius of the wetter area expands at a rate of [tex]4.244\times 10^{-3}[/tex] milimeters per second when radius is 150 milimeters.
Explanation:
From Geometry we remember that area of a circle is described by this expression:
[tex]A =\pi\cdot r^{2}[/tex] (Eq. 1)
Where:
[tex]r[/tex] - Radius of the circle, measured in milimeters.
[tex]A[/tex] - Area of the circle, measured in square milimeters.
Then, the rate of change of the area in time is derived by concept of rate of change, that is:
[tex]\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}[/tex] (Eq. 2)
Where:
[tex]\frac{dr}{dt}[/tex] - Rate of change of radius in time, measured in milimeters per second.
[tex]\frac{dA}{dt}[/tex] - Rate of change of area in time, measured in square milimeters per second.
Now the rate of change of radius in time is cleared within equation above:
[tex]\frac{dr}{dt} = \left(\frac{1}{2\pi\cdot r}\right)\cdot \frac{dA}{dt}[/tex]
If we know that [tex]r = 150\,mm[/tex] and [tex]\frac{dA}{dt} = 4\,\frac{mm^{2}}{s}[/tex], then the rate of change of radius in time is:
[tex]\frac{dr}{dt} = \left[\frac{1}{2\pi\cdot (150\,m)} \right] \cdot \left(4\,\frac{mm^{2}}{s} \right)[/tex]
[tex]\frac{dr}{dt}\approx 4.244\times 10^{-3}\,\frac{mm}{s}[/tex]
The radius of the wetter area expands at a rate of [tex]4.244\times 10^{-3}[/tex] milimeters per second when radius is 150 milimeters.
