Answer:
The rate of area of the disturbed water changing is 12.56 [tex]\frac{ft^{2} }{sec}[/tex]
Explanation:
Given:
Radius increasing rate [tex]\frac{dr}{dt} = 6[/tex] [tex]\frac{in}{sec}[/tex]
Radius [tex]r = 4[/tex] ft
Now we convert radius increasing rate into feet per second,
[tex]\frac{dr}{dt} = \frac{6}{12} \frac{ft}{sec}[/tex]
Here we have to find total area rate [tex]\frac{dA}{dt}[/tex]
[tex]A = \pi r^{2}[/tex]
[tex]\frac{dA}{dt} = \frac{d(\pi r^{2} )}{dt}[/tex]
[tex]\frac{dA}{dt} =\pi \frac{d( r^{2} )}{dt}[/tex]
[tex]\frac{dA}{dt} = 2\pi r\frac{dr }{dt}[/tex]
[tex]\frac{dA}{dt} = 2\pi \times 4 \times \frac{6}{12}[/tex]
[tex]\frac{dA}{dt} = 12.56[/tex] [tex]\frac{ft^{2} }{sec}[/tex]
Therefore, the rate of area of the disturbed water changing is 12.56 [tex]\frac{ft^{2} }{sec}[/tex]
