Answer: [tex]\frac{4\pi}{3} \text{ meter per hour}[/tex]
Step-by-step explanation:
The circular oil slick is expanding at a rate of [tex]2 m^2/h[/tex]
Let A be the area of the circular oil slick,
So, the changes in A with respect to time (t),
[tex]\frac{dA}{dt} = 2[/tex]
[tex]\frac{d(\pi r^2)}{dt} = 2[/tex]
[tex]2\pi r\frac{dr}{dt} = 2[/tex]
[tex]\frac{dr}{dt} = \frac{1}{\pi r}[/tex]
Also, the change in diameter with respect to time(t),
[tex]\frac{d}{dt} (2 r) = 2 \frac{dr}{dt}[/tex]
[tex]\frac{d}{dt} (2 r) = 2 \times \frac{1}{\pi r}[/tex]
[tex]\frac{d}{dt} (2 r) = \frac{2}{\pi r}[/tex]
For r = 1.5 m,
[tex]\frac{d}{dt} ( 2 r)]_{r=1.5} = \frac{2}{\pi \times 1.5}=\frac{20}{\pi \times 15}=\frac{4\pi }{3}\text{ meter per hour}[/tex]