[tex] \frac{dr}{dt} = 6.79 \: \frac{m}{s} [/tex]
Step by Step Explanation:
The area a is defined as
[tex]a = \pi {r}^{2} [/tex]
Now take derivative of the above expression with respect to time:
[tex] \frac{da}{dt} = 2\pi r \frac{dr}{dt} [/tex]
Solving for dr/dt, we then get
[tex] \frac{dr}{dt} = \frac{1}{2\pi r} \frac{da}{dt} [/tex]
Note that da/dt = 900 m^2/s and when the area of the oil slick is 1400 m^2, the radius r is r = √(1400/π) = 21.1 m. Therefore,the rate at which the radius is increasing dr/dt is
[tex] \frac{dr}{dt} = \frac{1}{2\pi(21.1 \: m)}(900 \frac{ {m}^{2} }{s} ) = 6.79 \frac{m}{s} [/tex]
