Let's solve this step by step.

a) Express the radius of the spill, $r$, as a power function of time, $t$.

The problem states that the radius of the circular oil slick increases at a rate of 200 meters per hour. This implies:

$\frac{dr}{dt} = 200 \, \text{m/h}.$

To find $r(t)$, integrate the rate of change with respect to time:

$r(t) = \int 200 \, dt.$

The integral gives:

$r(t) = 200t + C,$ where $C$ is the initial radius at $t = 0$. Since no initial radius is given, we assume $C = 0$. Therefore:

$r(t) = 200t.$

Thus, the radius of the oil spill as a function of time is:

$r(t) = 200t.$

b) Express the area of the spill, $A$, as a power function of time, $t$.

The oil spill is circular, so the area $A$ of a circle is given by:

$A = \pi r^2.$

Substitute $r(t) = 200t$ into this equation:

$A(t) = \pi (200t)^2.$

Simplify:

$A(t) = \pi (200^2) t^2.$

Since $200^2 = 40,000$, we get:

$A(t) = 40,000\pi t^2.$

Final Answers:

a) The radius as a function of time:

$r(t) = 200t.$

b) The area as a function of time:

$A(t) = 40,000\pi t^2.$

Would you like a detailed explanation of any step? Let me know!

Related Questions:

1. What would happen if the rate of change of the radius was not constant?
2. How does the area change if the radius increases at a rate of $100 \, \text{m/h}$ instead?
3. What is the area of the spill after 3 hours?
4. How do you find the instantaneous rate of change of the area with respect to time?
5. How would you graph $r(t)$ and $A(t)$ for $t \geq 0$?

Tip: The derivative of the area $A$ with respect to time helps to determine how fast the area is growing.