Let's go step-by-step to answer each part of the problem:

Given

• Radius of the puddle as a function of time: $r(t) = \sqrt{t}$

(a) Area of the Puddle as a Function of Time

The area $A$ of a circle is given by:

$A = \pi r^2$

Substitute $r(t) = \sqrt{t}$:

$A(t) = \pi (\sqrt{t})^2 = \pi t$

So, the area of the puddle as a function of time is:

$\boxed{A(t) = \pi t}$

(b) Average Rate of Change (AROC) of the Area with Respect to Time Between $t = 0$ and $t = 16$

The AROC of the area with respect to time between two points $t_1$ and $t_2$ is given by:

$\text{AROC} = \frac{A(t_2) - A(t_1)}{t_2 - t_1}$

For $t_1 = 0$ and $t_2 = 16$:

$A(0) = \pi \cdot 0 = 0, \quad A(16) = \pi \cdot 16 = 16\pi$

So, the AROC is:

$\text{AROC} = \frac{16\pi - 0}{16 - 0} = \frac{16\pi}{16} = \pi$

The AROC of the area with respect to time between $t = 0$ and $t = 16$ is:

$\boxed{\pi \, \text{cm}^2/\text{s}}$

(c) AROC of the Area with Respect to the Radius Between $t = 0$ and $t = 16$

We first need to find the radius at $t = 0$ and $t = 16$:

$r(0) = \sqrt{0} = 0, \quad r(16) = \sqrt{16} = 4$

The AROC of the area with respect to the radius between these radii is:

$\text{AROC} = \frac{A(r(16)) - A(r(0))}{r(16) - r(0)} = \frac{16\pi - 0}{4 - 0} = \frac{16\pi}{4} = 4\pi$

The AROC of the area with respect to the radius between $t = 0$ and $t = 16$ is:

$\boxed{4\pi \, \text{cm}}$

(d) AROC of the Area with Respect to the Circumference of the Puddle Between $t = 0$ and $t = 16$

The circumference $C$ of a circle is given by:

$C = 2\pi r$

At $t = 0$ and $t = 16$, the radii are $r(0) = 0$ and $r(16) = 4$. Thus, the circumferences are:

$C(0) = 2\pi \cdot 0 = 0, \quad C(16) = 2\pi \cdot 4 = 8\pi$

The AROC of the area with respect to the circumference is:

$\text{AROC} = \frac{A(C(16)) - A(C(0))}{C(16) - C(0)} = \frac{16\pi - 0}{8\pi - 0} = \frac{16\pi}{8\pi} = 2$

The AROC of the area with respect to the circumference of the puddle between $t = 0$ and $t = 16$ is:

$\boxed{2 \, \text{cm}}$

Summary of Answers

(a) $A(t) = \pi t$

(b) AROC with respect to time: $\pi \, \text{cm}^2/\text{s}$

(c) AROC with respect to the radius: $4\pi \, \text{cm}$

(d) AROC with respect to the circumference: $2 \, \text{cm}$

Would you like more details on any part of the solution, or do you have any questions?

Relative Questions:

1. How would the AROC change if the radius function was different?
2. What is the physical significance of AROC in this context?
3. How would you find the instantaneous rate of change of the area with respect to time?
4. How does the relationship between circumference and area change as the radius increases?
5. What happens to the AROC values if the time interval changes?

Tip:

When working with rates of change, always consider the function's domain and the behavior of the function at the endpoints to understand the implications of your results better.