Let's go step by step to solve this problem.

1. Finding the Area of the Puddle when the Radius is 10 cm

The area $A$ of a circular puddle is given by: $A = \pi r^2$ We are asked to find the area when the radius $r = 10 \, \text{cm}$.

Substitute $r = 10 \, \text{cm}$ into the equation for $A$: $A = \pi (10)^2 = 100\pi \, \text{cm}^2$

Thus, the area of the puddle when the radius is 10 cm is: $A = 100\pi \, \text{cm}^2$

2. Units of $\frac{dA}{dr}$

The derivative $\frac{dA}{dr} = 2\pi r$ represents how the area $A$ changes with respect to the radius $r$.

• The area $A$ is measured in square centimeters $\text{cm}^2$.
• The radius $r$ is measured in centimeters $\text{cm}$.

So the units of $\frac{dA}{dr}$ are: $\frac{\text{cm}^2}{\text{cm}} = \text{cm}$

Thus, the units of $\frac{dA}{dr}$ are centimeters.

3. Finding the Average Rate of Change in the Area

The average rate of change of the area is the change in area divided by the change in the radius. To calculate this, we need the radius when the area reaches $400\pi \, \text{cm}^2$.

The area is given by $A = \pi r^2$. Set $A = 400\pi \, \text{cm}^2$ and solve for $r$: $400\pi = \pi r^2$ $r^2 = 400$ $r = \sqrt{400} = 20 \, \text{cm}$

Thus, when the area is $400\pi \, \text{cm}^2$, the radius is 20 cm.

The leak starts when the radius is 0 cm, so the average rate of change of the area as the radius increases from 0 to 20 cm is: $\frac{\Delta A}{\Delta r} = \frac{A(20) - A(0)}{20 - 0}$ Since $A(20) = 400\pi \, \text{cm}^2$ and $A(0) = 0$, we have: $\frac{\Delta A}{\Delta r} = \frac{400\pi - 0}{20} = \frac{400\pi}{20} = 20\pi \, \text{cm}$

Thus, the average rate of change of the area is: $20\pi \, \text{cm}$

4. Finding the Instantaneous Rate of Change When the Area is $400\pi \, \text{cm}^2$

The instantaneous rate of change of the area is given by the derivative $\frac{dA}{dr} = 2\pi r$. We need to find this when the area is $400\pi \, \text{cm}^2$, which corresponds to $r = 20 \, \text{cm}$ (as we found earlier).

Substitute $r = 20 \, \text{cm}$ into the derivative: $\frac{dA}{dr} = 2\pi(20) = 40\pi \, \text{cm}$

Thus, the instantaneous rate of change of the area when the area is $400\pi \, \text{cm}^2$ is: $40\pi \, \text{cm}$

Summary of Results:

1. The area of the puddle when the radius is 10 cm is $100\pi \, \text{cm}^2$.
2. The units of $\frac{dA}{dr}$ are centimeters ($\text{cm}$).
3. The average rate of change of the area from the time the leak begins until the area is $400\pi \, \text{cm}^2$ is $20\pi \, \text{cm}$.
4. The instantaneous rate of change of the area when the area is $400\pi \, \text{cm}^2$ is $40\pi \, \text{cm}$.

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Would you like more details on any part of the solution?

Here are 5 related questions to expand your understanding:

1. How does the rate of change of the area change as the radius increases?
2. What would the area be when the radius is doubled from 10 cm to 20 cm?
3. How does the average rate of change compare to the instantaneous rate of change at different points?
4. What would happen to the average rate of change if the area reached $900\pi \, \text{cm}^2$?
5. Can you find the area of the puddle when the radius is tripled?

Tip: The average rate of change gives you an overall view of how fast a quantity changes over an interval, while the instantaneous rate tells you how fast it's changing at a specific moment!