We are asked to find the rate of change of the radius of a circle when the area of the circle is increasing at a constant rate.

Step 1: Relationship Between Area and Radius

The area $A$ of a circle is given by the formula: $A = \pi r^2$ where $r$ is the radius.

Step 2: Differentiate with Respect to Time

We want to relate the rate of change of the area $\frac{dA}{dt}$ to the rate of change of the radius $\frac{dr}{dt}$. To do this, we'll differentiate both sides of the area equation with respect to time $t$: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$ Here, $\frac{dA}{dt}$ is the rate of change of the area, and $\frac{dr}{dt}$ is the rate of change of the radius.

Step 3: Plug in the Known Values

We are given:

• $\frac{dA}{dt} = 178$ square feet per second (rate of change of area),
• $r = 44$ feet (the radius at the instant we're interested in).

Substitute these values into the differentiated equation: $178 = 2\pi (44) \frac{dr}{dt}$

Step 4: Solve for $\frac{dr}{dt}$

Now, solve for $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{178}{2\pi(44)}$ $\frac{dr}{dt} = \frac{178}{88\pi}$ $\frac{dr}{dt} \approx \frac{178}{276.46} \approx 0.644 \text{ feet per second}$

Thus, the rate of change of the radius is approximately 0.644 feet per second.

Would you like more details or explanations on any part of this solution?

Related Questions:

1. How would the result change if the area increased at a different constant rate?
2. What happens to the rate of change of the radius as the radius increases further?
3. How does the formula for the circumference of a circle relate to this problem?
4. How can we