We are given that the radius of a circle is increasing at a constant rate of $\frac{dr}{dt} = 7$ feet per minute, and we are asked to find the rate of change of the area of the circle at the instant when the radius $r = 44$ feet.

Steps:

1. Formula for the area of a circle: $A = \pi r^2$

2. Differentiate the area formula with respect to time $t$ to find the rate of change of area with respect to time ($\frac{dA}{dt}$): $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

3. Substitute the given values:

   • $r = 44$ feet
   • $\frac{dr}{dt} = 7$ feet per minute

   Substituting these into the differentiated formula: $\frac{dA}{dt} = 2\pi (44)(7)$

4. Calculate the rate of change: $\frac{dA}{dt} = 2\pi (308) = 616\pi$ Using $\pi \approx 3.1416$: $\frac{dA}{dt} \approx 616 \times 3.1416 = 1935.838 \text{ square feet per minute}$

Thus, the rate of change of the area is approximately $1935.838$ square feet per minute.

Would you like more details, or do you have any questions?

Related Questions:

1. How does the rate of change of the radius affect the rate of change of the area?
2. What happens to the rate of change of the area as the radius increases?
3. How would the rate of change of area differ if the radius were decreasing?
4. Can you express the rate of change of the area as a function of the radius?
5. How would the result change if the radius were given in a different unit of measurement?

Tip:

Always remember to differentiate implicitly with respect to time when dealing with rates of change in related quantities.