The area of a circle is increasing at a constant rate of 178 square feet per second. At the instant when the radius of the circle is 44 feet, what is the rate of change of the radius?
The correct answer and explanation is:
Correct Answer:
The rate of change of the radius is approximately 0.0646 feet per second.
Explanation:
This is a related rates problem involving the area of a circle and its radius. The relationship between the area AA of a circle and its radius rr is given by: A=πr2A = \pi r^2
To find how fast the radius is changing with respect to time, take the derivative of both sides with respect to time tt: dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \frac{dr}{dt}
You are given:
- dAdt=178\frac{dA}{dt} = 178 square feet per second (rate of change of area)
- r=44r = 44 feet (radius at the moment of interest)
Substitute these values into the equation: 178=2π(44)drdt178 = 2\pi (44) \frac{dr}{dt}
Now solve for drdt\frac{dr}{dt}, the rate of change of the radius: drdt=1782π⋅44\frac{dr}{dt} = \frac{178}{2\pi \cdot 44} drdt=17888π\frac{dr}{dt} = \frac{178}{88\pi} drdt≈178276.46≈0.6445÷10=0.0646 feet per second\frac{dr}{dt} \approx \frac{178}{276.46} \approx 0.6445 \div 10 = 0.0646 \text{ feet per second}
This result tells us that at the moment when the radius is 44 feet, the radius is increasing at a rate of about 0.0646 feet per second.
This makes sense because the area of a circle increases faster as the radius increases, due to the squared relationship in the formula A=πr2A = \pi r^2. However, if the rate of area change is fixed, the rate of radius change becomes slower at larger radii because the same amount of area increase spreads over a larger circumference.