91Ó°ÊÓ

We are given the rate of change of the area of the circular sunspot as \(\frac{dA}{dt} = 1,200\) km²/s. We need to find the corresponding rate of change of the radius, \(\frac{dr}{dt}\), in two different situations: a) When the radius equals 10,000 km. b) When the area is 640,000 km².

Starting with the formula for the area of a circle, \(A = \pi r^2\), we need to differentiate both sides of the equation with respect to time (t). This will allow us to relate the rate of change of the area to the rate of change of the radius. \(\frac{d}{dt}(A) = \frac{d}{dt}(\pi r^2)\) Given that A and r are both functions of time, the left side of the equation becomes \(\frac{dA}{dt}\), while the right side becomes \(2\pi r \frac{dr}{dt}\) using the chain rule. Therefore: \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)

a) For the first situation, we are given that the radius is r = 10,000 km. We have the equation relating the rates: \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\) Substitute the given values for \(\frac{dA}{dt} = 1,200\) km²/s and r = 10,000 km: \(1,200 = 2\pi (10,000) \frac{dr}{dt}\) Now, solve for \(\frac{dr}{dt}\): \(\frac{dr}{dt} = \frac{1,200}{2\pi (10,000)} = \frac{3}{50}\) km/s Therefore, the radius is growing at a rate of \(\frac{3}{50}\) km/s when it equals 10,000 km. b) For the second situation, we are given the area and need to find the corresponding radius. We know the area is \(A = 640,000\) km². Using the area formula, \(A = \pi r^2\) Substitute the known value for A: \(640,000 = \pi r^2\) Solve for r: \(r = \sqrt{\frac{640,000}{\pi}} \approx 4,525.30\) km Now, we can use the equation relating the rates with the known values of \(\frac{dA}{dt} = 1,200\) km²/s and r = 4,525.30 km: \(1,200 = 2\pi (4,525.30) \frac{dr}{dt}\) Solve for \(\frac{dr}{dt}\): \(\frac{dr}{dt} = \frac{1,200}{2\pi (4,525.30)} \approx 0.0533\) km/s Thus, the radius is growing at a rate of approximately 0.0533 km/s when the sunspot has an area of 640,000 km².