00:01
For this problem, we are told that we have a circle with an area which is decreasing at a rate of 2 centimeters squared per minute.
00:09
So we have da by dt equals negative 2 cm squared per minute.
00:13
And we are asked, how fast is the radius of the circle changing? so dr by dt equals question mark, when the area is 100 centimeters squared.
00:27
So what we have is that the area of a circle is equal to pi times the radius squared.
00:34
We want the radius as a function of area.
00:37
So we'd have that r squared equals a over pi, which in turn means that r is going to be the square root of a over pi, where we know, and i'll write this down explicitly, a, the area is a function of time.
00:52
Having that can then take the derivative of this with respect to time, in which case we'll have 1 over 2 times the square root of a over pi times da by dt...