The area of a circle is changing at a rate of [tex]$5 cm^2 s^{-1}$[/tex]. Find the rate of change of the circumference at an instant when the radius is 2 cm.
Answer (1)
The area of the circle is changing at a rate of d t d A = 5 c m 2 / s and the radius is r = 2 cm.
Differentiate the area formula A = π r 2 with respect to time to get d t d A = 2 π r d t d r .
Solve for d t d r to find d t d r = 4 π 5 .
Differentiate the circumference formula C = 2 π r with respect to time to get d t d C = 2 π d t d r , and substitute d t d r to find the rate of change of the circumference: d t d C = 2 5 .
The rate of change of the circumference is 2 5 cm/s.
Explanation
Problem Setup We are given that the area of a circle is changing at a rate of 5 c m 2 / s , which means d t d A = 5 . We are also given that the radius at a particular instant is 2 cm, so r = 2 . We want to find the rate of change of the circumference, which is d t d C .
Relating Area and Radius The area of a circle is given by the formula A = π r 2 . To relate the rate of change of the area to the rate of change of the radius, we differentiate both sides of the equation with respect to time t . This gives us d t d A = d t d ( π r 2 ) = 2 π r d t d r .
Finding dr/dt We are given that d t d A = 5 and r = 2 . Substituting these values into the equation above, we get 5 = 2 π ( 2 ) d t d r = 4 π d t d r . Solving for d t d r , we find d t d r = 4 π 5 .
Relating Circumference and Radius The circumference of a circle is given by the formula C = 2 π r . To find the rate of change of the circumference, we differentiate both sides of the equation with respect to time t . This gives us d t d C = d t d ( 2 π r ) = 2 π d t d r .
Finding dC/dt Now, we substitute the value of d t d r we found earlier into this equation: d t d C = 2 π ( 4 π 5 ) = 4 π 10 π = 2 5 .
Final Answer Therefore, the rate of change of the circumference is 2 5 cm/s.
Examples
Imagine you're designing a circular fountain. You need to control how quickly the water surface area expands. Knowing the rate at which the area changes and the current radius, you can calculate how fast the circumference (the edge of the water) is growing. This helps in designing the fountain's water flow and edge features to maintain a visually appealing and balanced design.
