(a) Find the average rate of change of the area of a circle with respect to its radius [tex]$r$[/tex] as [tex]$r$[/tex] changes from 5 to each of the following.
(i) 5 to 6
(ii) 5 to 5.5
(iii) 5 to 5.1
(b) Find the instantaneous rate of change when [tex]$r=5$[/tex].
Answer (1)
Calculate the average rate of change using the formula r 2 − r 1 A ( r 2 ) − A ( r 1 ) for the intervals [ 5 , 6 ] , [ 5 , 5.5 ] , and [ 5 , 5.1 ] , which yields 11 π , 10.5 π , and 10.1 π respectively.
Determine the instantaneous rate of change by finding the derivative of the area function A ( r ) = π r 2 , which gives d r d A = 2 π r .
Evaluate the instantaneous rate of change at r = 5 , resulting in 10 π .
The average rates of change are 11 π , 10.5 π , 10.1 π and the instantaneous rate of change at r = 5 is 10 π .
Explanation
Problem Analysis We are given the formula for the area of a circle, A = π r 2 , and we need to find the average rate of change of the area with respect to the radius for different intervals and the instantaneous rate of change at a specific point.
Average Rate of Change Formula (a) The average rate of change of a function A ( r ) over an interval [ r 1 , r 2 ] is given by the formula: r 2 − r 1 A ( r 2 ) − A ( r 1 ) We will use this formula to calculate the average rate of change for the given intervals.
Average Rate from 5 to 6 (i) For the interval [ 5 , 6 ] , we have r 1 = 5 and r 2 = 6 . Thus, the average rate of change is: 6 − 5 A ( 6 ) − A ( 5 ) = 1 π ( 6 2 ) − π ( 5 2 ) = 1 36 π − 25 π = 11 π 11 π ≈ 34.5575
Average Rate from 5 to 5.5 (ii) For the interval [ 5 , 5.5 ] , we have r 1 = 5 and r 2 = 5.5 . Thus, the average rate of change is: 5.5 − 5 A ( 5.5 ) − A ( 5 ) = 0.5 π ( 5. 5 2 ) − π ( 5 2 ) = 0.5 30.25 π − 25 π = 0.5 5.25 π = 10.5 π 10.5 π ≈ 32.9867
Average Rate from 5 to 5.1 (iii) For the interval [ 5 , 5.1 ] , we have r 1 = 5 and r 2 = 5.1 . Thus, the average rate of change is: 5.1 − 5 A ( 5.1 ) − A ( 5 ) = 0.1 π ( 5. 1 2 ) − π ( 5 2 ) = 0.1 26.01 π − 25 π = 0.1 1.01 π = 10.1 π 10.1 π ≈ 31.7301
Instantaneous Rate of Change (b) The instantaneous rate of change of the area A with respect to the radius r is given by the derivative d r d A . Since A = π r 2 , we have: d r d A = 2 π r At r = 5 , the instantaneous rate of change is: A ′ ( 5 ) = 2 π ( 5 ) = 10 π 10 π ≈ 31.4159
Final Answer Therefore, the average rates of change for the given intervals are 11 π , 10.5 π , and 10.1 π , and the instantaneous rate of change at r = 5 is 10 π .
Examples
Understanding rates of change is crucial in many fields. For instance, in urban planning, knowing how the area of a park changes with respect to its radius helps in optimizing green spaces. If a city wants to increase the size of a circular park, calculating the rate of change allows planners to predict how much additional area they gain for each unit increase in the radius, aiding in efficient resource allocation and design decisions. This ensures the park meets the recreational needs of the community while staying within budget and space constraints.
