At time t=0 s, the radius of a circle is equal to 15 cm. The radius of the circle increases at a rate of 0.5 cm/s. The rate of change of area at t=20 s is equal to mπ cm^2/s, where m is a positive integer. What is the value of m?
if dr/dt = .5 cm/s (given)
and at t=0 , r = 15
when t = 20
r = 15 + 20(.5) = 25 cm
A = πr^2
dA/dt = 2πr dr/dt
mπ = 2π(25)(.5)
m = 25
To find the rate of change of area, we can use the formula for the area of a circle: A = πr^2, where A is the area and r is the radius.
Given that the radius is increasing at a rate of 0.5 cm/s, we can differentiate the formula for the area with respect to time (t) to find the rate of change of area with respect to time.
dA/dt = 2πr(dr/dt)
Here, dA/dt represents the rate of change of area with respect to time, and dr/dt represents the rate of change of the radius with respect to time.
At time t=20 s, the radius is given as 15 cm. So, we can substitute the values into the formula:
dA/dt = 2π(15)(0.5)
= 15π cm^2/s
The rate of change of area at t=20 s is 15π cm^2/s.
Since we are asked to express the rate of change of area in the form of mπ cm^2/s, we can see that m = 15.
Therefore, the value of m is 15.