Answer:
0.29 mi / hr
Step-by-step explanation:
Given that:
Area (A) = 15mi²
Area increases at a constant rate of 4mi²/ hr
Area(A) of a circle is given as πr² - - - (1)
Derivative of A with respect to time ;
A = πr²
dA/dt = 2πr(dr/dt) - - - (2)
Rate of change of Area with time (dA/dt) = 4mi²/hr
From (1) :
A = πr²
A = 15mi²
15 = πr²
15/π = r²
r = √(15/π)
From (2) :
dA/dt = 2πr(dr/dt)
4 = 2π√(15/π)dr/dt
dr/dt = 4/2π√(15/π)
= 0.29134
= 0.29 mi / hr
