Asked by Ayanfeoluwa
The area of a circle is increasing at the rate of 3cm/sec find the rate change of the circumference when the radius is 2cm
Answers
Answered by
oobleck
A = πr^2
dA/dt = 2πr dr/dt
3 = 2π*2 dr/dt
dr/dt = 3/(4π)
C = 2πr
dC/dt = 2π dr/dt = 2π * 3/(4π) = 3/2 cm/s
or,
C = 2πr = 2π√(A/π) = 2√(πA)
so, when r=2, A = 4π
dC/dt = (2√π)/(2√A) dA/dt = (2√π)/(2√(4π)) * 3 = 3/2
dA/dt = 2πr dr/dt
3 = 2π*2 dr/dt
dr/dt = 3/(4π)
C = 2πr
dC/dt = 2π dr/dt = 2π * 3/(4π) = 3/2 cm/s
or,
C = 2πr = 2π√(A/π) = 2√(πA)
so, when r=2, A = 4π
dC/dt = (2√π)/(2√A) dA/dt = (2√π)/(2√(4π)) * 3 = 3/2
Answered by
oobleck
well duh. This is so much easier. since C = 2πr,
dA/dt = 2πr dr/dt = r * 2π dr/dt = r dC/dt
dA/dt = 2πr dr/dt = r * 2π dr/dt = r dC/dt