V = (4/3)πr^3
dV/dt = 4π r^2 dr/dt
when r = 7 , dV/dt = 35
35 = 4π(49) dr/dt
dr/dt = 35/(196π)
A = 4πr^2
dA/dt = 8π r dr/dt
when r = 7, dr/dt = 35/(196π)
dA/dt = 8π)(7)(35/(196π) )
= 1960π/(196π)
= 10 cm^2/min
dA/dt = 8π(7)(35) = 1960π cm^2/min
A spherical ballon is being inflated at the rate of 35cc/min. the rate of increase in the surfece area(un cm²/min.) of the ballon when its diameter is 14cm is:
1. 10 * under root 10
2. under root 10
3. 100
4. 10
1 answer
1 answer