Question
A spherical balloon is being inflated and its radius is changing at the rate of 2m/minute. When the radius is 4m, find the rate at which the:
(a).volume is increasing
V= 3/4 pie r^2
(b). surface area is increasing
A = 4 pie r^2
(a).volume is increasing
V= 3/4 pie r^2
(b). surface area is increasing
A = 4 pie r^2
Answers
Your formula for volume is incorrect.
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
so when r = 4
dV/dt = 4π(16)(2) = 72π cm^3/minute
A = 4πr^2
dA/dt = 8πr dr/dt
when r=4
dA/dt = 8π(4)(2) = 64π cm^2/min
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
so when r = 4
dV/dt = 4π(16)(2) = 72π cm^3/minute
A = 4πr^2
dA/dt = 8πr dr/dt
when r=4
dA/dt = 8π(4)(2) = 64π cm^2/min
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