Question
Air is being pumped into a spherical balloon so that its volume increases at a rate of 80cm^3/s. How fast is the surface area of the balloon increasing when its radius is 7cm?
Answers
V = (4/3)pi(r^3)
dV/dt = 4pi(r^2)dr/dt
when t=7
80 = 4pi(49)dr/dt
dr/dt = 80/(196pi)
SA = 4pi(r^2)
d(SA)/dt = 8pi(r)dr/dt
= 8pi(7)*80/(196pi)
etc
dV/dt = 4pi(r^2)dr/dt
when t=7
80 = 4pi(49)dr/dt
dr/dt = 80/(196pi)
SA = 4pi(r^2)
d(SA)/dt = 8pi(r)dr/dt
= 8pi(7)*80/(196pi)
etc
nmnm
22.85 cubic cm per second
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