A spherical balloon is inflated so that its volume is increasing at the rate of 2 ft3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.7 feet?

Question

anonymous · Answer

v = 4/3*pi*r^3

dv/dt = 4*pi*r^2*dr/dt

when d = 1.7, r = 0.85, and dv/dt = 2:

2 = 4*pi*(0.85)^2*dr/dt

thus dr/dt = 1/(2pi*(0.85)^2)

anonymous · Answer

= 0.2203 ft/min

anonymous · Answer

.2202836583 was my initial answer too, but apparently wrong. I pretty much did the same thing you did

anonymous · Answer

ohhh, lol, you're asked for the rate of change of the diameter, not the radius

anonymous · Answer

wait, nvm. it's 2/(2*pi*((.85)^2))

anonymous · Answer

hence .440567

anonymous · Answer

we both read the question incorrectly :)

anonymous · Answer

yes, you're asked for d/dt(2r) = 2 * dr/dt

anonymous · Answer

yeah, I blame the distractions...