A spherical balloon is inflated so that its volume is increasing at the rate of 2 ft^{3}/min How rapidly is the diameter of the balloon increasing when the diameter is 1.3 feet?
The diameter is increasing at _______ ft/min
SOLUTION :
Volume of ballon, V = 4/3 π r^3 = 4/3 π (d/2)^3
=> V = π / 6 * d^3
Differentiating w.r.t. t :
=> dV/dt = π/6 * 3 * d^2 d/dt (d) = π/2 * d^2 d/dt (d)
dV/dt at d = 1.3 ft. Is 2 ft^3 / min
=> 2 = π /2 * (1.3)^2 * d/dt (d)
=> 2 = 2.6546 d/dt (d)
=> d/dt (d) = 2 / 2.6546 = 0.7534 ft / min
So, diameter is increasing at the rate of 0.7534 ft / min when diameter is 1.3 ft . (ANSWER).
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