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A spherical balloon is inflated so that its volume is increasing at the rate of 2 ft3/min

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 1.3 feet? 


The diameter is increasing at _______ ft/min

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Answer #1

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).


answered by: Tulsiram Garg
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