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A spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. How fast is the radius of the balloon increasing at the instant the radius is 40 centimeters

User Santos
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1 Answer

4 votes

Answer:

0.245cm/min

Step-by-step explanation:

The volume of the spherical balloon is expressed as V = 4/3πr³ where r is the radius of the spherical balloon. If the spherical balloon is inflated with gas at the rate of 500 cubic centimetres per minute then dV/dt = 500cm³.

Using chain rule to express dV/dt;

dV/dt = dV/dr*dr/dt

dr/dt is the rate at which the radius of the gallon is increasing.

From the formula, dV/dr = 3(4/3πr^3-1))

dV/dr = 4πr²

dV/dt = 4πr² *dr/dt

500 = 4πr² *dr/dt

If radius r = 40;

500 = 4π(40)² *dr/dt

500 = 6400π*dr/dt

dr/dt = 500/6400π

dr/dt = 5/64π

dr/dt = 0.245cm/min

Hence, the radius of the balloon is increasing at the rate of 0.245cm/min

User Bbjay
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