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gas is escaping from a sperical ballom 4 cm3 per minute .how fast is the surface area shrinking when radius is 24cm

User Vitality
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1 Answer

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Answer:

dS/dt = 0.33 cm²/min

Step-by-step explanation:

Volume of sphere is given by;

V = (4/3)πr³

We are told that gas is escaping from a spherical balloon 4 cm³ per minute. Thus; dV/dt = 4 cm³/min

Now, let's find the rate at which the radius is changing.

dr/dt = (dV/dt) ÷ (dV/dr)

Now, dV/dr = 4πr²

Thus;

dr/dt = 4/(4πr²)

dr/dt = 1/(πr²)

Now, surface area of sphere is given by;

S = 4πr²

Thus, dS/dr = 8πr

Now, let's find the rate at t which area is shrinking.

dS/dt = (dS/dr) × (dr/dt)

Plugging relevant terms, we have;

dS/dt = 8πr × 1/(πr²)

dS/dt = 8/r

At r = 24 cm, we have;

dS/dt = 8/24

dS/dt = 1/3

dS/dt = 0.33 cm²/min

User Runningbirds
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