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(1 point) A spherical balloon is inflated so that its volume is increasing at the rate of 2.1 ft/min . How rapidly is the radius of the balloon increasing when the radius is 0.65 feet?

User Kliment Ru
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1 Answer

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To solve this problem, we can use the formula for the volume of a sphere:

V = (4/3)πr³

where V is the volume and r is the radius of the balloon.

Given that the volume is increasing at a rate of 2.1 ft/min, we can express this as:

dV/dt = 2.1 ft/min

We need to find how rapidly the radius (dr/dt) is changing when the radius is 0.65 feet.

To relate the rate of change of volume to the rate of change of radius, we can differentiate the volume formula with respect to time (t):

dV/dt = (dV/dr) * (dr/dt)

We know dV/dt = 2.1 ft/min. Now, let's find (dV/dr):

V = (4/3)πr³

Differentiating both sides with respect to r:

dV/dr = 4πr²

Now, we can substitute the values into the formula:

2.1 ft/min = (4π(0.65)²) * (dr/dt)

Simplifying:

2.1 ft/min = 4π(0.4225) * (dr/dt)

Now, solve for (dr/dt):

(dr/dt) = (2.1 ft/min) / (4π(0.4225))

(dr/dt) ≈ 0.396 ft/min

Therefore, when the radius is 0.65 feet, the radius of the balloon is increasing at a rate of approximately 0.396 ft/min.

User Adrian Panasiuk
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