Question

A spherical balloon is inflated so that its volume is increasing at the rate of 2.5 ft3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.3 feet?

Answer 1

about 0.94 ft/min

Explanation:

You want the rate of increase of the diameter of a balloon when it is 1.3 feet and the volume is increasing at 2.5 ft³/min.

Volume

The volume of a sphere is given by ...

V = 4/3πr³

In terms of diameter, this is ...

V = 4/3π(d/2)³ = (π/6)d³

Rate of change

The rate of change of volume is ...

V' = (π/6)(3d²)·d'

Solving for d' gives ...

d' = 2V'/(πd²)

Using the given values, we have ...

d' = 2(2.5 ft³/min)/(π(1.3 ft)²) ≈ 0.9417 ft/min

The diameter is increasing at about 0.94 ft/min.

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Additional comment

The attached calculator display shows the answer to 10 digits. We have rounded to 2 significant figures because that is the precision of the given values. You may use the calculator result to round to whatever precision you are asked for.

<95141404393>

Answer 2

.9417 ft/min

Explanation:

Sphere volume = 4/3 pi r^3

dv/dr = 4/3 pi * 3 r ^2 = 4 pi r^2

dv/dr * dr/dt = dv/dt = 4 pi r^2 dr/dt

2.5 = 4 pi (1.3/2)^2 dr/dt (Diameter is 1.3 , radius is 1.3/2)

dr/dt = .4709 ft/min

THIS answer is how fast the RADIUS is changing

the DIAMETER is changeing TWICE this:

2 x .4709 ft/min = .9417 ft / min = diameter/dt