Recall that the volume V of a sphere of radius r can be computed using the formula. If the radius of a spherical balloon is increasing at a rate of 5 cm/min, how fast is its volume increasing when the radius is 10 cm?

Explanation:

Given that,

A sphere has a volume of V and a radius r

Volume of sphere cam be determine using the formula

V = 4/3 πr³

V = 4πr³ / 3.

If the radius of the sphere is increasing by

dr / dt = 5cm / min

How fast is the volume increasing when r = 10cm

dV / dt =?

From V = 4πr³ / 3

We can calculate dV/dr

dV/dr = 12πr² / 3

dV/dr = 4πr²

Then,

We want to find dV/dt

Using chain rule

dV/dt = dV/dr × dr / dt

dV/dt = 4πr² × 5

dV/dt = 20πr²

So, at r = 10

dV/dt { 20π × 10²

dV/dt = 6283.19 cm/min

The rate at which the volume increase is 6283.19 cm/min

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

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